A165889 Irregular triangle T(n, k) = [x^k]( p(n, x) ), where p(n, x) = (1-x)^(2*n+4)*( Sum_{j >= 0} j^(n+1)*x^j )^2/x^2, read by rows.
1, 1, 2, 1, 1, 8, 18, 8, 1, 1, 22, 143, 244, 143, 22, 1, 1, 52, 808, 3484, 5710, 3484, 808, 52, 1, 1, 114, 3853, 35032, 125746, 188908, 125746, 35032, 3853, 114, 1, 1, 240, 16782, 290672, 2000703, 6040992, 8702820, 6040992, 2000703, 290672, 16782, 240, 1
Offset: 0
Examples
Irregular triangle begins as: 1; 1, 2, 1; 1, 8, 18, 8, 1; 1, 22, 143, 244, 143, 22, 1; 1, 52, 808, 3484, 5710, 3484, 808, 52, 1; 1, 114, 3853, 35032, 125746, 188908, 125746, 35032, 3853, 114, 1;
Links
- G. C. Greubel, Rows n = 0..50 of the irregular triangle, flattened
Programs
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Mathematica
p[n_, x_]:= p[n, x]= (1/x^2)*(1-x)^(2*n+4)*Sum[k^(n+1)*x^k, {k, 0, Infinity}]^2; Table[CoefficientList[p[n, x], x], {n,0,12}]//Flatten (* modified by G. C. Greubel, Mar 09 2022 *)
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Sage
def p(n,x): return (1/x^2)*(1-x)^(2*n+4)*sum( j^(n+1)*x^j for j in (0..2*n+4) )^2 def T(n,k): return ( p(n,x) ).series(x, 2*n+1).list()[k] flatten([[T(n,k) for k in (0..2*n)] for n in (0..12)])
Formula
T(n, k) = [x^k]( p(n, x) ), where p(n, x) = (1-x)^(2*n+4)*( Sum_{j >= 0} j^(n+1)*x^j )^2/x^2.
T(n, k) = [x^k]( p(n, x) ), where p(n, x) = (1-x)^(2*n+4)*( PolyLog(-n-1, x)/x)^2.
T(n, n-k) = T(n, k). - G. C. Greubel, Mar 09 2022
Extensions
Edited by G. C. Greubel, Mar 09 2022