Massimo Nocentini

September 27, 2016: refactoring and sync


Abstract
In this notebook we study the *Pascal triangle*, locking at it recursively: using it's $A$-sequence, we perform a series of unfolding using the main recurrence relation, where subscripts dependends on *two* dimensions, as rewriting rule. This is a natural enhancement to the case of recurrences where subscripts have *one* dimension only, as the *Fibonacci* sequence; on the other hand, Pascal array is a deeply studied and well know triangle, yet simple, toy we can play with.


In [64]:
%run "../src/start_session.py"
%run "../src/recurrences.py"

In [50]:
import oeis

Stirling array $\mathcal{S}$, of the second kind

This notebook studies the Riordan array $\mathcal{P}$, aka the Pascal triangle, defined according to the following definition:

$$\mathcal{P}=\left(\frac{1}{1-t}, \frac{t}{1-t}\right)$$

with $A$-sequence $A(t)=1+t$ and $Z$-sequence $Z(t)=1$.


In [3]:
s = oeis.oeis_search(id=48993)


*

In [4]:
s()


Out[4]:

Results for query: https://oeis.org/search?q=id%3AA048993&start=0&fmt=json


A048993: Triangle of Stirling numbers of 2nd kind, S(n,k), n >= 0, 0<=k<=n.

by N. J. A. Sloane, Dec 11 1999

Keywords: nonn,tabl,nice

Data:

$$ \begin{array}{c|ccccccccccc} n, k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 0 & 1 & & & & & & & & & \\1 & 0 & 1 & & & & & & & & \\2 & 0 & 1 & 1 & & & & & & & \\3 & 0 & 1 & 3 & 1 & & & & & & \\4 & 0 & 1 & 7 & 6 & 1 & & & & & \\5 & 0 & 1 & 15 & 25 & 10 & 1 & & & & \\6 & 0 & 1 & 31 & 90 & 65 & 15 & 1 & & & \\7 & 0 & 1 & 63 & 301 & 350 & 140 & 21 & 1 & & \\8 & 0 & 1 & 127 & 966 & 1701 & 1050 & 266 & 28 & 1 & \\9 & 0 & 1 & 255 & 3025 & 7770 & 6951 & 2646 & 462 & 36 & 1 \end{array} $$

Comments:

  • Also known as Stirling set numbers.
  • S(n,k) enumerates partitions of an n-set into k nonempty subsets.
  • The o.g.f. for the sequence of diagonal k (k=0 for the main diagonal) is G(k,x)= ((x^k)/(1-x)^(2k+1))sum(A008517(k,m+1)*x^m,m=0..k-1). A008517 is the second-order Eulerian triangle. - Wolfdieter Lang, Oct 14 2005.
  • From Philippe Deléham, Nov 14 2007:
    • Sum_{k, 0<=k<=n}S(n,k)*x^k = B_n(x), where B_n(x) = Bell polynomials. The first few Bell polynomials are:
    • B_0(x) = 1;
    • B_1(x) = 0 + x;
    • B_2(x) = 0 + x + x^2;
    • B_3(x) = 0 + x + 3x^2 + x^3;
    • B_4(x) = 0 + x + 7x^2 + 6x^3 + x^4;
    • B_5(x) = 0 + x + 15x^2 + 25x^3 + 10x^4 + x^5;
    • B_6(x) = 0 + x + 31x^2 + 90x^3 + 65x^4 + 15x^5 + x^6;
  • This is the Sheffer triangle (1, exp(x) - 1), an exponential (binomial) convolution triangle. The a-sequence is given by A006232/A006233 (Cauchy sequence). The z-sequence is the zero sequence. See the link under A006232 for the definition and use of these sequences. The row sums give A000110 (Bell), and the alternating row sums give A000587 (see the Philippe Deléham formulas and crossreferences below). - Wolfdieter Lang, Oct 16 2014
  • Also the inverse Bell transform of the factorial numbers (A000142). For the definition of the Bell transform see A264428 and for cross-references A265604. - Peter Luschny, Dec 31 2015

Formulae:

  • S(n, k) = k*S(n-1, k) + S(n-1, k-1), n>0; S(0, k) = 0, k>0; S(0, 0)=1.
  • Equals [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ..] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is Deléham's operator defined in A084938.
  • Sum_{k = 0..n} x^k*S(n, k) = A213170(n), A000587(n), A000007(n), A000110(n), A001861(n), A027710(n), A078944(n), A144180(n), A144223(n), A144263(n) respectively for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7. - Philippe Deléham, May 09 2004, Feb 16 2013
  • S(n, k) = sum{i=0..k, (-1)^(k+i)binomial(k, i)i^n/k!}. - Paul Barry, Aug 05 2004
  • Sum(k*S(n,k), k=0..n)=B(n+1)-B(n), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Nov 01 2006
  • Equals the inverse binomial transform of A008277. - Gary W. Adamson, Jan 29 2008
  • G.f.: 1/(1-xy/(1-x/(1-xy/(1-2x/(1-xy/1-3x/(1-xy/(1-4x/(1-xy/(1-5x/(1-... (continued fraction equivalent to Deleham DELTA construction). - Paul Barry, Dec 06 2009
  • G.f.: 1/Q(0), where Q(k) = 1 -(y+k)x - (k+1)y*x^2/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
  • Inverse of padded A008275 (padded just as A048993 = padded A008277). - Tom Copeland, Apr 25 2014
  • E.g.f. for the row polynomials s(n,x) = sum(S(n,k)x^k, k=0..n) is exp(x(exp(z)-1)) (Sheffer property). E.g.f. for the k-th column sequence with k leading zeros is ((exp(x)-1)^k)/k! (Sheffer property). - Wolfdieter Lang, Oct 16 2014

Cross references:

Links:

References:

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 310.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Springer, p. 92.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 244.
  • J. Riordan, An Introduction to Combinatorial Analysis, p. 48.

In [3]:
from sympy.functions.combinatorial.numbers import stirling

In [76]:
s = IndexedBase('s')
n, k = symbols('n k')

stirling_recurrence_spec = recurrence_spec(recurrence_eq=Eq(s[n+1, k+1], s[n, k] + (k+1)*s[n, k+1]), 
                                           recurrence_symbol=s, 
                                           variables=[n,k])

In [77]:
stirling_recurrence_spec


Out[77]:
$\left(\Theta, \Gamma\right)_{n,k}^{s}$ where:
  • $\Theta = \left\{ s_{n + 1,k + 1} = \left(k + 1\right) s_{n,k + 1} + s_{n,k} \right\}$
  • $\Gamma = \left\{\begin{array}{c}\end{array}\right\}$

In [78]:
unfolded = stirling_recurrence_spec.unfold(depth=1)

In [79]:
unfolded


Out[79]:
$\left(\Theta, \Gamma\right)_{n,k}^{s}$ where:
  • $\Theta = \left\{ s_{n + 1,k + 1} = k \left(\left(k + 1\right) s_{n - 1,k + 1} + s_{n - 1,k}\right) + k s_{n - 1,k} + \left(k + 1\right) s_{n - 1,k + 1} + s_{n - 1,k} + s_{n - 1,k - 1} \right\}$
  • $\Gamma = \left\{\begin{array}{c}s_{n,k} = k s_{n - 1,k} + s_{n - 1,k - 1}\\s_{n,k + 1} = \left(k + 1\right) s_{n - 1,k + 1} + s_{n - 1,k}\end{array}\right\}$

In [81]:
instantiated = unfolded.instantiate(strategy=raw(substitutions={n:9,k:4}))
instantiated


Out[81]:
$\left(\Theta, \Gamma\right)_{n,k}^{s}$ where:
  • $\Theta = \left\{ s_{10,5} = s_{8,3} + 9 s_{8,4} + 25 s_{8,5} \right\}$
  • $\Gamma = \left\{\begin{array}{c}s_{9,4} = s_{8,3} + 4 s_{8,4}\\s_{9,5} = s_{8,4} + 5 s_{8,5}\end{array}\right\}$

In [58]:
known_binomials = {s[n,k]:stirling(n,k) for n in [10,9,8] for k in range(2,6)}

checked = instantiated.instantiate(strategy=raw(substitutions=known_binomials))
checked


Out[58]:
$\left(\Theta, \Gamma\right)_{n,k}^{s}$ where:
  • $\Theta = \left\{ \mathrm{True} \right\}$
  • $\Gamma = \left\{\begin{array}{c}\end{array}\right\}$

In [59]:
checked.subsume()


Out[59]:
$\left(\Theta, \Gamma\right)_{n,k}^{s}$ where:
  • $\Theta = \left\{ \mathrm{True} \right\}$
  • $\Gamma = \left\{\begin{array}{c}\end{array}\right\}$

In [82]:
based_recurrence_spec = unfolded.instantiate(strategy=based(arity=doubly_indexed()))
based_recurrence_spec


Out[82]:
$\left(\Theta, \Gamma\right)_{n,k}^{s}$ where:
  • $\Theta = \left\{ s_{4,2} = s_{2,0} + 3 s_{2,1} + 4 s_{2,2} \right\}$
  • $\Gamma = \left\{\begin{array}{c}s_{3,1} = s_{2,0} + s_{2,1}\\s_{3,2} = s_{2,1} + 2 s_{2,2}\end{array}\right\}$

In [83]:
based_recurrence_spec.subsume()


Out[83]:
$\left(\Theta, \Gamma\right)_{n,k}^{s}$ where:
  • $\Theta = \left\{ s_{4,2} = s_{2,0} + 3 s_{2,1} + 4 s_{2,2} \right\}$
  • $\Gamma = \left\{\begin{array}{c}s_{3,1} = s_{2,0} + s_{2,1}\\s_{3,2} = s_{2,1} + 2 s_{2,2}\end{array}\right\}$

In [66]:
ipython_latex_description(rec_spec=stirling_recurrence_spec, depths=range(9), based_instantiation=False,
                          arity=doubly_indexed())


Out[66]:
\begin{array}{c}s_{n + 1,k + 1} = k s_{n,k + 1} + s_{n,k} + s_{n,k + 1}\\ s_{n + 1,k + 1} = k^{2} s_{n - 1,k + 1} + 2 k s_{n - 1,k} + 2 k s_{n - 1,k + 1} + s_{n - 1,k} + s_{n - 1,k - 1} + s_{n - 1,k + 1}\\ s_{n + 1,k + 1} = k^{3} s_{n - 2,k + 1} + 3 k^{2} s_{n - 2,k} + 3 k^{2} s_{n - 2,k + 1} + 3 k s_{n - 2,k} + 3 k s_{n - 2,k - 1} + 3 k s_{n - 2,k + 1} + s_{n - 2,k} + s_{n - 2,k - 2} + s_{n - 2,k + 1}\\ s_{n + 1,k + 1} = k^{4} s_{n - 3,k + 1} + 4 k^{3} s_{n - 3,k} + 4 k^{3} s_{n - 3,k + 1} + 6 k^{2} s_{n - 3,k} + 6 k^{2} s_{n - 3,k - 1} + 6 k^{2} s_{n - 3,k + 1} + 4 k s_{n - 3,k} + 4 k s_{n - 3,k - 2} + 4 k s_{n - 3,k + 1} + s_{n - 3,k} + s_{n - 3,k - 3} - 2 s_{n - 3,k - 2} + s_{n - 3,k - 1} + s_{n - 3,k + 1}\\ s_{n + 1,k + 1} = k^{5} s_{n - 4,k + 1} + 5 k^{4} s_{n - 4,k} + 5 k^{4} s_{n - 4,k + 1} + 10 k^{3} s_{n - 4,k} + 10 k^{3} s_{n - 4,k - 1} + 10 k^{3} s_{n - 4,k + 1} + 10 k^{2} s_{n - 4,k} + 10 k^{2} s_{n - 4,k - 2} + 10 k^{2} s_{n - 4,k + 1} + 5 k s_{n - 4,k} + 5 k s_{n - 4,k - 3} - 10 k s_{n - 4,k - 2} + 5 k s_{n - 4,k - 1} + 5 k s_{n - 4,k + 1} + s_{n - 4,k} + s_{n - 4,k - 4} - 5 s_{n - 4,k - 3} + 5 s_{n - 4,k - 2} + s_{n - 4,k + 1}\\ s_{n + 1,k + 1} = k^{6} s_{n - 5,k + 1} + 6 k^{5} s_{n - 5,k} + 6 k^{5} s_{n - 5,k + 1} + 15 k^{4} s_{n - 5,k} + 15 k^{4} s_{n - 5,k - 1} + 15 k^{4} s_{n - 5,k + 1} + 20 k^{3} s_{n - 5,k} + 20 k^{3} s_{n - 5,k - 2} + 20 k^{3} s_{n - 5,k + 1} + 15 k^{2} s_{n - 5,k} + 15 k^{2} s_{n - 5,k - 3} - 30 k^{2} s_{n - 5,k - 2} + 15 k^{2} s_{n - 5,k - 1} + 15 k^{2} s_{n - 5,k + 1} + 6 k s_{n - 5,k} + 6 k s_{n - 5,k - 4} - 30 k s_{n - 5,k - 3} + 30 k s_{n - 5,k - 2} + 6 k s_{n - 5,k + 1} + s_{n - 5,k} + s_{n - 5,k - 5} - 9 s_{n - 5,k - 4} + 20 s_{n - 5,k - 3} - 10 s_{n - 5,k - 2} + s_{n - 5,k - 1} + s_{n - 5,k + 1}\\ s_{n + 1,k + 1} = k^{7} s_{n - 6,k + 1} + 7 k^{6} s_{n - 6,k} + 7 k^{6} s_{n - 6,k + 1} + 21 k^{5} s_{n - 6,k} + 21 k^{5} s_{n - 6,k - 1} + 21 k^{5} s_{n - 6,k + 1} + 35 k^{4} s_{n - 6,k} + 35 k^{4} s_{n - 6,k - 2} + 35 k^{4} s_{n - 6,k + 1} + 35 k^{3} s_{n - 6,k} + 35 k^{3} s_{n - 6,k - 3} - 70 k^{3} s_{n - 6,k - 2} + 35 k^{3} s_{n - 6,k - 1} + 35 k^{3} s_{n - 6,k + 1} + 21 k^{2} s_{n - 6,k} + 21 k^{2} s_{n - 6,k - 4} - 105 k^{2} s_{n - 6,k - 3} + 105 k^{2} s_{n - 6,k - 2} + 21 k^{2} s_{n - 6,k + 1} + 7 k s_{n - 6,k} + 7 k s_{n - 6,k - 5} - 63 k s_{n - 6,k - 4} + 140 k s_{n - 6,k - 3} - 70 k s_{n - 6,k - 2} + 7 k s_{n - 6,k - 1} + 7 k s_{n - 6,k + 1} + s_{n - 6,k} + s_{n - 6,k - 6} - 14 s_{n - 6,k - 5} + 56 s_{n - 6,k - 4} - 70 s_{n - 6,k - 3} + 21 s_{n - 6,k - 2} + s_{n - 6,k + 1}\\ s_{n + 1,k + 1} = k^{8} s_{n - 7,k + 1} + 8 k^{7} s_{n - 7,k} + 8 k^{7} s_{n - 7,k + 1} + 28 k^{6} s_{n - 7,k} + 28 k^{6} s_{n - 7,k - 1} + 28 k^{6} s_{n - 7,k + 1} + 56 k^{5} s_{n - 7,k} + 56 k^{5} s_{n - 7,k - 2} + 56 k^{5} s_{n - 7,k + 1} + 70 k^{4} s_{n - 7,k} + 70 k^{4} s_{n - 7,k - 3} - 140 k^{4} s_{n - 7,k - 2} + 70 k^{4} s_{n - 7,k - 1} + 70 k^{4} s_{n - 7,k + 1} + 56 k^{3} s_{n - 7,k} + 56 k^{3} s_{n - 7,k - 4} - 280 k^{3} s_{n - 7,k - 3} + 280 k^{3} s_{n - 7,k - 2} + 56 k^{3} s_{n - 7,k + 1} + 28 k^{2} s_{n - 7,k} + 28 k^{2} s_{n - 7,k - 5} - 252 k^{2} s_{n - 7,k - 4} + 560 k^{2} s_{n - 7,k - 3} - 280 k^{2} s_{n - 7,k - 2} + 28 k^{2} s_{n - 7,k - 1} + 28 k^{2} s_{n - 7,k + 1} + 8 k s_{n - 7,k} + 8 k s_{n - 7,k - 6} - 112 k s_{n - 7,k - 5} + 448 k s_{n - 7,k - 4} - 560 k s_{n - 7,k - 3} + 168 k s_{n - 7,k - 2} + 8 k s_{n - 7,k + 1} + s_{n - 7,k} + s_{n - 7,k - 7} - 20 s_{n - 7,k - 6} + 126 s_{n - 7,k - 5} - 294 s_{n - 7,k - 4} + 231 s_{n - 7,k - 3} - 42 s_{n - 7,k - 2} + s_{n - 7,k - 1} + s_{n - 7,k + 1}\\ s_{n + 1,k + 1} = k^{9} s_{n - 8,k + 1} + 9 k^{8} s_{n - 8,k} + 9 k^{8} s_{n - 8,k + 1} + 36 k^{7} s_{n - 8,k} + 36 k^{7} s_{n - 8,k - 1} + 36 k^{7} s_{n - 8,k + 1} + 84 k^{6} s_{n - 8,k} + 84 k^{6} s_{n - 8,k - 2} + 84 k^{6} s_{n - 8,k + 1} + 126 k^{5} s_{n - 8,k} + 126 k^{5} s_{n - 8,k - 3} - 252 k^{5} s_{n - 8,k - 2} + 126 k^{5} s_{n - 8,k - 1} + 126 k^{5} s_{n - 8,k + 1} + 126 k^{4} s_{n - 8,k} + 126 k^{4} s_{n - 8,k - 4} - 630 k^{4} s_{n - 8,k - 3} + 630 k^{4} s_{n - 8,k - 2} + 126 k^{4} s_{n - 8,k + 1} + 84 k^{3} s_{n - 8,k} + 84 k^{3} s_{n - 8,k - 5} - 756 k^{3} s_{n - 8,k - 4} + 1680 k^{3} s_{n - 8,k - 3} - 840 k^{3} s_{n - 8,k - 2} + 84 k^{3} s_{n - 8,k - 1} + 84 k^{3} s_{n - 8,k + 1} + 36 k^{2} s_{n - 8,k} + 36 k^{2} s_{n - 8,k - 6} - 504 k^{2} s_{n - 8,k - 5} + 2016 k^{2} s_{n - 8,k - 4} - 2520 k^{2} s_{n - 8,k - 3} + 756 k^{2} s_{n - 8,k - 2} + 36 k^{2} s_{n - 8,k + 1} + 9 k s_{n - 8,k} + 9 k s_{n - 8,k - 7} - 180 k s_{n - 8,k - 6} + 1134 k s_{n - 8,k - 5} - 2646 k s_{n - 8,k - 4} + 2079 k s_{n - 8,k - 3} - 378 k s_{n - 8,k - 2} + 9 k s_{n - 8,k - 1} + 9 k s_{n - 8,k + 1} + s_{n - 8,k} + s_{n - 8,k - 8} - 27 s_{n - 8,k - 7} + 246 s_{n - 8,k - 6} - 924 s_{n - 8,k - 5} + 1407 s_{n - 8,k - 4} - 735 s_{n - 8,k - 3} + 85 s_{n - 8,k - 2} + s_{n - 8,k + 1}\\\end{array}

In [36]:
s = oeis.oeis_search(seq=[1,1,4,3,1,27,19,6,1,256,175,55,10,1])


*

In [37]:
s()


Out[37]:

Results for query: https://oeis.org/search?q=1%2C+1%2C+4%2C+3%2C+1%2C+27%2C+19%2C+6%2C+1%2C+256%2C+175%2C+55%2C+10%2C+1&start=0&fmt=json


A039621: Triangle of Lehmer-Comtet numbers of 2nd kind.

by Len Smiley

Keywords: tabl,sign

Data:

$$ \begin{array}{c|ccccccccccc} n, k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 0 & 1 & & & & & & & & & \\1 & -1 & 1 & & & & & & & & \\2 & 4 & -3 & 1 & & & & & & & \\3 & -27 & 19 & -6 & 1 & & & & & & \\4 & 256 & -175 & 55 & -10 & 1 & & & & & \\5 & -3125 & 2101 & -660 & 125 & -15 & 1 & & & & \\6 & 46656 & -31031 & 9751 & -1890 & 245 & -21 & 1 & & & \\7 & -823543 & 543607 & -170898 & 33621 & -4550 & 434 & -28 & 1 & & \\8 & 16777216 & -11012415 & 3463615 & -688506 & 95781 & -9702 & 714 & -36 & 1 & \\9 & -387420489 \end{array} $$

Comments:

  • Also the Bell transform of (-n)^n adding 1,0,0,0,... as column 0. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 16 2016

Formulae:

  • (k-1)!a(n, k) = Sum_{i=0..k-1}((-1)^(n-k-i)binomial(k-1, i)*(n-i-1)^(n-1)).

Cross references:

Links:


In [40]:
s = oeis.oeis_search(id=264428)


*

In [41]:
s()


Out[41]:

Results for query: https://oeis.org/search?q=id%3AA264428&start=0&fmt=json


A264428: Triangle read by rows, Bell transform of Bell numbers.

by Peter Luschny, Nov 13 2015

Keywords: nonn,tabl

Data:

$$ \begin{array}{c|ccccccccccc} n, k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 0 & 1 & & & & & & & & & \\1 & 0 & 1 & & & & & & & & \\2 & 0 & 1 & 1 & & & & & & & \\3 & 0 & 2 & 3 & 1 & & & & & & \\4 & 0 & 5 & 11 & 6 & 1 & & & & & \\5 & 0 & 15 & 45 & 35 & 10 & 1 & & & & \\6 & 0 & 52 & 205 & 210 & 85 & 15 & 1 & & & \\7 & 0 & 203 & 1029 & 1330 & 700 & 175 & 21 & 1 & & \\8 & 0 & 877 & 5635 & 8946 & 5845 & 1890 & 322 & 28 & 1 & \\9 & 0 & 4140 & 33387 & 63917 & 50358 & 20055 & 4410 & 546 & 36 & 1 \end{array} $$

Comments:

  • Consider the sequence S0 -> T0 -> S1 -> T1 -> S2 -> T2 -> ... Here Sn -> Tn indicates the Bell transform mapping a sequence Sn to a triangle Tn as defined in the link and Tn -> S{n+1} the operator associating a triangle with the sequence of its row sums. If
  • S0 = A000012 = <1,1,1,...> then
  • T0 = A048993 # Stirling subset numbers,
  • S1 = A000110 # Bell numbers,
  • T1 = A264428 # Bell transform of Bell numbers,
  • S2 = A187761 # second-order Bell numbers,
  • T2 = A264430 # Bell transform of second-order Bell numbers,
  • S3 = A264432 # third-order Bell numbers.
  • This construction is closely related to permutations trees and A179455. Sn is A179455_col(n+1) prepended by A179455_diag(k) = k! for k <= n. In other words, Sn 'converges' to n! for n -> inf.
  • Given a sequence (s(n))n>=0 with s(0) = 0 and with e.g.f. B(x) = Sum_{n >= 1} s(n)*x^n/n!, then the Bell matrix associated with s(n) equals the exponential Riordan array [1, B(x)] belonging to the Lagrange subgroup of the exponential Riordan group. Omitting the first row and column from the Bell matrix produces the exponential Riordan array [d/dx(B(x)), B(x)] belonging to the Derivative subgroup of the exponential Riordan group. - Peter Bala, Jun 07 2016

Formulae:

  • From Peter Bala, Jun 07 2016:
    • E.g.f. exp(tB(x)), where B(x) = Integral_{u = 0..x} exp(exp(u) - 1) du = x + x^2/2! + 2x^3/3! + 5x^4/4! + 15x^5/5! + 52*x^6/6! + ....
    • Row polynomial recurrence: R(n+1,t) = tSum_{k = 0 ..n} binomial(n,k)Bell(k)* R(n-k,t) with R(0,t) = 1.

Cross references:

Links: