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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

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Author

Johannes W. Meijer, Aug 13 2009

Keywords

Comments

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

Examples

			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.
		

Crossrefs

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.

Programs

  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    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
    
  • PARI
    \\ 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

Formula

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