A128813 Triangle of coefficients of (x+1)*(x+3)*(x+6)*...*(x+n(n+1)/2).

1, 1, 1, 1, 4, 3, 1, 10, 27, 18, 1, 20, 127, 288, 180, 1, 35, 427, 2193, 4500, 2700, 1, 56, 1162, 11160, 50553, 97200, 56700, 1, 84, 2730, 43696, 363033, 1512684, 2778300, 1587600, 1, 120, 5754, 141976, 1936089, 14581872, 57234924, 101606400, 57153600

Offset

0,5

Links

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Formula

a(0,0)=1, a(1,0)=1, a(1,1)=1, a(i,j)=i*(i+1)/2*a(i-1,j-1)+a(i-1,j), j=0..i-1.

a(i,i) = i*(i+1)/2*a(i-1,i-1).

a(n,n) = Product_{k=1..n} k*(k+1)/2 = A006472(n+1)

Sum_{m=0..n} a(n,m) = Product_{k=1..n} k*(k+1)/2+1 = A128814(n).

Example

(x+1)(x+3)(x+6)=x^3+10x^2+27x+18, so a(3,j)=1, 10, 27, 18.
The triangle begins:
  1
  1, 1
  1, 4,  3
  1, 10, 27, 18
  1, 20, 127, 288, 180
  1, 35, 427, 2193, 4500, 2700
  1, 56, 1162, 11160, 50553, 97200, 56700

Maple

for n from 1 to 9 do b[n]:=n*(n+1)/2 od: a[0, 0]:=1:a[1, 0]:=1:a[1, 1]:=1:for i from 2 to 9 do a[i, 0]:=1:for j from 1 to i-1 do a[i, j]:=b[i]*a[i-1, j-1]+ a[i-1, j] od:a[i, i]:=b[i]*a[i-1, i-1] od: seq(seq(a[i, j], j=0..i), i=0..9);

Mathematica

Flatten[Table[Reverse[CoefficientList[Expand[Times@@Table[x+(n(n+1))/2, {n, i}]], x]], {i, 0, 9}]] (* Harvey P. Dale, Nov 11 2011 *)

Prog

(PARI) row(n) = Vec(prod(i=1, n, (x+i*(i+1)/2))); \\ Michel Marcus, Mar 18 2023

Crossrefs

Cf. A006472 (right diagonal), A128814 (row sums).

Sequence in context: A039758 A157894 A172106 * A109062 A112493 A370609

Adjacent sequences: A128810 A128811 A128812 * A128814 A128815 A128816

Keyword

easy,tabl,nonn

Author

Miklos Kristof, Apr 10 2007

Status

approved