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%I #21 Mar 18 2023 06:13:35
%S 1,1,1,1,4,3,1,10,27,18,1,20,127,288,180,1,35,427,2193,4500,2700,1,56,
%T 1162,11160,50553,97200,56700,1,84,2730,43696,363033,1512684,2778300,
%U 1587600,1,120,5754,141976,1936089,14581872,57234924,101606400,57153600
%N Triangle of coefficients of (x+1)*(x+3)*(x+6)*...*(x+n(n+1)/2).
%H Harvey P. Dale, <a href="/A128813/b128813.txt">Table of n, a(n) for n = 0..1000</a>
%F 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.
%F a(i,i) = i*(i+1)/2*a(i-1,i-1).
%F a(n,n) = Product_{k=1..n} k*(k+1)/2 = A006472(n+1)
%F Sum_{m=0..n} a(n,m) = Product_{k=1..n} k*(k+1)/2+1 = A128814(n).
%e (x+1)(x+3)(x+6)=x^3+10x^2+27x+18, so a(3,j)=1, 10, 27, 18.
%e The triangle begins:
%e 1
%e 1, 1
%e 1, 4, 3
%e 1, 10, 27, 18
%e 1, 20, 127, 288, 180
%e 1, 35, 427, 2193, 4500, 2700
%e 1, 56, 1162, 11160, 50553, 97200, 56700
%p 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);
%t Flatten[Table[Reverse[CoefficientList[Expand[Times@@Table[x+(n(n+1))/2,{n,i}]],x]],{i,0,9}]] (* _Harvey P. Dale_, Nov 11 2011 *)
%o (PARI) row(n) = Vec(prod(i=1, n, (x+i*(i+1)/2))); \\ _Michel Marcus_, Mar 18 2023
%Y Cf. A006472 (right diagonal), A128814 (row sums).
%K easy,tabl,nonn
%O 0,5
%A _Miklos Kristof_, Apr 10 2007
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