A248925 Triangle in which row n consists of the coefficients in Sum_{m=0..n} x^m * Product_{k=m+1..n} (1-k*x), as read by rows.

1, 1, 0, 1, -2, 1, 1, -5, 7, -2, 1, -9, 27, -30, 9, 1, -14, 72, -165, 159, -44, 1, -20, 156, -597, 1149, -998, 265, 1, -27, 296, -1689, 5328, -9041, 7251, -1854, 1, -35, 512, -4057, 18840, -51665, 79579, -59862, 14833, 1, -44, 827, -8665, 55353, -221225, 544564, -776073, 553591, -133496

Offset: 0

Author

Paul D. Hanna, Oct 16 2014

Keywords

• sign
• tabl

Comments

If m=n, we have Sum_{k=0..n} A008277(n, k) = A000110(n) = Sum_{j=0..n} T(n,j)*A008277(2n-j,n) where A000110(n) is the n-th Bell number. - Robert A. Russell, Apr 08 2018

Examples

			Triangle begins:
1;
1, 0;
1, -2, 1;
1, -5, 7, -2;
1, -9, 27, -30, 9;
1, -14, 72, -165, 159, -44;
1, -20, 156, -597, 1149, -998, 265;
1, -27, 296, -1689, 5328, -9041, 7251, -1854;
1, -35, 512, -4057, 18840, -51665, 79579, -59862, 14833;
1, -44, 827, -8665, 55353, -221225, 544564, -776073, 553591, -133496;
1, -54, 1267, -16935, 142003, -774755, 2756814, -6221713, 8314321, -5669406, 1334961; ...
Generating method for row n:
n=0: 1 = 1;
n=1: 1 + 0*x = (1-x) * ( 1 + x/(1-x) );
n=2: 1 - 2*x + x^2 = (1-x)*(1-2*x) * ( 1 + x/(1-x) + x^2/((1-x)*(1-2*x)) );
n=3: 1 - 5*x + 7*x^2 - 2*x^3 = (1-x)*(1-2*x)*(1-3*x) * ( 1 + x/(1-x) + x^2/((1-x)*(1-2*x)) + x^3/((1-x)*(1-2*x)*(1-3*x)) );
n=4: 1 - 9*x + 27*x^2 - 30*x^3 + 9*x^4 = (1-x)*(1-2*x)*(1-3*x)*(1-4*x) * ( 1 + x/(1-x) + x^2/((1-x)*(1-2*x)) + x^3/((1-x)*(1-2*x)*(1-3*x)) + x^4/((1-x)*(1-2*x)*(1-3*x)*(1-4*x)) ); ...
Compare the row g.f.s to the o.g.f. of Bell numbers (A000110):
B(x) = 1 + x/(1-x) + x^2/((1-x)*(1-2*x)) + x^3/((1-x)*(1-2*x)*(1-3*x)) + x^4/((1-x)*(1-2*x)*(1-3*x)*(1-4*x)) +...
Central terms of triangle begin:
[1, -2, 27, -597, 18840, -774755, 39320575, -2375828028, 166592007731, -13300276081039, 1191315248017730, ...].
		

Crossrefs

Cf. A000166, A002467, A000110.

Programs

Mathematica

Table[LinearSolve[Table[StirlingS2[m+j, n], {m, 0, n}, {j, n, 0, -1}],
  Table[Sum[StirlingS2[m, j], {j, 0, n}], {m, 0, n}]], {n, 0, 20}]
  // Flatten (* Robert A. Russell, Mar 30 2018 *)
Table[PadRight[CoefficientList[Sum[x^m*Product[1-j*x, {j, m+1, n}],
  {m, 0, n}], x], n+1], {n, 0, 20}] // Flatten (* Robert A. Russell, Apr 08 2018 *)
T[n_, 0] := T[n,0] = 1;
T[n_, k_] := T[n,k] = If[kRobert A. Russell, Apr 25 2018 *)

PARI

{T(n,k)=polcoeff(sum(m=0,n, x^m*prod(j=m+1,n,1-j*x)), k)}
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

Formula

Right border equals A000166, the subfactorial numbers.

Row sums equal A000166 (shift right 1 place).

Row sums of unsigned terms yields A002467(n) = n! - A000166(n).

Sum_{k=0..n} A008277(m, k) = Sum_{j=0..n} T(n, j)*A008277(m+n-j, n) where A008277(m, k) are Stirling subset numbers. - Robert A. Russell, Mar 30 2018

T(n,0) = 1.

For k>0, T(n,k) = [k==n] + [kRobert A. Russell, Apr 25 2018