A130524 -id:A130524 - cp's OEIS frontend

A163936 Triangle related to the o.g.f.s. of the right-hand columns of A130534 (E(x,m=1,n)).

1, 1, 0, 2, 1, 0, 6, 8, 1, 0, 24, 58, 22, 1, 0, 120, 444, 328, 52, 1, 0, 720, 3708, 4400, 1452, 114, 1, 0, 5040, 33984, 58140, 32120, 5610, 240, 1, 0, 40320, 341136, 785304, 644020, 195800, 19950, 494, 1, 0, 362880, 3733920, 11026296, 12440064, 5765500, 1062500

Offset: 1

Johannes W. Meijer, Aug 13 2009

The asymptotic expansions of the higher-order exponential integral E(x,m=1,n) lead to triangle A130524, see A163931 for information on E(x,m,n). The o.g.f.s. of the right-hand columns of triangle A130534 have a nice structure: gf(p) = W1(z,p)/(1-z)^(2*p-1) with p = 1 for the first right-hand column, p = 2 for the second right-hand column, etc. The coefficients of the W1(z,p) polynomials lead to the triangle given above, n >= 1 and 1 <= m <= n. Our triangle is the same as A112007 with an extra right-hand column, see also the second Eulerian triangle A008517. The row sums of our triangle lead to A001147.
We observe that the row sums of the triangles A163936 (m=1), A163937 (m=2), A163938 (m=3) and A163939 (m=4) for z=1 lead to A001147, A001147 (minus a(0)), A001879 and A000457 which are the first four left-hand columns of the triangle of the Bessel coefficients A001497 or, if one wishes, the right-hand columns of A001498. We checked this phenomenon for a few more values of m and found that this pattern persists: m = 5 leads to A001880, m=6 to A001881, m=7 to A038121 and m=8 to A130563 which are the next left- (right-) hand columns of A001497 (A001498). An interesting phenomenon.
If one assumes the triangle not (1,1) based but (0,0) based, one has T(n, k) = E2(n, n-k), where E2(n, k) are the second-order Eulerian numbers A340556. - Peter Luschny, Feb 12 2021

			Triangle starts:
[ 1]      1;
[ 2]      1,       0;
[ 3]      2,       1,      0;
[ 4]      6,       8,      1,      0;
[ 5]     24,      58,     22,      1,      0;
[ 6]    120,     444,    328,     52,      1,     0;
[ 7]    720,    3708,   4400,   1452,    114,     1,   0;
[ 8]   5040,   33984,  58140,  32120,   5610,   240,   1,  0;
[ 9]  40320,  341136, 785304, 644020, 195800, 19950, 494,  1, 0;
The first few W1(z,p) polynomials are
W1(z,p=1) = 1/(1-z);
W1(z,p=2) = (1 + 0*z)/(1-z)^3;
W1(z,p=3) = (2 + 1*z + 0*z^2)/(1-z)^5;
W1(z,p=4) = (6 + 8*z + 1*z^2 + 0*z^3)/(1-z)^7.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Row sums equal A001147.
A000142, A002538, A002539, A112008, A112485 are the first few left hand columns.
A000007, A000012, A005803(n+2), A004301, A006260 are the first few right hand columns.
Cf. A163931 (E(x,m,n)), A048994 (Stirling1) and A008517 (Euler).
Cf. A112007, A163937 (E(x,m=2,n)), A163938 (E(x,m=3,n)) and A163939 (E(x,m=4,n)).
Cf. A001497 (Bessel), A001498 (Bessel), A001147 (m=1), A001147 (m=2), A001879 (m=3) and A000457 (m=4), A001880 (m=5), A001881 (m=6) and A038121 (m=7).
Cf. A340556.

with(combinat): a := proc(n, m): add((-1)^(n+k+1)*binomial(2*n-1, k)*stirling1(m+n-k-1, m-k), k=0..m-1) end: seq(seq(a(n, m), m=1..n), n=1..9);  # Johannes W. Meijer, revised Nov 27 2012

Table[Sum[(-1)^(n + k + 1)*Binomial[2*n - 1, k]*StirlingS1[m + n - k - 1, m - k], {k, 0, m - 1}], {n, 1, 10}, {m, 1, n}] // Flatten (* G. C. Greubel, Aug 13 2017 *)

for(n=1,10, for(m=1,n, print1(sum(k=0,m-1,(-1)^(n+k+1)* binomial(2*n-1,k)*stirling(m+n-k-1,m-k, 1)), ", "))) \\ G. C. Greubel, Aug 13 2017

\\ assuming offset = 0:
E2poly(n,x) = if(n == 0, 1, x*(x-1)^(2*n)*deriv((1-x)^(1-2*n)*E2poly(n-1,x)));
{ for(n = 0, 9, print(Vec(E2poly(n,x)))) } \\ Peter Luschny, Feb 12 2021

a(n, m) = Sum_{k=0..(m-1)} (-1)^(n+k+1)*binomial(2*n-1,k)*Stirling1(m+n-k-1,m-k), for 1 <= m <= n.

Assuming offset = 0 the T(n, k) are the coefficients of recursively defined polynomials. T(n, k) = [x^k] x^n*E2poly(n, 1/x), where E2poly(n, x) = x*(x - 1)^(2*n)*d_{x}((1 - x)^(1 - 2*n)*E2poly(n - 1, x))) for n >= 1 and E2poly(0, x) = 1. - Peter Luschny, Feb 12 2021

A130523 Square array, read by antidiagonals, where row n+1 equals the partial sums of the previous row after removing the n-th term from row n for n>=0, with row 0 equal to all 1's.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 5, 8, 4, 1, 1, 6, 18, 13, 5, 1, 1, 7, 24, 37, 19, 6, 1, 1, 8, 31, 87, 63, 26, 7, 1, 1, 9, 39, 118, 184, 97, 34, 8, 1, 1, 10, 48, 157, 442, 324, 140, 43, 9, 1, 1, 11, 58, 205, 599, 959, 517, 193, 53, 10, 1, 1, 12, 69, 263, 804, 2332, 1733, 774, 257, 64, 11

Offset: 0

Paul D. Hanna, Jun 02 2007, Jun 06 2007

The g.f. of n-th lower diagonal equals D(x)*F(x)*C(x)^n and the g.f. of n-th upper diagonal equals D(x)*F(x)^n, where D(x) is g.f. of main diagonal (A007857), C(x) is g.f. of Catalan numbers (A000108) and F(x) is g.f. of ternary numbers (A001764).

			Square array begins:
  (1), 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...;
  1, (2), 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, ...;
  1, 4, (8), 13, 19, 26, 34, 43, 53, 64, 76, 89, 103, 118, 134, ...;
  1, 5, 18, (37), 63, 97, 140, 193, 257, 333, 422, 525, 643, 777, ...;
  1, 6, 24, 87, (184), 324, 517, 774, 1107, 1529, 2054, 2697, 3474, ...;
  1, 7, 31, 118, 442, (959), 1733, 2840, 4369, 6423, 9120, 12594, ...;
  1, 8, 39, 157, 599, 2332, (5172), 9541, 15964, 25084, 37678, ...;
  1, 9, 48, 205, 804, 3136, 12677, (28641), 53725, 91403, 146077, ...;
  1, 10, 58, 263, 1067, 4203, 16880, 70605, (162008), 308085, ...;
  1, 11, 69, 332, 1399, 5602, 22482, 93087, 401172, (932503), ...;
  ...
For each row, remove the term along the diagonal (in parenthesis here),
and then take partial sums to obtain the next row.

Cf. Diagonals: A007857, A130524, A130525; related: A000108, A001764.

{T(n,k) = if(n<0||k<0,0,if(n==0,1,if(n>k+1, T(n,k-1) + T(n-1,k), T(n,k-1) + T(n-1,k+1))))}
for(n=0,10,for(k=0,10,print1(T(n,k),", "));print(""))

/* Using Formula for G.F.: */
{T(n,k) = local(m=max(n,k)+1,C,F,D); C=subst(Ser(vector(m,r,binomial(2*r-2,r-1)/r)),x,x*y); F=subst(Ser(vector(m,r,binomial(3*r-3,r-1)/(2*r-1))),x,x*y); D=1/(1-x*y*C*F-x*y*F^2);A=D*(1/(1-y*F) + x*C*F/(1-x*C)); polcoeff(polcoeff(A+O(x^m),n,x)+O(y^m),k,y)}
for(n=0,10,for(k=0,10,print1(T(n,k),", "));print(""))

G.f.: A(x,y) = D(x*y)*( 1/(1 - y*F(x*y)) + x*C(x*y)*F(x*y)/(1 - x*C(x*y)) ), where D(x) = 1/(1 - x*C(x)*F(x) - x*F(x)^2) is the g.f. of the main diagonal (A007857), C(x) = 1 + x*C(x)^2 is the g.f. of Catalan numbers (A000108) and F(x) = 1 + x*F(x)^3 is the g.f. of ternary numbers (A001764).

A130525 Diagonal immediately above the main diagonal of square array A130523.

Original entry on oeis.org

1, 3, 13, 63, 324, 1733, 9541, 53725, 308085, 1793528, 10574165, 63018105, 379061652, 2298508911, 14035748542, 86240951745, 532812883413, 3307967729867, 20627845299471, 129141164822496, 811394148828087, 5114638998643903, 32336393838083539, 205000199138736499, 1302892014385402691

Offset: 0

Paul D. Hanna, Jun 02 2007

nonn

Cf. A130523; diagonals: A007857, A130524; related: A000108, A001764.

{a(n) = my(C,F,D); C=Ser(vector(n+1,r,binomial(2*r-2,r-1)/r)); F=Ser(vector(n+1,r,binomial(3*r-3,r-1)/(2*r-1))); D=1/(1-x*C*F-x*F^2); polcoef(D*F+x*O(x^n),n,x)}
for(n=0,25,print1(a(n),", "))

G.f.: A(x) = D(x)*F(x), where D(x) = 1/(1 - x*C(x)*F(x) - x*F(x)^2) is the g.f. of the main diagonal (A007857), C(x) = 1 + x*C(x)^2 is the g.f. of Catalan numbers (A000108) and F(x) = 1 + x*F(x)^3 is the g.f. of ternary numbers (A001764).

Edited and corrected by Paul D. Hanna, Jan 27 2025

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A163936 Triangle related to the o.g.f.s. of the right-hand columns of A130534 (E(x,m=1,n)).

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A130523 Square array, read by antidiagonals, where row n+1 equals the partial sums of the previous row after removing the n-th term from row n for n>=0, with row 0 equal to all 1's.

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A130525 Diagonal immediately above the main diagonal of square array A130523.

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