A137597 -id:A137597 - cp's OEIS frontend

A035009 STIRLING transform of [1,1,2,4,8,16,32,...].

1, 1, 3, 11, 47, 227, 1215, 7107, 44959, 305091, 2206399, 16913987, 136823263, 1163490499, 10366252031, 96491364675, 935976996127, 9440144423875, 98800604237119, 1071092025420867, 12008090971866207, 139014305916844739, 1659578039401022079, 20405708646650507075

Offset: 0

N. J. A. Sloane

Numerators of sequence that shifts left one place under 1/2 order binomial transform. (Denominators are 2^(n-1) for n > 0.) - Franklin T. Adams-Watters, Jul 31 2005
Row sums of triangle A137597 starting (1, 3, 11, 47, 227, ...). - Gary W. Adamson, Jan 29 2008
From Gary W. Adamson, Jul 22 2011: (Start)
a(n)/2^(n-1) = upper left term in M^n, M = an infinite square production matrix in which a column of (1/2, 1/2, 1/2, ...) is appended to the right of Pascal's triangle, as follows:
  1,  1/2,  0,   0,   0,   0,  ...
  1,   1,  1/2,  0,   0,   0,  ...
  1,   2,   1,  1/2,  0,   0,  ...
  1,   3,   3,   1,  1/2,  0,  ...
  1,   4,   6,   4,   1,  1/2, ..., etc.
(End)
From Bruno Berselli, Mar 20 2013: (Start)
Note that, for t=A222391:
a(1)*t = Sum_{n >= 1} 1  /(Gamma(n/2)*Gamma((n+1)/2)),
a(2)*t = Sum_{n >= 1} n  /(Gamma(n/2)*Gamma((n+1)/2)),
a(3)*t = Sum_{n >= 1} n^2/(Gamma(n/2)*Gamma((n+1)/2)),
a(4)*t = Sum_{n >= 1} n^3/(Gamma(n/2)*Gamma((n+1)/2)),
a(5)*t = Sum_{n >= 1} n^4/(Gamma(n/2)*Gamma((n+1)/2)),
a(6)*t = Sum_{n >= 1} n^5/(Gamma(n/2)*Gamma((n+1)/2)), etc.
(End)
Except for the initial term, the main diagonal of A129340. - Peter Bala, Apr 14 2017

			Given the production matrix M, upper left term of M^5 = a(5)/2^4 = 227/16.

Seiichi Manyama, Table of n, a(n) for n = 0..558
Paul Barry, Eulerian-Dowling Polynomials as Moments, Using Riordan Arrays, arXiv:1702.04007 [math.CO], 2017.
Mikhail Khovanov, Victor Ostrik and Yakov Kononov, Two-dimensional topological theories, rational functions and their tensor envelopes, arXiv:2011.14758 [math.QA], 2020.
Toufik Mansour and Mark Shattuck, A recurrence related to the Bell Numbers, Integers 11 (2011), #A67.

Cf. A000110, A011782, A137597, A222391, A129340.

A035009 := proc(n) local a,b,i;
a := [seq(2,i=1..n-1)]; b := [seq(1,i=1..n-1)];
exp(-x)*hypergeom(a,b,x); round(evalf(subs(x=2,%), 10+2*n)) end:
seq(A035009(n),n=0..19);  # Peter Luschny, Mar 30 2011
# second Maple program:
b:= proc(n, m) option remember;
     `if`(n=0, ceil(2^(m-1)), m*b(n-1, m)+b(n-1, m+1))
    end:
a:= n-> b(n, 0):
seq(a(n), n=0..25);  # Alois P. Heinz, Aug 03 2021

1/(2*E^2)*Sum[(i + j)^n/(i!*j!), {i, 0, Infinity}, {j, 0, Infinity}] (* Starting from the 2nd term *) (* Vladimir Reshetnikov, Dec 31 2008 *)
Join[{1}, Table[BellB[n, 2]/2, {n, 1, 25}]] (* Vaclav Kotesovec, Jun 26 2022 *)

x='x+O('x^99); Vec(serlaplace((1 + exp(2*exp(x)-2))/2)) \\ Joerg Arndt, Apr 01 2011

a(n) = (1/2)*A001861(n), n > 0.
E.g.f.: (1 + exp(2*exp(x)-2))/2. - Emeric Deutsch, Feb 09 2002
a(n+1) = 1 + 2*Sum_{j=1..n} binomial(n, j)*a(j). - Jon Perry, Apr 25 2005
Define f_1(x), f_2(x), ... such that f_1(x)=e^x and for n=2,3,... f_{n+1}(x) = (d/dx)(x*f_n(x)). Then a(n) = e^(-2)*f_n(2). - Milan Janjic, May 30 2008
G.f.: 1 + x/(Q(0) - 2*x) where Q(k) =  1 - x*(k+1)/( 1 - 2*x/Q(k+1) ); (continued fraction ). - Sergei N. Gladkovskii, Mar 22 2013
G.f.: 1/Q(0), where Q(k)= 1 - x - 2*x/(1 - x*(2*k+1)/(1 - x - 2*x/(1 - x*(2*k+2)/Q(k+1)))); (continued fraction). - Sergei N. Gladkovskii, May 13 2013
G.f.: 1 + Sum_{k>=1} 2^(k-1)*x^k/Product_{j=1..k} (1 - j*x). - Ilya Gutkovskiy, Jun 19 2018

A269951 Triangle read by rows, T(n,k) = Sum_{j=0..n} (-1)^(n-j)C(-j,-n)S1(j,k), S1 the Stirling cycle numbers A132393, for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 5, 5, 1, 0, 16, 23, 9, 1, 0, 65, 116, 65, 14, 1, 0, 326, 669, 470, 145, 20, 1, 0, 1957, 4429, 3634, 1415, 280, 27, 1, 0, 13700, 33375, 30681, 14084, 3535, 490, 35, 1, 0, 109601, 283072, 284066, 147532, 43939, 7756, 798, 44, 1

Offset: 0

Peter Luschny, Apr 10 2016

			Triangle begins:
  1;
  0,   1;
  0,   2,   1;
  0,   5,   5,   1;
  0,  16,  23,   9,   1;
  0,  65, 116,  65,  14,  1;
  0, 326, 669, 470, 145, 20, 1;

Peter Luschny, Extensions of the binomial

A000522 (col. 1), A073596 (col. 2), A000096 (diag. n-1), A241765 (diag. n-2).
A001339 (row sums), A137597 (unsigned matrix inverse).
Cf. A132393, A269952.

A269951 := (n,k) -> add((-1)^(n-j)*binomial(-j,-n)*abs(Stirling1(j,k)), j=0..n):
seq(seq(A269951(n,k), k=0..n), n=0..9);

Flatten[ Table[ Sum[(-1)^(n-j) Binomial[-j,-n] Abs[StirlingS1[j,k]], {j,0,n}], {n,0,9}, {k,0,n}]]

A126350 Triangle read by rows: matrix product of the binomial coefficients with the Stirling numbers of the second kind.

Original entry on oeis.org

1, 1, 2, 1, 5, 5, 1, 9, 22, 15, 1, 14, 61, 99, 52, 1, 20, 135, 385, 471, 203, 1, 27, 260, 1140, 2416, 2386, 877, 1, 35, 455, 2835, 9156, 15470, 12867, 4140, 1, 44, 742, 6230, 28441, 72590, 102215, 73681, 21147

Offset: 1

Thomas Wieder, Dec 29 2006

Many well-known integer sequences arise from such a matrix product of combinatorial coefficients. In the present case we have as the first row (not surprisingly) A000110 = Bell or exponential numbers: ways of placing n labeled balls into n indistinguishable boxes . As second row we have A033452 = "STIRLING" transform of squares A000290. As the column sums we have 1, 3, 11, 47, 227, 1215, 7107, 44959, 305091 which is A035009 = STIRLING transform of [1,1,2,4,8,16,32, ...].

			Matrix begins:
1 2 5 15 52 203  877  4140  21147
0 1 5 22 99 471 2386 12867  73681
0 0 1  9 61 385 2416 15470 102215
0 0 0  1 14 135 1140  9156  72590
0 0 0  0 1   20  260  2835  28441
0 0 0  0 0    1   27   455   6230
0 0 0  0 0    0    1    35    742
0 0 0  0 0    0    0     1     44
0 0 0  0 0    0    0     0      1

Cf. A039810, A039814, A126351, A054654, A126353.

Cf. A137597.

T:= (n, k)-> add(Stirling2(n, j)*binomial(j-1, n-k), j=n-k+1..n):
seq(seq(T(n, k), k=1..n), n=1..10);  # Alois P. Heinz, Sep 03 2019

T[dim_] := T[dim] = Module[{M}, M[n_, n_] = 1; M[, ] = 0; Do[M[n, k] = M[n-1, k-1] + (k+2) M[n-1, k] + (k+1) M[n-1, k+1], {n, 0, dim-1}, {k, 0, n-1}]; Array[M, {dim, dim}, {0, 0}]];
dim = 9;
Table[T[dim][[n]][[1 ;; n]] // Reverse, {n, 1, dim}] (* Jean-François Alcover, Jun 27 2019, from Sage *)

def A126350_triangle(dim): # rows in reversed order
    M = matrix(ZZ,dim,dim)
    for n in (0..dim-1): M[n,n] = 1
    for n in (1..dim-1):
        for k in (0..n-1):
            M[n,k] = M[n-1,k-1]+(k+2)*M[n-1,k]+(k+1)*M[n-1,k+1]
    return M
A126350_triangle(9) # Peter Luschny, Sep 19 2012

(In Maple notation:) Matrix product A.B of matrix A[i,j]:=binomial(j-1,i-1) with i = 1 to p+1, j = 1 to p+1, p=8 and of matrix B[i,j]:=stirling2(j,i) with i from 1 to d, j from 1 to d, d=9.

cp's OEIS Frontend

A035009 STIRLING transform of [1,1,2,4,8,16,32,...].

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A269951 Triangle read by rows, T(n,k) = Sum_{j=0..n} (-1)^(n-j)C(-j,-n)S1(j,k), S1 the Stirling cycle numbers A132393, for n>=0 and 0<=k<=n.

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A126350 Triangle read by rows: matrix product of the binomial coefficients with the Stirling numbers of the second kind.

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Author

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Programs

Maple

Mathematica

Sage

Formula

cp's OEIS Frontend

A035009 STIRLING transform of [1,1,2,4,8,16,32,...].

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A269951 Triangle read by rows, T(n,k) = Sum_{j=0..n} (-1)^(n-j)*C(-j,-n)*S1(j,k), S1 the Stirling cycle numbers A132393, for n>=0 and 0<=k<=n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

A126350 Triangle read by rows: matrix product of the binomial coefficients with the Stirling numbers of the second kind.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Sage

Formula

A269951 Triangle read by rows, T(n,k) = Sum_{j=0..n} (-1)^(n-j)C(-j,-n)S1(j,k), S1 the Stirling cycle numbers A132393, for n>=0 and 0<=k<=n.