A pretty printer for OEIS search results

In [1]:

    

from pprinting import search


    

In [15]:

    

%run ../src/


    

    




ERROR:root:File `'../src/.py'` not found.

In [15]:

    

from functools import partial

# the following is a simple hack because if we try to use the cache it raises an exception :S
search = partial(search, cache_info={'cache_dir':'./fetched/', 'cache_first':False})


    

In [16]:

    

searchable = search(A_id='A000045')


    

    




∙

In [17]:

    

searchable(data_only=True, comment=lambda i,c: "binomial" in c)


    

    
Out[17]:




Results for query: https://oeis.org/search?fmt=json&start=0&q=id%3AA000045
A000045: Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.

by N. J. A. Sloane, 1964
Keywords: nonn,core,nice,easy,hear,changed
Data:
$$
\begin{array}{c|ccccccccccccccc }
    n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\
    \hline
    A000045(n) & 0 & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144 & 233 & 377
\end{array}
$$

In [18]:

    

searchable = search(A_id='A000045')


    

    




∙

In [19]:

    

searchable(comment=lambda i,c: "binomial" in c, formula=lambda i,f:False)


    

    
Out[19]:




Results for query: https://oeis.org/search?fmt=json&start=0&q=id%3AA000045
A000045: Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.

by N. J. A. Sloane, 1964
Keywords: nonn,core,nice,easy,hear,changed
Data:
$$
\begin{array}{c|ccccccccccccccc }
    n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\
    \hline
    A000045(n) & 0 & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144 & 233 & 377
\end{array}
$$
Comments:

F(n+2) = Sum_{k=0..n} binomial(floor((n+k)/2),k), row sums of A046854. - Paul
Barry, Mar 11 2003
The sequence F(n) is the binomial transformation of the alternating sequence
(-1)^(n-1)F(n), whereas the sequence F(n+1) is the binomial transformation of
the alternating sequence (-1)^nF(n-1). Both of these facts follow easily from
the equalities a(n;1)=F(n+1) and b(n;1)=F(n) where a(n;d) and b(n;d) are
so-called "delta-Fibonacci" numbers as defined in comments to A014445 (see
also the papers of Witula et al.). - Roman Witula, Jul 24 2012

Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and
C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108,
with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers,
A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the
Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)]
= (x - x^2) / (1 - x + x^2) (cf. A057078).
BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan
numbers, with BTC^(-1)(x) = P[Cinv(x)].
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 +
... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence
A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and
Fib(x) = -BTC^(-1)(-x).
Generalizing to P(x,t) = x /(1 + tx) and Pinv(x,t) = x /(1 - tx) =
-P(-x,t) gives other relations to lattice paths, such as the o.g.f. for
A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1].



Cross references:

Cf. A039834 (signed Fibonacci numbers), A001690 (complement), A000213,
A000288, A000322, A000383, A060455, A030186, A020695, A020701, A071679,
A099731, A100492, A094216, A094638, A000108, A101399, A101400, A001611,
A000071, A157725, A001911, A157726, A006327, A157727, A157728, A157729,
A167616, A059929, A144152, A152063, A114690, A003893, A000032, A060441,
A000930, A003269, A000957, A057078, A007317, A091867, A104597, A249548,
A262342, A001060, A022095, A072649.
First row of arrays A103323, A172236, A234357. Second row of arrays A099390,
A048887, and A092921 (k-generalized Fibonacci numbers).
Cf. A001175 (Pisano periods), A001177 (Entry points), A001176 (number of zeros
in a fundamental period).
Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809,
A109906, A111006, A114197, A162741, A228074.
Boustrophedon transforms: A000738, A000744.
Powers: A103323, A105317, A254719.
Numbers of prime factors: A022307 and A038575.
Cf. A163733.

In [20]:

    

searchable = search(seq=[1,1,2,5,14,42, 132, 429], max_results=4)


    

    




∙

In [24]:

    

searchable(data_only=True)


    

    
Out[24]:




Results for query: https://oeis.org/search?fmt=json&start=0&q=1%2C+1%2C+2%2C+5%2C+14%2C+42%2C+132%2C+429
A000108: Catalan numbers: C(n) = binomial(2n,n)/(n+1) = (2n)!/(n!(n+1)!). Also called Segner numbers.

by N. J. A. Sloane
Keywords: core,nonn,easy,eigen,nice
Data:
$$
\begin{array}{c|ccccccccccccccc }
    n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\
    \hline
    A000108(n) & 1 & 1 & 2 & 5 & 14 & 42 & 132 & 429 & 1430 & 4862 & 16796 & 58786 & 208012 & 742900 & 2674440
\end{array}
$$
A120588: G.f. satisfies: 3*A(x) = 2 + x + A(x)^2, with a(0) = 1.

by Paul D. Hanna, Jun 16 2006, Jan 24 2008
Keywords: nonn,easy
Data:
$$
\begin{array}{c|ccccccccccccccc }
    n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\
    \hline
    A120588(n) & 1 & 1 & 1 & 2 & 5 & 14 & 42 & 132 & 429 & 1430 & 4862 & 16796 & 58786 & 208012 & 742900
\end{array}
$$
A211216: Expansion of (1-8*x+21*x^2-20*x^3+5*x^4)/(1-9*x+28*x^2-35*x^3+15*x^4-x^5).

by Bruno Berselli, May 11 2012
Keywords: nonn,easy
Data:
$$
\begin{array}{c|ccccccccccccccc }
    n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\
    \hline
    A211216(n) & 1 & 1 & 2 & 5 & 14 & 42 & 132 & 429 & 1430 & 4862 & 16795 & 58766 & 207783 & 740924 & 2660139
\end{array}
$$
A033191: Binomial transform of [ 1, 0, 1, 1, 3, 6, 15, 36, 91, 231, 595,... ], which is essentially binomial(fibonacci(k) + 1, 2).

by Simon P. Norton (simon(AT)dpmms.cam.ac.uk)
Keywords: nonn,easy
Data:
$$
\begin{array}{c|ccccccccccccccc }
    n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\
    \hline
    A033191(n) & 1 & 1 & 2 & 5 & 14 & 42 & 132 & 429 & 1430 & 4861 & 16778 & 58598 & 206516 & 732825 & 2613834
\end{array}
$$

In [23]:

    

searchable(comment=lambda i,c: i in range(10))


    

    
Out[23]:




Results for query: https://oeis.org/search?fmt=json&start=0&q=1%2C+1%2C+2%2C+5%2C+14%2C+42%2C+132%2C+429
A000108: Catalan numbers: C(n) = binomial(2n,n)/(n+1) = (2n)!/(n!(n+1)!). Also called Segner numbers.

by N. J. A. Sloane
Keywords: core,nonn,easy,eigen,nice
Data:
$$
\begin{array}{c|ccccccccccccccc }
    n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\
    \hline
    A000108(n) & 1 & 1 & 2 & 5 & 14 & 42 & 132 & 429 & 1430 & 4862 & 16796 & 58786 & 208012 & 742900 & 2674440
\end{array}
$$
Comments:

The solution to Schröder's first problem. A very large number of combinatorial
interpretations are known - see references, esp. Stanley, Enumerative
Combinatorics, Volume 2. This is probably the longest entry in the OEIS, and
rightly so.
Number of ways to insert n pairs of parentheses in a word of n+1 letters.
E.g., for n=2 there are 2 ways: ((ab)c) or (a(bc)); for n=3 there are 5 ways:
((ab)(cd)), (((ab)c)d), ((a(bc))d), (a((bc)d)), (a(b(cd))).
Consider all the binomial(2n,n) paths on squared paper that (i) start at (0,
0), (ii) end at (2n, 0) and (iii) at each step, either make a (+1,+1) step or
a (+1,-1) step. Then the number of such paths that never go below the x-axis
(Dyck paths) is C(n). [Chung-Feller]
Number of noncrossing partitions of the n-set. For example, of the 15 set
partitions of the 4-set, only [{13},{24}] is crossing, so there are a(4)=14
noncrossing partitions of 4 elements. - Joerg Arndt, Jul 11 2011
a(n-1) is the number of ways of expressing an n-cycle in the symmetric group
S_n as a product of n-1 transpositions
(u_1,v_1)(u_2,v_2)...*(u{n-1},v{n-1}) where u_k <= u_j and v_k <= v_j for
k < j; see example.  If the condition is dropped, one obtains A000272. -
Joerg Arndt and Greg Stevenson, Jul 11 2011
a(n) is the number of ordered rooted trees with n nodes, not including the
root. See the Conway-Guy reference where these rooted ordered trees are called
plane bushes. See also the Bergeron et al. reference, Example 4, p. 167. -
Wolfdieter Lang, Aug 07 2007
As shown in the paper from Beineke and Pippert (1971), a(n-2)=D(n) is the
number of labeled dissections of a disk, related to the number
R(n)=A001761(n-2) of labeled planar 2-trees having n vertices and rooted at a
given exterior edge, by the formula D(n)=R(n)/(n-2)!. - M. F. Hasler, Feb 22
2012
Shifts one place left when convolved with itself.
For n >= 1 a(n) is also the number of rooted bicolored unicellular maps of
genus 0 on n edges. - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 15 2001
Ways of joining 2n points on a circle to form n nonintersecting chords. (If no
such restriction imposed, then ways of forming n chords is given by
(2n-1)!!=(2n)!/n!2^n=A001147(n).)

Formulae:

a(n) = A000984(n)/(n+1) = binomial(2n, n)/(n+1) = (2n)!/(n!*(n+1)!).
a(n) = binomial(2n, n) - binomial(2n, n-1).
a(n) = Sum_{k=0..n-1} a(k)a(n-1-k).
a(n) = Product_{k=2..n} (1 + n/k), if n>1.
G.f.: A(x) = (1 - sqrt(1 - 4x)) / (2x). G.f. A(x) satisfies A = 1 + x*A^2.
a(n+1) = Sum_{i} binomial(n, 2i)2^(n-2i)a(i). - Touchard
2(2n-1)a(n-1) = (n+1)a(n).
It is known that a(n) is odd if and only if n=2^k-1, k=0, 1, 2, 3, ... -
Emeric Deutsch, Aug 04 2002, corrected by M. F. Hasler, Nov 08 2015
Using the Stirling approximation in A000142 we get the asymptotic expansion
a(n) ~ 4^n / (sqrt(Pi  n)  (n + 1)). - Dan Fux (dan.fux(AT)OpenGaia.com or
danfux(AT)OpenGaia.com), Apr 13 2001
Integral representation: a(n) = int(x^nsqrt((4-x)/x), x=0..4)/(2Pi). -
Karol A. Penson, Apr 12 2001
E.g.f.: exp(2x)(I_0(2x)-I_1(2x)), where I_n is Bessel function. - Karol
A. Penson, Oct 07 2001
Polygorial(n, 6)/Polygorial(n, 3). - Daniel Dockery (peritus(AT)gmail.com),
Jun 24 2003
G.f. A(x) satisfies ((A(x) + A(-x)) / 2)^2 = A(4*x^2). - Michael Somos, Jun
27, 2003
G.f. A(x) satisfies Sum{k>=1} k(A(x)-1)^k = Sum{n >= 1} 4^{n-1}*x^n. -
Shapiro, Woan, Getu
a(n+m) = Sum_{k} A039599(n, k)*A039599(m, k). - Philippe Deléham, Dec 22
2003
a(n+1) = (1/(n+1))Sum_{k=0..n} a(n-k)binomial(2k+1, k+1). - Philippe
Deléham, Jan 24 2004
a(n) = Sum_{k>=0} A008313(n, k)^2. - Philippe Deléham, Feb 14 2004
a(m+n+1) = Sum_{k>=0} A039598(m, k)*A039598(n, k). - Philippe Deléham, Feb
15 2004
C(n-1) = binomial(2n-2,n-1)/n = (1/n!)  [ n^(n-1) + ( binomial(n-2,1) +
binomial(n-2,2) )n^(n-2) + ( 2binomial(n-3,1) + 7binomial(n-3,2) +
8binomial(n-3,3) + 3binomial(n-3,4) )n^(n-3) + ( 6binomial(n-4,1) +
38binomial(n-4,2) + 93binomial(n-4,3) + 111binomial(n-4,4) +
65binomial(n-4,5) + 15binomial(n-4,6) )*n^(n-4) + ... ]. - André F.
Labossière, Nov 10 2004, corrected by M. F. Hasler, Nov 10 2015
a(n) = Sum_{k=0..n} (-1)^k2^(n-k)binomial(n, k)*binomial(k, floor(k/2)). -
Paul Barry, Jan 27 2005
Sum_{n>=0} 1/a(n) = 2 + 4*Pi/3^(5/2) = F(1,2;1/2;1/4) = 2.806133050770763...
(see L'Univers de Pi link). - Gerald McGarvey and Benoit Cloitre, Feb 13
2005
a(n) = Sum{k=0..floor(n/2)} ((n-2k+1)binomial(n, n-k)/(n-k+1))^2, which is
equivalent to: a(n) = Sum{k=0..n} A053121(n, k)^2, for n>=0. - Paul D.
Hanna, Apr 23 2005
a((m+n)/2) = Sum_{k>=0} A053121(m, k)*A053121(n, k) if m+n is even. -
Philippe Deléham, May 26 2005
E.g.f. Sum_{n>=0} a(n)  x^(2n) / (2n)! = BesselI(1, 2x) / x. - Michael
Somos, Jun 22 2005
Given g.f. A(x), then B(x) = x  A(x^3) satisfies 0 = f(x, B(X)) where f(u, v)
= u - v + (uv)^2 or B(x) = x + (x * B(x))^2 which implies B(-B(x)) = -x and
also (1 + B^3) / B^2 = (1 - x^3) / x^2. - Michael Somos, Jun 27 2005
a(n) = a(n-1)(4-6/(n+1)). a(n) = 2a(n-1)(8a(n-2)+a(n-1))/(10a(n-2)-a(n-1)).

Franklin T. Adams-Watters, Feb 08 2006


Sum_{k>=1} a(k)/4^k = 1. - Franklin T. Adams-Watters, Jun 28 2006
a(n) = A047996(2*n+1, n). - Philippe Deléham, Jul 25 2006
Binomial transform of A005043. - Philippe Deléham, Oct 20 2006
a(n) = Sum_{k=0..n} (-1)^k*A116395(n,k). - Philippe Deléham, Nov 07 2006
a(n) = (1/(s-n))Sum_{k=0..n} (-1)^k (k+s-n)binomial(s-n,k) *
binomial(s+n-k,s) with s a nonnegative free integer [H. W. Gould].
a(k) = Sum_{i=1..k} |A008276(i,k)| * (k-1)^(k-i) / k!. - André F.
Labossière, May 29 2007
a(n) = Sum_{k=0..n} A129818(n,k) * A007852(k+1). - Philippe Deléham, Jun 20
2007
a(n) = Sum_{k=0..n} A109466(n,k) * A127632(k). - Philippe Deléham, Jun 20
2007
Row sums of triangle A124926. - Gary W. Adamson, Oct 22 2007
Lim{n->infinity}(1+Sum{k=0..n}a(k)/A004171(k)) = 4/Pi. - Reinhard
Zumkeller, Aug 26 2008
a(n) = Sum{k=0..n} A120730(n,k)^2 and a(k+1) = Sum{n>=k} A120730(n,k). -
Philippe Deléham, Oct 18 2008
Given an integer t >= 1 and initial values u = [a_0, a1, ..., a{t-1}], we
may define an infinite sequence Phi(u) by setting an = a{n-1} + a0*a{n-1}

a1*a{n-2} + ... + a_{n-2}*a_1 for n >= t. For example, the present
sequence is Phi([1]) (also Phi([1,1])). - Gary W. Adamson, Oct 27 2008


a(n) = Sum_{l1=0..n+1} Sum{l2=0..n}...Sum{li=0..n-i}...Sum{l_n=0..1}
delta(l_1,l_2,...,l_i,...,l_n) where delta(l_1,l_2,...,l_i,...,l_n) = 0 if any
li < l(i+1) and l_(i+1) <> 0 for i=1..n-1 and delta(l_1,l_2,...,l_i,...,l_n)
= 1 otherwise. - Thomas Wieder, Feb 25 2009
Let A(x) be the g.f., then B(x)=xA(x) satisfies the differential equation
B'(x)-2B'(x)*B(x)-1=0. - Vladimir Kruchinin, Jan 18 2011
G.f.: 1/(1-x/(1-x/(1-x/(...)))) (continued fraction). - Joerg Arndt, Mar 18
2011
With F(x) = (1-2x-sqrt(1-4x))/(2*x) an o.g.f. in x for the Catalan series,
G(x) = x/(1+x)^2 is the compositional inverse of F (nulling the n=0 term). -
Tom Copeland, Sep 04 2011
With H(x) = 1/(dG(x)/dx) = (1+x)^3 / (1-x), the n-th Catalan number is given
by (1/n!)((H(x)d/dx)^n)x evaluated at x=0, i.e., F(x) = exp(xH(u)d/du)u,
evaluated at u = 0. Also, dF(x)/dx = H(F(x)), and H(x) is the o.g.f. for
A115291. - Tom Copeland, Sep 04 2011
With F(x) = {1-sqrt[1-4x]}/2 an o.g.f. in x for the Catalan series, G(x)=
x(1-x) is the compositional inverse and this relates the Catalan numbers to
the row sums of A125181. - Tom Copeland, Sep 30 2011
With H(x)=1/(dG(x)/dx)= 1/(1-2x), the n-th Catalan number (offset 1) is given
by (1/n!)((H(x)d/dx)^n)x evaluated at x=0, i.e., F(x) = exp(xH(u)d/du)u,
evaluated at u = 0. Also, dF(x)/dx = H(F(x)). - Tom Copeland, Sep 30 2011
G.f.: (1-sqrt(1-4x))/(2x)=G(0) where
G(k)=1+(4k+1)x/(k+1-2x(k+1)(4k+3)/(2x(4k+3)+(2k+3)/G(k+1)));
(continued fraction). - Sergei N. Gladkovskii, Nov 30 2011
E.g.f.: exp(2x)(BesselI(0,2x)-BesselI(1,2x))=G(0) where
G(k)=1+(4k+1)x/((k+1)(2k+1)-x(k+1)(2k+1)(4k+3)/(x(4k+3)+(k+1)(2*k+3)/G(k+1)));
(continued fraction). - Sergei N. Gladkovskii, Nov 30 2011
E.g.f.: Hypergeometric([1/2],[2],4*x) which coincides with the e.g.f. given
just above, and also by Karol A. Penson further above. - Wolfdieter Lang,
Jan 13 2012
a(n) = A208355(2n-1) = A208355(2n) for n > 0. - Reinhard Zumkeller, Mar 04
2012
G.f.: 1 + 2x/(U(0)-2x) where U(k)= k(4x+1) + 2x + 2 -
x(2k+3)(2*k+4)/U(k+1); (continued fraction, Euler's 1st kind, 1-step). -
Sergei N. Gladkovskii, Sep 20 2012
G.f.: hypergeom([1/2,1],[2],4*x). - Joerg Arndt, Apr 06 2013
Special values of Jacobi polynomials, in Maple notation: a(n) =
4^n*JacobiP(n,1,-1/2-n,-1)/(n+1). - Karol A. Penson, Jul 28 2013
For n > 0: a(n) = sum of row n in triangle A001263. - Reinhard Zumkeller,
Oct 10 2013
a(n) = binomial(2n,n-1)/n and a(n) mod n = binomial(2n,n) mod n = A059288(n).

Jonathan Sondow, Dec 14 2013


a(n-1) = sum(t1+2t2+...+ntn=n, (-1)^(1+t1+t2+...+tn)multinomial(t1+t2
+...+tn,t1,t2,...,tn)a(1)^t1a(2)^t2...*a(n)^tn). - Mircea Merca, Feb 27
2014
a(n) = sum_{k=1..n} binomial(n+k-1,n)/n if n>0. Alexander Adamchuk, Mar 25
2014
a(n) = -2^(2n+1)  binomial(n-1/2, -3/2). - Peter Luschny, May 06 2014
a(n) = (4*A000984(n)-A000984(n+1))/2. - Stanislav Sykora, Aug 09 2014
a(n) = A246458(n) * A246466(n). - Tom Edgar, Sep 02 2014
a(n) = (2n)![x^(2*n)]hypergeom([],[2],x^2). - Peter Luschny, Jan 31 2015
a(n) = 4^(n-1)*hypergeom([3/2, 1-n], [3], 1). - Peter Luschny, Feb 03 2015
a(2n) = 2A000150(2n); a(2n+1) = 2A000150(2n+1) + a(n). - John Bodeen, Jun
24 2015
a(n) = Sum{t=1, n+1} n^(t-1)*abs(stirling1(n+1, t)) / Sum{t=1, n+1}
abs(stirling1(n+1, t)), for n > 0, see (10) in Cereceda link. - Michel
Marcus, Oct 06 2015
a(n) ~ 4^(n-2)(128 + 160/N^2 + 84/N^4 + 715/N^6 -
10180/N^8)/(N^(3/2)Pi^(1/2)) where N = 4*n+3. - Peter Luschny, Oct 14 2015
a(n) = Sum_{k=1..floor((n+1)/2)} (-1)^(k-1)binomial(n+1-k,k)a(n-k) if n > 0;
and a(0) = 1. - David Pasino, Jun 29 2016
Sum_{n>=0} (-1)^n/a(n) = 14/25 - 24arccsch(2)/(25sqrt(5)) = 14/25 -
24A002390/(25sqrt(5)) = 0.353403708337278061333... - Ilya Gutkovskiy, Jun
30 2016
C(n) = (1/n)  Sum_{i+j+k=n-1} C(i)C(j)C(k)(k+1), n >= 1. - Yuchun Ji,
Feb 21 2016
C(n) = 1 + Sum(C(i)C(j)C(k), i+j+k<n-1). - Yuchun Ji, Sep 01 2016
a(n) = A001700(n) - A162551(n) = binomial(2n+1,n+1) - 2binomial(2*n,n-1). -
Taras Goy, Aug 09 2018

Cross references:

Cf. A000142, A000245, A000344, A000588, A000957, A000984, A001392, A001453,
A001791, A002057, A002420, A003046, A003517, A003518, A003519, A006480,
A008276, A008549, A014137, A014138, A014140, A022553, A024492, A032357,
A032443, A039599, A048990, A059288, A068875, A069640, A086117, A094216,
A094638, A094639, A098597, A099731, A119822, A120304, A124926, A129763,
A137697, A154559, A161581, A167892, A167893, A179277, A211611.
A row of A060854.
See A001003, A001190, A001699, A000081 for other ways to count parentheses.
Enumerates objects encoded by A014486.
A diagonal of any of the essentially equivalent arrays A009766, A030237,
A033184, A059365, A099039, A106566, A130020, A047072.
Cf. A051168 (diagonal of the square array described).
Cf. A033552, A176137 (partitions into Catalan numbers).
Cf. A000753, A000736 (Boustrophedon transforms).
Cf. A120303 (largest prime factor of Catalan number).
Cf. A121839 (reciprocal Catalan constant).
Cf. A038003, A119861, A119908, A120274, A120275 (odd Catalan number).
Cf. A002390 (decimal expansion of natural logarithm of golden ratio).
Coefficients of square root of the g.f. are A001795/A046161.
For a(n) mod 6 see A259667.
For a(n) in base 2 see A264663.
Cf. A006858, A091962, A078920, A123352 (Hankel transforms with first terms
omitted).

A120588: G.f. satisfies: 3*A(x) = 2 + x + A(x)^2, with a(0) = 1.

by Paul D. Hanna, Jun 16 2006, Jan 24 2008
Keywords: nonn,easy
Data:
$$
\begin{array}{c|ccccccccccccccc }
    n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\
    \hline
    A120588(n) & 1 & 1 & 1 & 2 & 5 & 14 & 42 & 132 & 429 & 1430 & 4862 & 16796 & 58786 & 208012 & 742900
\end{array}
$$
Comments:

This is essentially a duplicate of entry A000108, the Catalan numbers (a(n) =
A000108(n-1) for n>0).
In order for the g.f. of an integer sequence to satisfy a functional equation
of the form: rA(x) = c + bx + A(x)^n, where n > 1, it is necessary that the
sequence start with [1, d, mn(n-1)/2], where d divides mn(n-1)/2 (m>0) and
that the coefficients are given by r = n + d^2/m, c = r-1 and b = d^3/m. The
remaining terms may then be integer and still satisfy: a_n(k) = r*a(k), where
a_n(k) is the k-th term of the n-th self-convolution of the sequence.

Formulae:

G.f.: A(x) = 1 + Series_Reversion(1+3*x - (1+x)^2).
Lagrange Inversion yields g.f.: A(x) = Sum_{n>=0}
C(2n,n)/(n+1)(2+x)^(n+1)/3^(2*n+1).
G.f.: (3 - sqrt(1-4*x))/2. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009
a(n) = Sum_{k=1..n-1} a(k) * a(n-k) if n>1. - Michael Somos, Jul 23 2011
G.f.: 2 - G(0), where G(k)= 2x(2k+1) + k +1 - 2x(k+1)(2*k+3)/G(k+1);
(continued fraction). - Sergei N. Gladkovskii, Jul 14 2013
G.f.: 2 - G(0), where G(k)= 1 - x/G(k+1) ; (continued fraction). - Sergei N.
Gladkovskii, Jul 19 2013
a(n) ~ 2^(2n-2)/(sqrt(Pi)n^(3/2)). - Vaclav Kotesovec, Aug 19 2013
Given g.f. A(x), A001850(n-1) = coefficient of x^n in A(x)^n if n>0, the
derivative of log(A(x)) is the g.f. for A026641. - Michael Somos, May 18
2015
A(x) = (1 + 2Sum_{n >= 1} Catalan(n)x^n)/(1 + Sum{n >= 1} Catalan(n)x^n) =
(1 + 3/2Sum{n >= 1} binomial(2n,n)x^n )/(1 + Sum_{n >= 1}
binomial(2n,n)x^n). - Peter Bala, Sep 01 2016

Cross references:

Cf. A120589 (A(x)^2); A120590 - A120607.
Cf. A001850, A026641.

A211216: Expansion of (1-8*x+21*x^2-20*x^3+5*x^4)/(1-9*x+28*x^2-35*x^3+15*x^4-x^5).

by Bruno Berselli, May 11 2012
Keywords: nonn,easy
Data:
$$
\begin{array}{c|ccccccccccccccc }
    n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\
    \hline
    A211216(n) & 1 & 1 & 2 & 5 & 14 & 42 & 132 & 429 & 1430 & 4862 & 16795 & 58766 & 207783 & 740924 & 2660139
\end{array}
$$
Comments:

In the paper of Kitaev, Remmel and Tiefenbruck (see the Links section),
Q_(132)^(k,0,0,0)(x,0) represents a generating function depending on k and x.
For successive values ​​of k we have:
k=1, the g.f. of A000012: 1/(1-x);
k=2,      "      A011782: (1-x)/(1-2*x);
k=3,      "      A001519: (1-2x)/(1-3x+x^2);
k=4,      "      A124302: (1-3x+x^2)/(1-4x+3*x^2);
k=5,      "      A080937: (1-4x+3x^2)/(1-5x+6x^2-x^3);
k=6,      "      A024175: (1-5x+6x^2-x^3)/(1-6x+10x^2-4*x^3);
k=7,      "      A080938: (1-6x+10x^2-4x^3)/(1-7x+15x^2-10x^3+x^4);
k=8,      "      A033191: (1-7x+15x^2-10x^3+x^4)/(1-8x+21*x^2

Formulae:

G.f.: (1-3x+x^2)(1-5x+5x^2)/(1-9x+28x^2-35x^3+15x^4-x^5).
G.f.: 1/(1-x/(1-x/(1-x/(1-x/(1-x/(1-x/(1-x/(1-x/(1-x))))))))). - Philippe
Deléham, Mar 14 2013

Cross references:

Cf. A001519, A011782, A024175, A033191, A080937, A080938, A124302.
Cf. square array in A080934.

A033191: Binomial transform of [ 1, 0, 1, 1, 3, 6, 15, 36, 91, 231, 595,... ], which is essentially binomial(fibonacci(k) + 1, 2).

by Simon P. Norton (simon(AT)dpmms.cam.ac.uk)
Keywords: nonn,easy
Data:
$$
\begin{array}{c|ccccccccccccccc }
    n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\
    \hline
    A033191(n) & 1 & 1 & 2 & 5 & 14 & 42 & 132 & 429 & 1430 & 4861 & 16778 & 58598 & 206516 & 732825 & 2613834
\end{array}
$$
Comments:

Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 10 and |s(i) - s(i-1)|
= 1 for i = 1,2,....,2n, s(0) = 1, s(2n) = 1. - Herbert Kociemba, Jun 14
2004
The sequence 1,2,5,14,... has g.f. 1/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-2x)))) =
(1-6x+10x^2-4x^3)/(1-8x+21x^2-20x^3+5x^4), and is the second binomial
transform A001519 aerated. - Paul Barry, Dec 17 2009
Counts all paths of length (2*n), n>=0, starting and ending at the initial
node on the path graph P_9, see the Maple program. - Johannes W. Meijer, May
29 2010

Formulae:

G.f.: (1-7x+15x^2-10x^3+x^4)/(1-8x+21x^2-20x^3+5x^4). - Ralf Stephan, May 13
2003
a(n) = (1/5)sum(r, 1, 9, sin(rPi/10)^2(2cos(rPi/10))^(2n)), n>=1; a(n) =
8a(n-1)-21a(n-2)+20a(n-3)-5a(n-4), n>=5. - Herbert Kociemba, Jun 14 2004
G.f.: 1 / (1 - x / (1 - x / (1 - x / (1 - x / (1 - x / (1 - x / (1 - x / (1 -
x )))))))). - Michael Somos, May 12 2012

Cross references:

Cf. A033192.
Cf. A081567, A147748 and A178381.
Cf. A211216.

In [28]:

    

searchable = search(query="pascal", table=True)


    

    




∙

In [30]:

    

searchable(comment=lambda i,c: i in range(5))


    

    
Out[30]:




Results for query: https://oeis.org/search?fmt=json&start=0&q=pascal+keyword%3Atabl
A007318: Pascal's triangle read by rows: C(n,k) = binomial(n,k) = n!/(k!*(n-k)!), 0 <= k <= n.

by N. J. A. Sloane and Mira Bernstein, Apr 28 1994
Keywords: nonn,tabl,nice,easy,core,look,hear
Data:
$$
\begin{array}{c|ccccccccccc }
n, k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
0 & 1 &  &  &  &  &  &  &  &  & \\1 & 1 & 1 &  &  &  &  &  &  &  & \\2 & 1 & 2 & 1 &  &  &  &  &  &  & \\3 & 1 & 3 & 3 & 1 &  &  &  &  &  & \\4 & 1 & 4 & 6 & 4 & 1 &  &  &  &  & \\5 & 1 & 5 & 10 & 10 & 5 & 1 &  &  &  & \\6 & 1 & 6 & 15 & 20 & 15 & 6 & 1 &  &  & \\7 & 1 & 7 & 21 & 35 & 35 & 21 & 7 & 1 &  & \\8 & 1 & 8 & 28 & 56 & 70 & 56 & 28 & 8 & 1 & \\9 & 1 & 9 & 36 & 84 & 126 & 126 & 84 & 36 & 9 & 1
\end{array}
$$
Comments:

C(n,k) = number of k-element subsets of an n-element set.
Row n gives coefficients in expansion of (1+x)^n.
Binomial(n+k-1,n-1) is the number of ways of placing k indistinguishable balls
into n boxes (the "bars and stars" argument - see Feller).
Binomial(n-1,k-1) is the number of compositions (ordered partitions) of n with
k summands.
Binomial(n+k-1,k-1) is the number of weak compositions (ordered weak
partitions) of n into exactly k summands. - Juergen Will, Jan 23 2016

Formulae:

a(n, k) = C(n,k) = binomial(n, k).
C(n, k) = C(n-1, k) + C(n-1, k-1).
The triangle is symmetric: C(n,k) = C(n,n-k).
a(n+1, m) = a(n, m) + a(n, m-1), a(n, -1) := 0, a(n, m) := 0, n<m; a(0, 0)=1.
C(n, k)=n!/(k!(n-k)!) if 0<=k<=n, otherwise 0.
G.f.: 1/(1-y-xy) = Sum(C(n, k)x^k*y^n, n, k>=0)
G.f.: 1/(1-x-y) = Sum(C(n+k, k)x^ky^n, n, k>=0).
G.f. for row n: (1+x)^n = Sum_{k=0..n} C(n, k)x^k.
G.f. for column n: x^n/(1-x)^n.
E.g.f.: A(x, y) = exp(x+x*y).
E.g.f. for column n: x^n*exp(x)/n!.
In general the m-th power of A007318 is given by: T(0, 0) = 1, T(n, k) =
T(n-1, k-1) + m*T(n-1, k), where n is the row-index and k is the column; also
T(n, k) = m^(n-k) C(n, k).
Triangle T(n, k) read by rows; given by A000007 DELTA A000007, where DELTA is
Deléham's operator defined in A084938.
Let P(n+1) = the number of integer partitions of (n+1); let p(i) = the number
of parts of the i-th partition of (n+1); let d(i) = the number of different
parts of the i-th partition of (n+1); let m(i, j) = multiplicity of the j-th
part of the i-th partition of (n+1). Define the operator Sum{i=1..P(n+1),
p(i)=k+1} as the sum running from i=1 to i=P(n+1) but taking only partitions
with p(i)=(k+1) parts into account. Define the operator Product{j=1..d(i)} =
product running from j=1 to j=d(i). Then C(n, k) = Sum{p(i)=(k+1),
i=1..P(n+1)} p(i)! / [Product{j=1..d(i)} m(i, j)!]. E.g., C(5, 3) = 10
because n=6 has the following partitions with m=3 parts: (114), (123), (222).
For their multiplicities one has: (114): 3!/(2!1!) = 3; (123): 3!/(1!1!*1!)
= 6; (222): 3!/3! = 1. The sum is 3 + 6 + 1 = 10 = C(5, 3). - Thomas Wieder,
Jun 03 2005
C(n, k) = Sum_{j=0..k} = (-1)^jC(n+1+j, k-j)A000108(j). - Philippe
Deléham, Oct 10 2005
G.f.: 1 + x(1 + x) + x^3(1 + x)^2 + x^6(1 + x)^3 + ... . - Michael Somos,
Sep 16 2006
Sum{k=0..floor(n/2)} x^(n-k)*T(n-k,k) = A000007(n), A000045(n+1), A002605(n),
A030195(n+1), A057087(n), A057088(n), A057089(n), A057090(n), A057091(n),
A057092(n), A057093(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively.
Sum{k=0..floor(n/2)} (-1)^kx^(n-k)T(n-k,k) = A000007(n), A010892(n),
A009545(n+1), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n),
A057084(n), A057085(n+1), A057086(n), A084329(n+1) for x = 0, 1, 2, 3, 4, 5,
6, 7, 8, 9, 10, 20, respectively. - Philippe Deléham, Sep 16 2006
C(n,k) <= A062758(n) for n > 1. - Reinhard Zumkeller, Mar 04 2008
C(t+p-1, t) = Sum{i=0..t} C(i+p-2, i) = Sum{i=1..p} C(i+t-2, t-1). A
binomial number is the sum of its left parent and all its right ancestors,
which equals the sum of its right parent and all its left ancestors. - Lee
Naish (lee(AT)cs.mu.oz.au), Mar 07 2008
From Paul D. Hanna, Mar 24 2011:

Let A(x) = Sum_{n>=0} x^(n(n+1)/2)*(1+x)^n be the g.f. of the flattened
triangle:
A(x) = 1 + (x + x^2) + (x^3 + 2x^4 + x^5) + (x^6 + 3x^7 + 3*x^8 + x^9)
+...
then A(x) equals the series Sum{n>=0} (1+x)^nx^nProduct{k=1..n}
(1-(1+x)x^(2k-1))/(1-(1+x)x^(2k));
also, A(x) equals the continued fraction 1/(1- x(1+x)/(1+
x(1-x)(1+x)/(1- x^3(1+x)/(1+ x^2(1-x^2)(1+x)/(1- x^5(1+x)/(1+
x^3(1-x^3)(1+x)/(1- x^7(1+x)/(1+ x^4(1-x^4)(1+x)/(1- ...))))))))).
These formulas are due to (1) a q-series identity and (2) a partial
elliptic theta function expression.


Row n of the triangle is the result of applying the ConvOffs transform to the
first n terms of the natural numbers (1, 2, 3, ..., n). See A001263 or A214281
for a definition of this transformation. - Gary W. Adamson, Jul 12 2012
From L. Edson Jeffery, Aug 02 2012:

Row n (n >= 0) of the triangle is given by the n-th antidiagonal of the
infinite matrix P^n, where P = (p_{i,j}), i,j >= 0, is the production
matrix
0, 1,
1, 0, 1,
0, 1, 0, 1,
0, 0, 1, 0, 1,
0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 0, 1, 0, 1,
...


Row n of the triangle is also given by the n+1 coefficients of the polynomial
P_n(x) defined by the recurrence P_0(x) = 1, P_1(x) = x + 1, Pn(x) =
x*P{n-1}(x) + P_{n-2}(x), n > 1. - L. Edson Jeffery, Aug 12 2013
For a closed-form formula for arbitrary left and right borders of Pascal-like
triangles see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. -
Boris Putievskiy, Sep 04 2013
(1+x)^n = Sum{k=0..n} (-1)^(n-k)binomial(n,k)Sum{i=0..k}
k^(n-i)binomial(k,i)x^(n-i)/(n-i)!. - Vladimir Kruchinin, Oct 21 2013
E.g.f.: A(x,y) = exp(x+xy) = 1 + (x+yx)/( E(0)-(x+yx)), where E(k) = 1 +
(x+yx)/(1 + (k+1)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii,
Nov 08 2013
E.g.f.:  E(0) -1, where E(k) = 2 + x(1+y)/(2k+1 - x*(1+y)/E(k+1) );
(continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
G.f.: 1 + x(1+x)(1+x^2(1+x)/(W(0)-x^2-x^3)), where W(k) = 1 + (1+x)x^(k+2)

(1+x)*x^(k+3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec
24 2013


Sum{n>=0} C(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where
C(n,k) = 0. Also Sum{n>=0} C(n+k-1,k)/n! = e  A000262(k)/k!, and for k>=1
equals e  A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
Sum_{n>=k} 1/C(n,k) = k/(k-1) for k>=1. - Richard R. Forberg, Feb 10 2014
From Tom Copeland, Apr 17 and 26 2014:

Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n
and designate the result by A007318(x) = P(x). Then with :xD:^n =
x^n*(d/dx)^n and B(n,x), the Bell polynomials (A008277),
A) P(x)= exp(xdP) = exp[x(e^M-I)] = exp[M*B(.,x)] = (I+dP)^B(.,x)
with dP = A132440, M = A238385-I, and I = identity matrix, and
B) P(:xD:) = exp(dP:xD:) = exp[(e^M-I):xD:] = exp[MB(.,:xD:)] = exp[MxD]
= (I+dP)^(xD) with action P(:xD:)g(x) = exp(dP:xD:)g(x) = g[(I+dP)*x] (cf.
also A238363).
C) P(x)^y = P(yx). P(2x) = A038207(x) = exp[MB(.,2x)], the face vectors
of the n-dim hypercubes.
D) P(x) = [St2]exp(xM)[St1] = [St2](I+dP)^x*[St1]
E) = [St1]^(-1)(I+dP)^x[St1] = [St2](I+dP)^x[St2]^(-1)
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 and
exp(x*M) = (I+dP)^x = sum(k=0,..,infinity, C(x,k) dP^k).


From Peter Bala, Dec 21 2014:

Recurrence equation: T(n,k) = T(n-1,k)*(n + k)/(n - k) - T(n-1,k-1) for n
= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1. Note,
changing the minus sign in the recurrence to a plus sign gives a
recurrence for the square of the binomial coefficients - see A008459.


There is a relation between the e.g.f.'s of the rows and the diagonals of
the triangle, namely, exp(x)  e.g.f. for row n = e.g.f. for diagonal n.
For example, for n = 3 we have exp(x)(1 + 3x + 3x^2/2! + x^3/3!) = 1 +
4x + 10x^2/2! + 20x^3/3! + 35x^4/4! + .... This property holds more
generally for the Riordan arrays of the form ( f(x), x/(1 - x) ), where
f(x) is an o.g.f. of the form 1 + f_1x + f_2x^2 + .... See, for example,
A055248 and A106516.
Let P denote the present triangle. For k = 0,1,2,... define P(k) to be the
lower unit triangular block array
/I_k 0\
\ 0  P/ having the k X k identity matrix I_k as the upper left block; in
particular, P(0) = P. The infinite product P(0)P(1)P(2)..., which is
clearly well-defined, is equal to the triangle of Stirling numbers of the
second kind A008277. The infinite product in the reverse order, that is,
...P(2)P(1)P(0), is equal to the triangle of Stirling cycle numbers
A130534.


C(a+b,c) = Sum{k=0..a} C(a,k)*C(b,b-c+k). This is a generalization of
equation 1 from section 4.2.5 of the Prudnikov et al. reference, for a=b=c=n:
C(2n,n) = Sum{k=0..n} C(n,k)^2. See Links section for animation of new
formula. - Hermann Stamm-Wilbrandt, Aug 26 2015
The row polynomials of the Pascal matrix P(n,x) = (1+x)^n are related to the
Bernoulli polynomials Br(n,x) and their umbral compositional inverses Bv(n,x)
by the umbral relation P(n,x) = (-Br(.,-Bv(.,x)))^n = (-1)^n Br(n,-Bv(.,x)),
which translates into the matrix relation P = M  Br  M * Bv, where P is the
Pascal matrix, M is the diagonal matrix diag(1,-1,1,-1,...), Br is the matrix
for the coefficients of the Bernoulli polynomials, and Bv that for the umbral
inverse polynomials defined umbrally by Br(n,Bv(.,x)) = x^n = Bv(n,Br(.,x)).
Note M = M^(-1). - Tom Copeland, Sep 05 2015
1/(1-x)^k = (r(x)  r(x^2)  r(x^4) * ...) where r(x) = (1+x)^k. - Gary W.
Adamson, Oct 17 2016

Cross references:

Equals differences between consecutive terms of A102363. - David G. Williams
(davidwilliams(AT)Paxway.com), Jan 23 2006
Row sums give A000079 (powers of 2).
Cf. A083093 (triangle read mod 3), A214292 (first differences of rows).
Partial sums of rows give triangle A008949.
Infinite matrix squared: A038207, cubed: A027465.
Cf. A101164. If rows are sorted we get A061554 or A107430.
Another version: A108044.
Cf. A008277, A132311, A132312, A052216, A052217, A052218, A052219, A052220,
A052221, A052222, A052223, A144225, A202750, A211226, A047999, A026729,
A052553, A051920, A193242.
Triangle sums (see the comments): A000079 (Row1); A000007 (Row2); A000045
(Kn11 & Kn21); A000071 (Kn12 & Kn22); A001924 (Kn13 & Kn23); A014162 (Kn14 &
Kn24); A014166 (Kn15 & Kn25); A053739 (Kn16 & Kn26); A053295 (Kn17 & Kn27);
A053296 (Kn18 & Kn28); A053308 (Kn19 & Kn29); A053309 (Kn110 & Kn210); A001519
(Kn3 & Kn4); A011782 (Fi1 & Fi2); A000930 (Ca1 & Ca2); A052544 (Ca3 & Ca4);
A003269 (Gi1 & Gi2); A055988 (Gi3 & Gi4); A034943 (Ze1 & Ze2); A005251 (Ze3 &
Ze4). - Johannes W. Meijer, Sep 22 2010
Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809,
A109906, A111006, A114197, A162741, A228074, A228196, A228576.
Cf. A137948, A245334.
Cf. A085478, A258993.

A047999: Sierpiński's [Sierpinski's] triangle (or gasket): triangle, read by rows, formed by reading Pascal's triangle mod 2.

by N. J. A. Sloane
Keywords: nonn,tabl,easy,nice
Data:
$$
\begin{array}{c|ccccccccccc }
n, k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
0 & 1 &  &  &  &  &  &  &  &  & \\1 & 1 & 1 &  &  &  &  &  &  &  & \\2 & 1 & 0 & 1 &  &  &  &  &  &  & \\3 & 1 & 1 & 1 & 1 &  &  &  &  &  & \\4 & 1 & 0 & 0 & 0 & 1 &  &  &  &  & \\5 & 1 & 1 & 0 & 0 & 1 & 1 &  &  &  & \\6 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &  &  & \\7 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 &  & \\8 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & \\9 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1
\end{array}
$$
Comments:

Restored the alternative spelling of Sierpinski to facilitate searching for
this triangle using regular-expression matching commands in ASCII. - N. J. A.
Sloane, Jan 18 2016
Also triangle giving successive states of cellular automaton generated by
"Rule 60" and "Rule 102". - Hans Havermann, May 26 2002
Also triangle formed by reading triangle of Eulerian numbers (A008292) mod 2.

Philippe Deléham, Oct 02 2003


Self-inverse when regarded as an infinite lower triangular matrix over GF(2).
Start with [1], repeatedly apply the map 0 -> [00/00], 1 -> [10/11] [Allouche
and Berthe]

Formulae:

Lucas's Theorem is that T(n,k) = 1 if and only if the 1's in the binary
expansion of k are a subset of the 1's in the binary expansion of n; or
equivalently, k AND NOT n is zero, where AND and NOT are bitwise operators. -
Chai Wah Wu, Feb 09 2016 and N. J. A. Sloane, Feb 10 2016
Sum_{k>=0} T(n, k) = A001316(n) = 2^A000120(n).
T(n,k) = T(n-1,k-1) XOR T(n-1,k), 0<k<n; T(n,0) = T(n,n) = 1. - Reinhard
Zumkeller, Dec 13 2009
T(n,k) = (T(n-1,k-1) + T(n-1,k)) mod 2 = |T(n-1,k-1) - T(n-1,k)|, 0<k<n;
T(n,0) = T(n,n) = 1. - Rick L. Shepherd, Feb 23 2018
From Vladimir Shevelev, Dec 31 2013:

For polynomial {s_n(x)} we have
s_0(x)=1; for n>=1, s_n(x) = prod{i=1,...,A000120(n)}(x^(2^k_i) + 1),
if the binary expansion of n is n = sum{i=1,...,A000120(n)}2^k_i;
G.f. sum {n>=0} s_n(x)*z^n = prod{k>=0}(1 + (x^(2^k)+1)z^(2^k)) (0<z<1/x).
Let x>1, t>0 be real numbers. Then
sum{n>=0}1/s_n(x)^t = prod{k>=0}(1 + 1/(x^(2^k)+1)^t);
sum{n>=0}(-1)^A000120(n)/s_n(x)^t = prod{k>=0}(1 - 1/(x^(2^k)+1)^t).
In particular, for t=1, x>1, we have
sum{n>=0}(-1)^A000120(n)/s_n(x) = 1 - 1/x.



Cross references:

Sequences based on the triangles formed by reading Pascal's triangle mod m:
A047999 (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m
= 6), A095142 (m = 7), A034930(m = 8), A095143 (m = 9), A008975 (m = 10),
A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).
Other versions: A090971, A038183.
Cf. A007318, A054431, A001317, A008292, A083093, A034931, A034930, A008975,
A034932, A166360, A249133, A064194, A227133.
From Johannes W. Meijer, Jun 05 2011:

A106344 is a skew version of this triangle.
Triangle sums (see the comments): A001316 (Row1; Related to Row2), A002487
(Related to Kn11, Kn12, Kn13, Kn21, Kn22, Kn23), A007306 (Kn3, Kn4),
A060632 (Fi1, Fi2), A120562 (Ca1, Ca2), A112970 (Gi1, Gi2), A127830 (Ze3,
Ze4).



A029635: The (1,2)-Pascal triangle (or Lucas triangle) read by rows.

by Mohammad K. Azarian
Keywords: nonn,tabl,nice,easy
Data:
$$
\begin{array}{c|ccccccccccc }
n, k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
0 & 2 &  &  &  &  &  &  &  &  & \\1 & 1 & 2 &  &  &  &  &  &  &  & \\2 & 1 & 3 & 2 &  &  &  &  &  &  & \\3 & 1 & 4 & 5 & 2 &  &  &  &  &  & \\4 & 1 & 5 & 9 & 7 & 2 &  &  &  &  & \\5 & 1 & 6 & 14 & 16 & 9 & 2 &  &  &  & \\6 & 1 & 7 & 20 & 30 & 25 & 11 & 2 &  &  & \\7 & 1 & 8 & 27 & 50 & 55 & 36 & 13 & 2 &  & \\8 & 1 & 9 & 35 & 77 & 105 & 91 & 49 & 15 & 2 & \\9 & 1 & 10 & 44 & 112 & 182 & 196 & 140 & 64 & 17 & 2
\end{array}
$$
Comments:

This is also called Vieta's array. - N. J. A. Sloane, Nov 22 2017
Dropping the first term and changing the boundary conditions to T(n,1)=n,
T(n,n-1)=2 (n>=2), T(n,n)=1 yields the number of nonterminal symbols (which
generate strings of length k) in a certain context-free grammar in Chomsky
normal form that generates all permutations of n symbols. Summation over k
(1<=k<=n) results in A003945. For the number of productions of this grammar:
see A090327. Example: 1; 2, 1; 3, 2, 1; 4, 5, 2, 1; 5, 9, 7, 2, 1; 6, 14, 16,
9, 2, 1; In addition to the example of A090327 we have T(3,3)=#{S}=1,
T(3,2)=#{D,E}=2 and T(3,1)=#{A,B,C}=3. - Peter R. J. Asveld, Jan 29 2004
With T(0,0)=1 : Triangle T(n,k), read by rows, given by (1,0,0,0,0,0,0,0,)
DELTA (2,-1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

Philippe Deléham, Oct 10 2011


For n > 0: T(n,k) = A097207(n-1,k), 0 <= k < n. - Reinhard Zumkeller, Mar 12
2012
Much as the original Pascal triangle gives the Fibonacci numbers as sums of
its diagonals, this triangle gives the Lucas numbers (A000032) as sums of its
diagonals; see Posamentier & Lehmann (2007). - Alonso del Arte, Apr 09 2012

Formulae:

T(n,k) = T(n-1, k-1) + T(n-1, k) = C(n, k) + C(n-1, k-1) = C(n, k)*(n+k)/n =
A007318(n, k) + A007318(n-1, k-1) = A061896(n+k, k) but with T(0, 0)=1 and
T(1, 1)=2. Row sum is floor[3^2(n-1)] i.e., A003945. - Henry Bottomley, Apr
26 2002
G.f.: 1 + (1 + xy) / (1 - x - xy). - Michael Somos, Jul 15 2003
G.f. for n-th row: (x+2y)(x+y)^(n-1).
O.g.f. for row n: (1+x)/(1-x)^(n+1). The entries in row n are the nonzero
entries in column n of A053120 divided by 2^(n-1). - Peter Bala, Aug 14 2008
T(2n,n) - T(2n,n+1)= Catalan(n)= A000108(n). - Philippe Deléham, Mar 19 2009
With T(0,0) = 1, as in the Example section below, this is known as Vieta's
array. The LU factorization of the square array is given by Yang and Leida,
equation 20. - Peter Bala, Feb 11 2012
exp(x)  e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we
have exp(x)(1 + 4x + 5x^2/2! + 2x^3/3!) = 1 + 5x + 14x^2/2! + 30x^3/3!

55*x^4/4! + .... The same property holds more generally for Riordan arrays
of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014


For n>=1: T(n,0) + T(n,1) + T(n,2) = A000217(n+1). T(n,n-2) = (n-1)^2. - Bob
Selcoe, Mar 29 2016:

Cross references:

Cf. A007318, A034807, A061896, A029653 (row-reversed), A157000.
Sums along ascending diagonals give Lucas numbers, n>0.
Cf. A090327, A003945, A029638, A228196, A228576.
Cf. A000330.
Cf. A000217.
Cf. A293600.

A008459: Square the entries of Pascal's triangle.

by N. J. A. Sloane
Keywords: nonn,tabl,easy
Data:
$$
\begin{array}{c|ccccccccccc }
n, k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
0 & 1 &  &  &  &  &  &  &  &  & \\1 & 1 & 1 &  &  &  &  &  &  &  & \\2 & 1 & 4 & 1 &  &  &  &  &  &  & \\3 & 1 & 9 & 9 & 1 &  &  &  &  &  & \\4 & 1 & 16 & 36 & 16 & 1 &  &  &  &  & \\5 & 1 & 25 & 100 & 100 & 25 & 1 &  &  &  & \\6 & 1 & 36 & 225 & 400 & 225 & 36 & 1 &  &  & \\7 & 1 & 49 & 441 & 1225 & 1225 & 441 & 49 & 1 &  & \\8 & 1 & 64 & 784 & 3136 & 4900 & 3136 & 784 & 64 & 1 & \\9 & 1 & 81 & 1296 & 7056 & 15876 & 15876 & 7056 & 1296 & 81 & 1
\end{array}
$$
Comments:

Number of lattice paths from (0, 0) to (n, n) with steps (1, 0) and (0, 1),
having k right turns. - Emeric Deutsch, Nov 23 2003
Product of A007318 and A105868. - Paul Barry, Nov 15 2005
Number of partitions that fit in an n X n box with Durfee square k. -
Franklin T. Adams-Watters, Feb 20 2006
From Peter Bala, Oct 23 2008:

Narayana numbers of type B. Row n of this triangle is the h-vector of the
simplicial complex dual to an associahedron of type B_n (a cyclohedron)
[Fomin & Reading, p. 60]. See A063007 for the corresponding f-vectors for
associahedra of type B_n. See A001263 for the h-vectors for associahedra
of type A_n. The Hilbert transform of this triangular array is A108625
(see A145905 for the definition of this term).
Let A_n be the root lattice generated as a monoid by {e_i - e_j: 0 <= i, j
<= n + 1}. Let P(A_n) be the polytope formed by the convex hull of this
generating set. Then the rows of this array are the h-vectors of a
unimodular triangulation of P(A_n) [Ardila et al.]. A063007 is the
corresponding array of f-vectors for these type A_n polytopes. See A086645
for the array of h-vectors for type C_n polytopes and A108558 for the
array of h-vectors associated with type D_n polytopes.



Formulae:

E.g.f.: exp((1+y)x)BesselI(0, 2sqrt(y)x). - Vladeta Jovovic, Nov 17 2003
G.f.: 1/sqrt(1-2x-2xy+x^2-2x^2y+x^2y^2); g.f. for row n: (1-t)^n
P_n[(1+t)/(1-t)] where the P_n's are the Legendre polynomials. - Emeric
Deutsch, Nov 23 2003 [The original version of the bivariate g.f. has been
modified with the roles of x and y interchanged so that now x corresponds to n
and y to k. - Petros Hadjicostas, Oct 22 2017]
G.f. for column k is sum(C(k, j)^2*x^(k+j), j, 0, k)/(1-x)^(2k+1). - Paul
Barry, Nov 15 2005
Column k has g.f. x^kLegendre_P(k, (1+x)/(1-x))/(1-x)^(k+1) = x^ksum{j =
0..k, C(k, j)^2*x^j}/(1-x)^(2k+1). - Paul Barry, Nov 19 2005
Let E be the operator DxD, where D denotes the derivative operator d/dx.
Then 1/n!^2  E^n(1/(1-x)) = (row n generating polynomial)/(1-x)^(2n+1) =
sum_{k = 0..inf} binomial(n + k, k)^2x^k. For example, when n = 3 we have
1/3!^2E^3(1/(1-x)) = (1 + 9x + 9x^2 + x^3)/(1-x)^7 = 1/3!^2  sum_{k =
0..inf} [(k+1)(k+2)(k+3)]^2*x^k. - Peter Bala, Oct 23 2008
G.f.: A(x, y) = Sum_{n >= 0} (2n)!/n!^2  x^(2n)y^n/(1-x-x*y)^(2n+1). - Paul
D. Hanna, Oct 31 2010
From Peter Bala, Jul 24 2013:

Let E(y) = sum_{n >= 0} y^n/n!^2 = BesselJ(0, 2sqrt(-y)). Generating
function: E(y)E(xy) = 1 + (1 + x)y + (1 + 4x + x^2)y^2/2!^2 + (1 +
9x + 9x^2 + x^3)y^3/3!^2 + .... Cf. the unsigned version of A021009
with generating function exp(y)E(x*y).
The n-th power of this array has the generating function E(y)^nE(xy). In
particular, the matrix inverse A055133 has the generating function
E(x*y)/E(y).


T(n,k) = T(n-1,k)*(n+k)/(n-k)+T(n-1,k-1), T(n,0)=T(n,n)=1. - Vladimir
Kruchinin, Oct 18 2014
Observe that the recurrence T(n,k) = T(n-1,k)*(n+k)/(n-k) - T(n-1,k-1), for n
= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1 gives
Pascal's triangle A007318. - Peter Bala, Dec 21 2014


n-th row polynomial R(n,x) = [z^n] (1 + (1 + x)z + xz^2)^n. Note that
1/n[z^(n-1)] (1 + (1 + x)z + x*z^2)^n gives the row polynomials of A001263.

Peter Bala, Jun 24 2015


Binomial transform of A105868. If G(x,t) = 1/sqrt(1 - 2(1 + t)x + (1 -
t)^2x^2) denotes the o.g.f. of this array then 1 + xd/dx(log(G(x,t)) = 1 +
(1 + t)x + (1 + 6t + t^2)*x^2 + ... is the o.g.f. for A086645. - Peter
Bala, Sep 06 2015
T(n,k) = Sum_{i=0..n} C(n-i,k)C(n,i)C(n+i,i)*(-1)^(n-i-k). - Vladimir
Kruchinin, Jan 14 2018
T(n,k) = A007318(n,k)^2. - Sean A. Irvine, Mar 29 2018

Cross references:

Row sums are in A000984. Columns 0-3 are A000012, A000290, A000537, A001249.
Cf. A007318, A055133, A116647, A001263, A086645, A063007, A108558,
A108625(Hilbert transform), A145903, A181543, A086645 (logarithmic
derivative), A105868 (inverse binomial transform).

A029653: Numbers in (2,1)-Pascal triangle (by row).

by Mohammad K. Azarian
Keywords: nonn,tabl
Data:
$$
\begin{array}{c|ccccccccccc }
n, k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
0 & 1 &  &  &  &  &  &  &  &  & \\1 & 2 & 1 &  &  &  &  &  &  &  & \\2 & 2 & 3 & 1 &  &  &  &  &  &  & \\3 & 2 & 5 & 4 & 1 &  &  &  &  &  & \\4 & 2 & 7 & 9 & 5 & 1 &  &  &  &  & \\5 & 2 & 9 & 16 & 14 & 6 & 1 &  &  &  & \\6 & 2 & 11 & 25 & 30 & 20 & 7 & 1 &  &  & \\7 & 2 & 13 & 36 & 55 & 50 & 27 & 8 & 1 &  & \\8 & 2 & 15 & 49 & 91 & 105 & 77 & 35 & 9 & 1 & \\9 & 2 & 17 & 64 & 140 & 196 & 182 & 112 & 44 & 10 & 1
\end{array}
$$
Comments:

Reverse of A029635. Row sums are A003945. Diagonal sums are Fib(n+2) =
sum_{k=0..floor(n/2)} (2n-3k)C(n-k,n-2k)/(n-k). - Paul Barry, Jan 30 2005
Riordan array ((1+x)/(1-x), x/(1-x)). The signed triangle (-1)^(n-k)T(n,k) or
((1-x)/(1+x), x/(1+x)) is the inverse of A055248. Row sums are A003945.
Diagonal sums are F(n+2). - Paul Barry, Feb 03 2005
Row sums = A003945: (1, 3, 6, 12, 24, 48, 96...) = (1, 3, 7, 15, 31, 63,
127...) - (0, 0, 1, 3, 7, 15, 31,...); where (1, 3, 7, 15,...) = A000225. -
Gary W. Adamson, Apr 22 2007
Triangle T(n,k), read by rows, given by (2,-1,0,0,0,0,0,0,0,...) DELTA
(1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. -
Philippe Deléham, Nov 17 2011
A029653 is jointly generated with A208510 as an array of coefficients of
polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1,
u(n,x)=u(n-1,x)+xv(n-1)x and v(n,x)=u(n-1,x)+xv(n-1,x)+1. See the
Mathematica section. - Clark Kimberling, Feb 28 2012

Formulae:

T(n, k) = C(n-2, k-1) + C(n-2, k) + C(n-1, k-1) + C(n-1, k) except for n=0.
G.f.: (1 + x + y + xy)/(1 - y - xy). - Ralf Stephan, May 17 2004
T(n, k) = (2n-k)*binomial(n, n-k)/n, n, k>0. - Paul Barry, Jan 30 2005
Sum_{k=0..n} T(n, k)*x^k gives A003945-A003954 for x = 1, 2, 3, 4, 5, 6, 7, 8,
9, 10. - Philippe Deléham, Jul 10 2005
T(n, k) = C(n-1, k) + C(n, k) . - Philippe Deléham, Jul 10 2005
Equals A097806  A007318, i.e., the pairwise operator  Pascal's Triangle as
infinite lower triangular matrices. - Gary W. Adamson, Apr 22 2007
From Peter Bala, Dec 27 2014:

exp(x)  e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3
we have exp(x)(2 + 5x + 4x^2/2! + x^3/3!) = 2 + 7x + 16x^2/2! +
30x^3/3! + 50x^4/4! + .... The same property holds more generally for
Riordan arrays of the form ( f(x), x/(1 - x) ).
Let M denote the lower unit triangular array with 1's on the main diagonal
and 1's everywhere else below the main diagonal except for the first
column which consists of the sequence [1,2,2,2,....]. For k = 0,1,2,...
define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/ having the k X k identity matrix I_k as the upper left block; in
particular, M(0) = M. Then the present triangle equals the infinite
product M(0)M(1)M(2)*... (which is clearly well-defined). See the
Example section.



Cross references:

(d, 1) Pascal triangles for d=3..10: A093560-5, A093644-5.
Cf. A007318, A003945, A208510, A228196, A228576.
Cf. A078812, A106195.

A074909: Running sum of Pascal's triangle (A007318), or beheaded Pascal's triangle read by beheaded rows.

by Wouter Meeussen, Oct 01 2002
Keywords: easy,nonn,tabl
Data:
$$
\begin{array}{c|ccccccccccc }
n, k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
0 & 1 &  &  &  &  &  &  &  &  & \\1 & 1 & 2 &  &  &  &  &  &  &  & \\2 & 1 & 3 & 3 &  &  &  &  &  &  & \\3 & 1 & 4 & 6 & 4 &  &  &  &  &  & \\4 & 1 & 5 & 10 & 10 & 5 &  &  &  &  & \\5 & 1 & 6 & 15 & 20 & 15 & 6 &  &  &  & \\6 & 1 & 7 & 21 & 35 & 35 & 21 & 7 &  &  & \\7 & 1 & 8 & 28 & 56 & 70 & 56 & 28 & 8 &  & \\8 & 1 & 9 & 36 & 84 & 126 & 126 & 84 & 36 & 9 & \\9 & 1 & 10 & 45 & 120 & 210 & 252 & 210 & 120 & 45 & 10
\end{array}
$$
Comments:

This sequence counts the "almost triangular" partitions of n. A partition is
triangular if it is of the form 0+1+2+...+k. Examples: 3=0+1+2, 6=0+1+2+3. An
"almost triangular" partition is a triangular partition with at most 1 added
to each of the parts. Examples: 7 = 1+1+2+3 = 0+2+2+3 = 0+1+3+3 = 0+1+2+4.
Thus a(7)=4. 8 = 1+2+2+3 = 1+1+3+3 = 1+1+2+4 = 0+2+3+3 = 0+2+2+4 = 0+1+3+4 so
a(8)=6. - Moshe Shmuel Newman, Dec 19 2002
The "almost triangular" partitions are the ones cycled by the operation of
"Bulgarian solitaire", as defined by Martin Gardner.
Start with A007318 - I (I = Identity matrix), then delete right border of
zeros. - Gary W. Adamson, Jun 15 2007
Also the number of increasing acyclic functions from {1..n-k+1} to {1..n+2}. A
function f is acyclic if for every subset B of the domain the image of B under
f does not equal B. For example, T(3,1)=4 since there are exactly 4 increasing
acyclic functions from {1,2,3} to {1,2,3,4,5}: f1={(1,2),(2,3),(3,4)},
f2={(1,2),(2,3),(3,5)}, f3={(1,2),(2,4),(3,5)} and f4={(1,3),(2,4),(4,5)}. -
Dennis P. Walsh, Mar 14 2008
Second Bernoulli polynomials are (from A164555 instead of A027641) B2(n,x) =
1; 1/2, 1; 1/6, 1, 1; 0, 1/2, 3/2, 1; -1/30, 0, 1, 2, 1; 0, -1/6, 0, 5/3, 5/2,
1; ... . Then (B2(n,x)/A002260) = 1; 1/2, 1/2; 1/6, 1/2, 1/3; 0, 1/4, 1/2,
1/4; -1/30, 0, 1/3, 1/2, 1/5; 0, -1/12, 0, 5/12, 1/2, 1/6; ... . See (from
Faulhaber 1631) Jacob Bernoulli Summae Potestatum (sum of powers) in A159688.
Inverse polynomials are 1; -1, 2; 1, -3, 3; -1, 4, -6, 4; ... = A074909 with
negative even diagonals. Reflected A053382/A053383 = reflected B(n,x) =
RB(n,x) = 1; -1/2, 1; 1/6, -1, 1; 0, 1/2, -3/2, 1; ... . A074909 is inverse of
RB(n,x)/A002260 = 1; -1/2, 1/2; 1/6, -1/2, 1/3; 0, 1/4, -1/2, 1/4; ... . -
Paul Curtz, Jun 21 2010

Formulae:

T(n, k) = Sum_{i=0..n} C(i, n-k) = C(n+1, k).
Row n has g.f. (1+x)^(n+1)-x^(n+1).
E.g.f.: ((1+x)e^t - x) e^(xt). The row polynomials p_n(x) satisfy dpn(x)/dx
= (n+1)*p(n-1)(x). - Tom Copeland, Jul 10 2018
T(n, k) = T(n-1, k-1) + T(n-1, k) for k: 0<k<n, T(n, 0)=1, T(n, n)=n. -
Reinhard Zumkeller, Apr 18 2005
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(0,0)=1,
T(1,0)=1, T(1,1)=2, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27
2013
G.f. for column k (with leading zeros): x^(k-1)*(1/(1-x)^(k+1)-1), k >= 0. -
Wolfdieter Lang, Nov 04 2014
Up(n, x+y) = (Up(.,x)+ y)^n = Sum_{k=0..n} binomial(n,k) Up(k,x)y^(n-k),
where Up(n,x) = ((x+1)^(n+1)-x^(n+1)) / (n+1) = P(n,x)/(n+1) with P(n,x) the
n-th row polynomial of this entry. dUp(n,x)/dx = n  Up(n-1,x) and dP(n,x)/dx
= (n+1)*P(n-1,x). - Tom Copeland, Nov 14 2014
The o.g.f. GF(x,t) = x / ((1-tx)(1-(1+t)x)) = x + (1+2t)x^2 +
(1+3t+3t^2)x^3 + ... has the inverse GFinv(x,t) =
(1+(1+2t)x-sqrt(1+(1+2t)*2x+x^2))/(2t(1+t)x) in x about 0, which generates the
row polynomials (mod row signs) of A033282. The reciprocal of the o.g.f.,
i.e., x/GF(x,t), gives the free cumulants (1, -(1+2t) , t(1+t) , 0, 0, ...)
associated with the moments defined by GFinv, and, in fact, these free
cumulants generate these moments through the noncrossing partitions of
A134264. The associated e.g.f. and relations to Grassmannians are described in
A248727, whose polynomials are the basis for an Appell sequence of polynomials
that are umbral compositional inverses of the Appell sequence formed from this
entry's polynomials (distinct from the one described in the comments above,
without the normalizing reciprocal). - Tom Copeland, Jan 07 2015

Cross references:

Cf. A007318, A181971, A228196, A228576.
Row sums are A000225, diagonal sums are A052952.
The number of acyclic functions is A058127.
Cf. A008292, A090582, A019538, A049019, A133314, A135278, A133437, A111785,
A126216, A134685, A133932, A248727, A033282, A134264.
Cf. A000027, A000217, A000292, A000332, A000389, A000579, A000580, A000581,
A000582.

A037027: Skew Fibonacci-Pascal triangle read by rows.

by Floor van Lamoen, Jan 01 1999
Keywords: easy,nonn,tabl
Data:
$$
\begin{array}{c|ccccccccccc }
n, k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
0 & 1 &  &  &  &  &  &  &  &  & \\1 & 1 & 1 &  &  &  &  &  &  &  & \\2 & 2 & 2 & 1 &  &  &  &  &  &  & \\3 & 3 & 5 & 3 & 1 &  &  &  &  &  & \\4 & 5 & 10 & 9 & 4 & 1 &  &  &  &  & \\5 & 8 & 20 & 22 & 14 & 5 & 1 &  &  &  & \\6 & 13 & 38 & 51 & 40 & 20 & 6 & 1 &  &  & \\7 & 21 & 71 & 111 & 105 & 65 & 27 & 7 & 1 &  & \\8 & 34 & 130 & 233 & 256 & 190 & 98 & 35 & 8 & 1 & \\9 & 55 & 235 & 474 & 594 & 511 & 315 & 140 & 44 & 9 & 1
\end{array}
$$
Comments:

T(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (0,1),
(1,0), (2,0). - Joerg Arndt, Jun 30 2011
T(n,k) is the number of lattice paths of length n, starting from the origin
and ending at (n,k), using horizontal steps H=(1,0), up steps U=(1,1) and down
steps D=(1,-1), never containing UUU, DD, HD. For instance, for n=4 and k=2,
we have the paths; HHUU, HUHU, HUUH, UHHU, UHUH, UUHH, UUDU, UDUU, UUUD. -
Emanuele Munarini, Mar 15 2011
Row sums form Pell numbers A000129, T(n,0) forms Fibonacci numbers A000045,
T(n,1) forms A001629. T(n+k,n-k) is polynomial sequence of degree k.
T(n,k) gives a convolved Fibonacci sequence (A001629, A001872, etc.).
As a Riordan array, this is (1/(1-x-x^2),x/(1-x-x^2)). An interesting
factorization is (1/(1-x^2),x/(1-x^2))*(1/(1-x),x/(1-x)) [abs(A049310) times
A007318]. Diagonal sums are the Jacobsthal numbers A001045(n+1). - Paul
Barry, Jul 28 2005

Formulae:

T(n, m) = T'(n-1, m)+T'(n-2, m)+T'(n-1, m-1), where T'(n, m) = T(n, m) for n
= 0 and 0< = m< = n and T'(n, m) = 0 otherwise.


G.f.: 1/(1 - y - y*z - y^2).
G.f. for k-th column: x/(1-x-x^2)^k.
T(n, m) = sum(binomial(m+k, m)*binomial(k, n-k-m), k=0..n-m), n>=m>=0, else 0.

Wolfdieter Lang, Jun 17 2002


T(n, m) = ((n-m+1)T(n, m-1) + 2(n+m)T(n-1, m-1))/(5m), n >= m >= 1; T(n,
0)= A000045(n+1); T(n, m)= 0 if n<m. - Wolfdieter Lang, Apr 12 2000
Chebyshev coefficient triangle (abs(A049310)) times Pascal's triangle
(A007318) as product of lower triangular matrices. T(n, k)=sum{k=0..n,
C((n+j)/2, j)(1+(-1)^(n+j))C(j, k)/2}. - Paul Barry, Dec 22 2004
Let R(n) = n-th row polynomial in x, with R(0)=1, then R(n+1)/R(n) equals the
continued fraction [1+x;1+x, ...(1+x) occurring (n+1) times..., 1+x] for n>=0.

Paul D. Hanna, Feb 27 2004


T(n,k) = sum{j=0..n, C(n-j,j)C(n-2j,k)}; in Egorychev notation,
T(n,k)=res_w(1-w-w^2)^(-k-1)*w^(-n+k+1). - Paul Barry, Sep 13 2006
Sum_{k=0..n} T(n,k)*x^k = A000045(n+1), A000129(n+1), A006190(n+1),
A001076(n+1), A052918(n), A005668(n+1), A054413(n), A041025(n), A099371(n+1),
A041041(n), A049666(n+1), A041061(n), A140455(n+1), A041085(n), A154597(n+1),
A041113(n) for n = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 respectively. -
Philippe Deléham, Nov 29 2009
T((m+1)n+r-1,mn+r-1)r/(mn+r) = Sum_{k=1..n}
k/nT((m+1)n-k-1,mn-1)(r+k,r), n>=m>1.
T(n-1,m-1) = m/nsum(k=1..n-m+1,kA000045(k)*T(n-k-1,m-2),k,1,n-m+1), n>=m>1.

Vladimir Kruchinin, Mar 17 2011


T(n,k) = binomial(n,k)*hypergeom([(k-n)/2, (k-n+1)/2], [-n], -4) for n>=1. -
Peter Luschny, Apr 25 2016

Cross references:

A038112(n)=T(2n, n). A038137 is reflected version. Maximal row entries:
A038149.
Diagonal differences are in A055830. Vertical sums are in A091186.
Cf. A007318, A049310, A000129, A155161, A122542, A059283, A228196, A228576.
Some other Fibonacci-Pascal triangles: A027926, A036355, A074829, A105809,
A109906, A111006, A114197, A162741, A228074.

A130595: Triangle read by rows: lower triangular matrix which is inverse to Pascal's triangle (A007318) regarded as a lower triangular matrix.

by Philippe Deléham, Jun 17 2007
Keywords: sign,nice,tabl
Data:
$$
\begin{array}{c|ccccccccccc }
n, k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
0 & 1 &  &  &  &  &  &  &  &  & \\1 & -1 & 1 &  &  &  &  &  &  &  & \\2 & 1 & -2 & 1 &  &  &  &  &  &  & \\3 & -1 & 3 & -3 & 1 &  &  &  &  &  & \\4 & 1 & -4 & 6 & -4 & 1 &  &  &  &  & \\5 & -1 & 5 & -10 & 10 & -5 & 1 &  &  &  & \\6 & 1 & -6 & 15 & -20 & 15 & -6 & 1 &  &  & \\7 & -1 & 7 & -21 & 35 & -35 & 21 & -7 & 1 &  & \\8 & 1 & -8 & 28 & -56 & 70 & -56 & 28 & -8 & 1 & \\9 & -1 & 9 & -36 & 84 & -126 & 126 & -84 & 36 & -9 & 1
\end{array}
$$
Comments:

Triangle T(n,k), read by rows, given by [-1,0,0,0,0,0,0,0,...] DELTA
[1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.
Coefficients of the polynomials generated by the e.g.f. exp(xt)exp(-t). -
Peter Luschny, Jul 13 2009
Riordan array (1/(1+x), x/(1+x)). - Philippe Deléham, Nov 29 2009
Multiplication of a sequence (written as column vector) by this matrix (to the
left) yields the inverse Binomial transform of the sequence. - M. F. Hasler,
Nov 01 2014
From Tom Copeland, Nov 16 2016:

Formulae:

T(n,k) = (-1)^(n-k)binomial(n,k) = (-1)^(n-k)A007318(n,k).
T(n,k) = T(n-1,k-1)-T(n-1,k). - Philippe Deléham, Oct 10 2011
G.f.: 1/(1+x-x*y). - R. J. Mathar, Aug 11 2015 [corrected by Anders
Claesson, Nov 28 2015]
Conjecture from Dale Gerdemann, Nov 28 2015: T(n,k) = (n-k+1)T(n-1,k-1) +
(k-1)T(n-1,k). Proof from Anders Claesson, Nov 29 2015: It follows from
T(n,k) = T(n-1,k-1) - T(n-1,k) and nT(n-1,k-1) = kT(n,k) that
(n-k+1)T(n-1,k-1) + (k-1)T(n-1,k) = nT(n-1,k-1) - (k-1)T(n-1,k-1) +
(k-1)T(n-1,k) = nT(n-1,k-1) - (k-1)(T(n-1,k-1) - T(n-1,k)) = nT(n-1,k-1) -
(k-1)T(n,k) = nT(n-1,k-1) - k*T(n,k) + T(n,k) = T(n,k). QED
(-1)^(n+1) Sum{k=1..n} T(n,k)/k = Sum{k=1..n} 1/k = H(n) where H(n) is the
n-th harmonic number. For a proof see link "Relation between binomial
coefficients and harmonic numbers". - Wolfgang Hintze, Oct 22 2016

Cross references:

Cf. A007318.

A132440: Infinitesimal Pascal matrix: generator (lower triangular matrix representation) of the Pascal matrix, the classical operator xDx, iterated Laguerre transforms, associated matrices of the list partition transform and general Euler transformation for sequences.

by Tom Copeland, Nov 13 2007, Nov 15 2007, Nov 22 2007, Dec 02 2007
Keywords: easy,nonn,tabl
Data:
$$
\begin{array}{c|ccccccccccc }
n, k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
0 & 0 &  &  &  &  &  &  &  &  & \\1 & 1 & 0 &  &  &  &  &  &  &  & \\2 & 0 & 2 & 0 &  &  &  &  &  &  & \\3 & 0 & 0 & 3 & 0 &  &  &  &  &  & \\4 & 0 & 0 & 0 & 4 & 0 &  &  &  &  & \\5 & 0 & 0 & 0 & 0 & 5 & 0 &  &  &  & \\6 & 0 & 0 & 0 & 0 & 0 & 6 & 0 &  &  & \\7 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & 0 &  & \\8 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & 0 & \\9 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 & 0
\end{array}
$$
Comments:

Let M(t) = exp(tT) = Lim_{n->infinity} (1 + tT/n)^n.
Pascal matrix = [ binomial(n,k) ] = M(1) = exp(T), truncating the series gives
the n X n submatrices.
Inverse Pascal matrix = M(-1) = exp(-T) = matrix for inverse binomial
transform.
A(j) = T^j / j! equals the matrix [bin(n,k) * delta(n-k-j)] where delta(n) = 1
if n=0 and vanishes otherwise (Kronecker delta); i.e., A(j) is a matrix with
all the terms 0 except for the j-th lower (or main for j=0) diagonal which
equals that of the Pascal triangle. Hence the A(j)'s form a linearly
independent basis for all matrices of the form [binomial(n,k) d(n-k)] which
include as a subset the invertible associated matrices of the list partition
transform (LPT) of A133314.
For sequences with b(0) = 1, umbrally,

Formulae:

T = log(P) with the Pascal matrix P:=A007318. This should be read as T_N =
log(P_N) with P_N the N X N matrix P, N>=2. Because P_N is lower triangular
with all diagonal elements 1, the series log(1_N-(1_N-P_N)) stops after N-1
terms because (1_N-P_N)^N is the 0_N-matrix. - Wolfdieter Lang, Oct 14 2010
Given a polynomial sequence p_n(x) with p_0(x)=1 and the lowering and raising
operators L and R defined by L pn(x) = n * p(n-1)(x) and R pn(x) =
p(n+1)(x), the matrix T represents the action of RLR in the p_n(x) basis.
For p_n(x) = x^n, L = D = d/dx and R = x. For p_n(x) = x^n/n!, L = DxD and R =
D^(-1). - Tom Copeland, Oct 25 2012
From Tom Copeland, Apr 26 2014:

A) T = exp(A238385-I) - I
B)   = [St1]P[St2] - I
C)   = [St1]P[St1]^(-1) - I
D)   = [St2]^(-1)P[St2] - I
E)   = [St2]^(-1)P[St1]^(-1) - I
where P=A007318, [St1]=padded A008275 just as [St2]=A048993=padded
A008277, and I=identity matrix.


From Robert Israel, Oct 02 2015:

G.f. Sum_{k >= 1} k x^((k+3/2)^2/2 - 17/8) is related to Jacobi theta
functions.
If 8*n+17 = y^2 is a square, then a(n) = (y-3)/2, otherwise a(n) = 0.



A135278: Triangle read by rows, giving the numbers T(n,m) = binomial(n+1, m+1); or, Pascal's triangle A007318 with its left-hand edge removed.

by Zerinvary Lajos, Dec 02 2007
Keywords: easy,nonn,tabl
Data:
$$
\begin{array}{c|ccccccccccc }
n, k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
0 & 1 &  &  &  &  &  &  &  &  & \\1 & 2 & 1 &  &  &  &  &  &  &  & \\2 & 3 & 3 & 1 &  &  &  &  &  &  & \\3 & 4 & 6 & 4 & 1 &  &  &  &  &  & \\4 & 5 & 10 & 10 & 5 & 1 &  &  &  &  & \\5 & 6 & 15 & 20 & 15 & 6 & 1 &  &  &  & \\6 & 7 & 21 & 35 & 35 & 21 & 7 & 1 &  &  & \\7 & 8 & 28 & 56 & 70 & 56 & 28 & 8 & 1 &  & \\8 & 9 & 36 & 84 & 126 & 126 & 84 & 36 & 9 & 1 & \\9 & 10 & 45 & 120 & 210 & 252 & 210 & 120 & 45 & 10 & 1
\end{array}
$$
Comments:

T(n,m) is the number of m-faces of a regular n-simplex.
An n-simplex is the n-dimensional analog of a triangle. Specifically, a
simplex is the convex hull of a set of (n + 1) affinely independent points in
some Euclidean space of dimension n or higher, i.e., a set of points such that
no m-plane contains more than (m + 1) of them. Such points are said to be in
general position.
Reversing the rows gives A074909, which as a linear sequence is essentially
the same as this.
From Tom Copeland, Dec 07 2007:

T(n,k) * (k+1)! = A068424. The comment on permuted words in A068424 shows
that T is related to combinations of letters defined by connectivity of
regular polytope simplexes.
If T is the diagonally-shifted Pascal matrix, binomial(n+m, k+m), for m=1,
then T is a fundamental type of matrix that is discussed in A133314 and
the following hold.
The infinitesimal matrix generator is given by A132681, so T = LM(1) of
A132681 with inverse LM(-1).
With a(k) = (-x)^k / k!, T * a = [ Laguerre(n,x,1) ], a vector array with
index n for the Laguerre polynomials of order 1. Other formulas for the
action of T are given in A132681.
T(n,k) = (1/n!) (D_x)^n (D_t)^k Gf(x,t) evaluated at x=t=0 with Gf(x,t) =
exp[ t * x/(1-x) ] / (1-x)^2.
[O.g.f. for T ] = 1 / { [ 1 - t  x/(1-x) ]  (1-x)^2 }. [ O.g.f. for row
sums ] = 1 / { (1-x) * (1-2x) }, giving A000225 (without a leading zero)
for the row sums. Alternating sign row sums are all 1. [Sign correction
noted by Vincent J. Matsko, Jul 19 2015]
O.g.f. for row polynomials = [ (1+q)**(n+1) - 1 ] / [ (1+q) -1 ] =
A(1,n+1,q) on page 15 of reference on Grassmann cells in A008292.



Formulae:

T(n, k) = Sum_{j=k..n} binomial(j,k) = binomial(n+1, k+1), n >= k >= 0, else

(Partial sum of column k of A007318 (Pascal), or summation on the upper
binomial index (Graham et al. (GKP), eq. (5.10). For the GKP reference see
A007318.) - Wolfdieter Lang, Aug 22 2012


E.g.f.: 1/x((1 + x)exp(t(1 + x)) - exp(t)) = 1 + (2 + x)t + (3 + 3x +
x^2)t^2/2! + .... The infinitesimal generator for this triangle has the
sequence [2,3,4,...] on the main subdiagonal and 0's elsewhere. - Peter
Bala, Jul 16 2013
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=1, T(1,0)=2,
T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
T(n,k) = A193862(n,k)/2^k. - Philippe Deléham, Jan 29 2014
G.f.: 1/((1-x)(1-x-xy)). - Philippe Deléham, Mar 13 2014
From Tom Copeland, Mar 26 2014:

[From Copeland's 2007 and 2008 comments]
A) O.g.f.: 1 / { [ 1 - t  x/(1-x) ]  (1-x)^2 } (same as Deleham's).
B) The infinitesimal generator for T is given in A132681 with m=1 (same as
Bala's), which makes connections to the ubiquitous associated Laguerre
polynomials of integer orders, for this case the Laguerre polynomials of
order one L(n,-t,1).
C) O.g.f. of row e.g.f.s: Sum_{n>=0} L(n,-t,1) x^n =
exp[tx/(1-x)]/(1-x)^2 = 1 + (2+t)x + (3+3t+t^2/2!)x^2 +
(4+6t+4t^2/2!+t^3/3!)x^3+ ... .
D) E.g.f. of row o.g.f.s: ((1+t)exp((1+t)x)-exp(x))/t (same as Bala's).
E) E.g.f. for T(n,k)*a(n-k): {(a+t) exp[(a+t)x] - a exp(a x)}/t, umbrally.
For example, for a(k)=2^k, the e.g.f. for the row o.g.f.s is {(2+t)
exp[(2+t)x] - 2 exp(2x)}/t.


From Tom Copeland, Apr 28 2014:

With different indexing
A) O.g.f. by row: [(1+t)^n-1]/t.
B) O.g.f. of row o.g.f.s: {1/[1-(1+t)*x] - 1/(1-x)}/t.
C) E.g.f. of row o.g.f.s: {exp[(1+t)*x]-exp(x)}/t.
These generating functions are related to row e.g.f.s of A111492.


From Tom Copeland, Sep 17 2014:
A) U(x,s,t)= x^2/[(1-tx)(1-(s+t)x)] = Sum_{n >= 0} F(n,s,t)x^(n+2) is a
generating function for bivariate row polynomials of T, e.g., F(2,s,t)= s^2 +
3st + 3t^2 (Buchstaber, 2008).
B) dU/dt=x^2 dU/dx with U(x,s,0)= x^2/(1-s*x) (Buchstaber, 2008).
C) U(x,s,t) = exp(tx^2d/dx)U(x,s,0) = U(x/(1-t*x),s,0).
D) U(x,s,t) = Sum[n >= 0, (tx)^n L(n,-:xD:,-1)] U(x,s,0), where
(:xD:)^k=x^k(d/dx)^k and L(n,x,-1) are the Laguerre polynomials of order -1,
related to normalized Lah numbers. (End)
E.g.f. satisfies the differential equation d/dt(e.g.f.(x,t)) =
(x+1)*e.g.f.(x,t) + exp(t). - Vincent J. Matsko, Jul 18 2015
The e.g.f. of the Norlund generalized Bernoulli (Appell) polynomials of order
m, NB(n,x;m), is given by exponentiation of the e.g.f. of the Bernoulli
numbers, i.e., multiple binomial self-convolutions of the Bernoulli numbers,
through the e.g.f. exp[NB(.,x;m)t] = (t/(e^t - 1))^(m+1) * e^(xt). Norlund
gave the relation to the factorials (x-1)!/(x-1-n)! = (x-1) ... (x-n) =
NB(n,x;n), so T(n,m) = NB(m+1,n+2;m+1)/(m+1)!. - Tom Copeland, Oct 01 2015

Cross references:

Cf. A007318, A014410, A228196.
Cf. Column sequences: A000027, A000217, A000292, A000332, A000389, A000579 -
A000582, A001287, A001288, A010965 - A011001, A017713 - A017764.
Cf. A000012, A008292.

In [60]:

    

from requests import get
from IPython.display import Markdown


    

In [61]:

    

code=45

payload = {"fmt": "json", 
           #"q": "id:A{:06d}".format(code)}
           "q": "fibonacci",
           "start": 0}

doc_result = get("https://oeis.org/search", params=payload,)
doc = doc_result.json()


    

In [62]:

    

doc_result.url


    

    
Out[62]:





'https://oeis.org/search?fmt=json&q=fibonacci&start=0'

In [63]:

    

doc.keys()


    

    
Out[63]:





dict_keys(['greeting', 'query', 'count', 'start', 'results'])

In [64]:

    

results = doc['results']
results[0]['comment'][:10]


    

    
Out[64]:





["Also sometimes called Lamé's sequence.",
 "F(n+2) = number of binary sequences of length n that have no consecutive 0's.",
 'F(n+2) = number of subsets of {1,2,...,n} that contain no consecutive integers.',
 'F(n+1) = number of tilings of a 2 X n rectangle by 2 X 1 dominoes.',
 'F(n+1) = number of matchings (i.e., Hosoya index) in a path graph on n vertices: F(5)=5 because the matchings of the path graph on the vertices A, B, C, D are the empty set, {AB}, {BC}, {CD} and {AB, CD}. - _Emeric Deutsch_, Jun 18 2001',
 'F(n) = number of compositions of n+1 with no part equal to 1. [Cayley, Grimaldi]',
 'Positive terms are the solutions to z = 2*x*y^4 + (x^2)*y^3 - 2*(x^3)*y^2 - y^5 - (x^4)*y + 2*y for x,y >= 0 (Ribenboim, page 193). When x=F(n), y=F(n + 1) and z>0 then z=F(n + 1).',
 'For Fibonacci search see Knuth, Vol. 3; Horowitz and Sahni; etc.',
 "F(n) is the diagonal sum of the entries in Pascal's triangle at 45 degrees slope. - _Amarnath Murthy_, Dec 29 2001",
 'F(n+1) is the number of perfect matchings in ladder graph L_n = P_2 X P_n. - Sharon Sela (sharonsela(AT)hotmail.com), May 19 2002']

In [65]:

    

Markdown(pretty_print(doc['results'][0], 
                      comment=lambda i,c: "binomial" in c,
                      formula=lambda i,f: "theorem" in f,
                      link=lambda i,l: "paul barry" in l,
                      reference=lambda i,r: "riordan" in r))


    

    
Out[65]:




A000045: Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.
by N. J. A. Sloane, 1964
Keywords: nonn,core,nice,easy,hear
Data:
$$
\begin{array}{c|ccccccccccccccc }
    n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\
    \hline
    A000045(n) & 0 & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144 & 233 & 377
\end{array}
$$
Comments:

Fib(n+2) = Sum_{k=0..n} binomial(floor((n+k)/2),k), row sums of A046854. - Paul Barry, Mar 11 2003
The sequence F(n) is the binomial transformation of the alternating sequence (-1)^(n-1)F(n), whereas the sequence F(n+1) is the binomial transformation of the alternating sequence (-1)^nF(n-1). Both of these facts follow easily from the equalities a(n;1)=F(n+1) and b(n;1)=F(n) where a(n;d) and b(n;d) are so-called "delta-Fibonacci" numbers as defined in comments to A014445 (see also the papers of Witula et al.). - Roman Witula, Jul 24 2012
From Tom Copeland, Nov 02 2014: 

Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078).
BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).
Generalizing to P(x,t) = x /(1 + tx) and Pinv(x,t) = x /(1 - tx) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1].



Formulae:

A sufficient condition for F(m) to be divisible by a prime p is (p - 1) divides m, if p == 1 or 4 (mod 5); (p + 1) divides m, if p == 2 or 3 (mod 5); or 5 divides m, if p = 5. (This is essentially Theorem 180 in Hardy and Wright.) - Fred W. Helenius (fredh(AT)ix.netcom.com), Jun 29 2001
Kurmang. Aziz. Rashid, Feb 21 2004, makes 4 conjectures and gives 3 theorems:
Theorem 1: for n>=0, {F(n+3)^ 2 - F(n+1)^ 2}/F(n+2)={F(n+3)+ F(n+1)}.
Theorem 2: for n>=0, F(n+10) = 11*F(n+5) + F(n).
Theorem 3: for n>=6, F(n) = 4*F(n-3) + F(n-6).

Cross references:

Cf. A039834 (signed Fibonacci numbers), A001690 (complement), A000213, A000288, A000322, A000383, A060455, A030186, A020695, A020701, A071679, A099731, A100492, A094216, A094638, A000108, A101399, A101400, A001611, A000071, A157725, A001911, A157726, A006327, A157727, A157728, A157729, A167616, A059929, A144152, A152063, A114690, A003893, A000032, A060441, A000930, A003269, A000957, A057078, A007317, A091867, A104597, A249548, A262342, A001060, A022095.
First row of arrays A103323, A234357. Second row of arrays A099390, A048887, and A092921 (k-generalized Fibonacci numbers).
a(n) = A094718(4, n). a(n) = A101220(0, j, n).
a(n) = A090888(0, n+1) = A118654(0, n+1) = A118654(1, n-1) = A109754(0, n) = A109754(1, n-1), for n > 0.
Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074.
Boustrophedon transforms: A000738, A000744.
Powers: A103323, A105317, A254719.
Numbers of prime factors: A022307 and A038575.
Cf. A163733.

Links:

Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

References:

J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, Princeton, NJ, 1978.