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A128813 Triangle of coefficients of (x+1)(x+3)(x+6)...(x+n(n+1)/2). 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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

0,5

COMMENTS

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

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*(k+1)/2, k=1..n) = 1, 1, 3, 18, 180, 2700, 56700,.. Sum(a(n,m), m=0..n)= Product(k*(k+1)/2+1, k=1..n) = 1, 2, 8, 56, 616, 9856,..

EXAMPLE

(x+1)(x+3)(x+6)=x^3+10x^2+27x+18, so a(3,j)=1, 10, 27, 18

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 *)

CROSSREFS

Cf. A007318, A094638, A000142.

Sequence in context: A109692 A157894 A172106 * A109062 A112493 A010305

Adjacent sequences:  A128810 A128811 A128812 * A128814 A128815 A128816

KEYWORD

easy,tabl,nonn

AUTHOR

Miklos Kristof, Apr 10 2007

STATUS

approved

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Last modified February 18 03:43 EST 2018. Contains 299298 sequences. (Running on oeis4.)