A142073 Irregular triangle, T(n, k) = coefficients of p(x, n), where p(x, n) = (1-2*x)^(n+1) * Sum_{j>=0} j^n*(x/(1-x))^j, read by rows.

1, 1, -1, 1, -1, 1, 1, -4, 2, 1, 7, -16, 8, 1, 21, -28, -26, 48, -16, 1, 51, 32, -356, 408, -136, 1, 113, 492, -1774, 1072, 912, -1088, 272, 1, 239, 2592, -5008, -6656, 20736, -15872, 3968, 1, 493, 10628, -50, -94432, 154528, -57856, -45056, 39680, -7936, 1, 1003, 38768, 108820, -621352, 455608, 848384, -1538816, 884480, -176896

Offset: 0

Author

Roger L. Bagula and Gary W. Adamson, Sep 15 2008

Keywords

• sign
• tabf

Comments

Except for n=0, the row sums are zero.

Examples

			Irregular triangle begins as:
  1;
  1,  -1;
  1,  -1;
  1,   1,    -4,     2;
  1,   7,   -16,     8;
  1,  21,   -28,   -26,     48,    -16;
  1,  51,    32,  -356,    408,   -136;
  1, 113,   492, -1774,   1072,    912,  -1088,    272;
  1, 239,  2592, -5008,  -6656,  20736, -15872,   3968;
  1, 493, 10628,   -50, -94432, 154528, -57856, -45056, 39680, -7936;
		

Links

• G. C. Greubel, Rows n = 0..50 of the irregular triangle, flattened

Crossrefs

Cf. A008292, A141720.

Programs

Magma

m:=12;
R:=PowerSeriesRing(Integers(), m+2);
b:= func< n | n eq 0 select 0 else 2*Floor((n+1)/2) -1 >;
Eulerian:= func< n,k | (&+[(-1)^j*Binomial(n+1,j)*(k-j)^n: j in [0..k]]) >;
p:= func< n,x | (&+[Eulerian(n,j)*(x-1)^j: j in [0..n]]) >;
T:= func< n,k | Coefficient(R!( p(n,x) ), k) >;
[T(n,n-k): k in [0..b(n)], n in [0..m]]; // G. C. Greubel, May 26 2024

Mathematica

p[x_, n_]= If[n==0, 1, (1-2*x)^(n+1)*Sum[k^n*(x/(1-x))^k, {k,0, Infinity}]/x];
Table[CoefficientList[p[x,n], x], {n,0,12}]//Flatten

SageMath

m=12
def b(n): return 2*int((n+1)/2) - 1 + int(n==0)
def Eulerian(n, k): return sum((-1)^j*binomial(n+1, j)*(k-j)^n for j in range(k+1))
def p(x,n): return sum(Eulerian(n,j)*(x-1)^j for j in range(n+1))
def T(n,k): return ( p(x,n) ).series(x, n+1).list()[k]
flatten([[T(n,n-k) for k in range(b(n)+1)] for n in range(m+1)]) # G. C. Greubel, May 26 2024

Formula

T(n, k) = [x^k]( p(x, n) ), where p(x, n) = (1-2*x)^(n+1) * Sum_{j>=0} j^n*(x/(1-x))^j, or p(x, n) = (1-2*x)^(n+1)*PolyLog(-n, x/(1-x))/x.

T(n, k) = [x^k]( f(x, n) ), where f(x, n) = Sum_{j=0..n} Eulerian(n, j)*(x-1)^j. - Mourad Rahmani (mrahmani(AT)usthb.dz), Aug 29 2010

Extensions

Edited by G. C. Greubel, May 26 2024