A141697 T(n,k) = (q*Sum_{j=0..k+1} (-1)^j*binomial(n+1, j)*(k+1-j)^n - p*binomial(n-1, k))/2 where p=12 and q=14.

1, 1, 1, 1, 16, 1, 1, 59, 59, 1, 1, 158, 426, 158, 1, 1, 369, 2054, 2054, 369, 1, 1, 804, 8247, 16792, 8247, 804, 1, 1, 1687, 29925, 109123, 109123, 29925, 1687, 1, 1, 3466, 102088, 617302, 1092910, 617302, 102088, 3466, 1, 1, 7037, 334664, 3185840, 9171722, 9171722, 3185840, 334664, 7037, 1

Offset: 1

Author

Roger L. Bagula, Sep 11 2008

Keywords

• nonn
• tabl
• easy

Comments

Row n is made of coefficients from 7*(1 - x)^(n+1) * polylog(-n,x)/x - 6*(1 + x)^(n-1). - Thomas Baruchel, Jun 03 2018

Examples

			Triangle begins:
  1;
  1,    1;
  1,   16,      1;
  1,   59,     59,       1;
  1,  158,    426,     158,       1;
  1,  369,   2054,    2054,     369,       1;
  1,  804,   8247,   16792,    8247,     804,       1;
  1, 1687,  29925,  109123,  109123,   29925,    1687,      1;
		

Links

• G. C. Greubel, Rows n = 1..100 of triangle, flattened
• Thomas Baruchel, A conjectured formula for the polylogarithm of a negative integer order, Mathematics Stack Exchange question, Jun 04 2018.

Crossrefs

Cf. Eulerian numbers (A008292) and Pascal's triangle (A007318).

Cf. A141696.

Programs

Magma

[ 7*(&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) - 6*Binomial(n-1,k): k in [0..n-1], n in [1..10]]; // G. C. Greubel, Nov 13 2019

Maple

T:= proc(n, k): 7*add((-1)^j*binomial(n+1, j)*(k-j+1)^n, j = 0..k+1) - 6*binomial(n-1, k); end proc; seq(seq(T(n,k), k=0..n-1), n=1..10); # G. C. Greubel, Nov 13 2019

Mathematica

i=12; l=14; Table[Table[(l*Sum[(-1)^j*Binomial[n+1, j](k+1-j)^n, {j, 0, k+1}] - i*Binomial[n-1, k])/2, {k,0,n-1}], {n,10}]//Flatten

PARI

T(n,k) = 7*sum(j=0, k+1, (-1)^j*binomial(n+1,j)*(k-j+1)^n) - 6* binomial(n-1,k);
for(n=1,10, for(k=0,n-1, print1(T(n,k), ", "))) \\ G. C. Greubel, Jun 03 2018

PARI

row(n) = Vec(7*(1 - x)^(n+1)*polylog(-n,x)/x - 6*(1 + x)^(n-1)); \\ Michel Marcus, Jun 08 2018

Sage

[[ 7*sum( (-1)^j*binomial(n+1,j)*(k-j+1)^n for j in (0..k+1)) - 6*binomial(n-1,k) for k in (0..n-1)] for n in (1..10)] # G. C. Greubel, Nov 13 2019

Formula

p=12; q=14; T(n,k) = (q*Sum_{j=0..k+1} (-1)^j*binomial(n+1, j)*(k+1-j)^n - p*binomial(n-1, k))/2.

a(n) = 3*A168524(n) - 2*A154337(n). - Thomas Baruchel, Jun 08 2018

Extensions

Edited by G. C. Greubel, Nov 13 2019