A165889 - OEIS

%I #7 Mar 10 2022 01:49:04

%S 1,1,2,1,1,8,18,8,1,1,22,143,244,143,22,1,1,52,808,3484,5710,3484,808,

%T 52,1,1,114,3853,35032,125746,188908,125746,35032,3853,114,1,1,240,

%U 16782,290672,2000703,6040992,8702820,6040992,2000703,290672,16782,240,1

%N 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.

%H G. C. Greubel, <a href="/A165889/b165889.txt">Rows n = 0..50 of the irregular triangle, flattened</a>

%F 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.

%F T(n, k) = [x^k]( p(n, x) ), where p(n, x) = (1-x)^(2*n+4)*( PolyLog(-n-1, x)/x)^2.

%F T(n, n-k) = T(n, k). - _G. C. Greubel_, Mar 09 2022

%e Irregular triangle begins as:

%e   1;

%e   1,   2,    1;

%e   1,   8,   18,     8,      1;

%e   1,  22,  143,   244,    143,     22,      1;

%e   1,  52,  808,  3484,   5710,   3484,    808,    52,    1;

%e   1, 114, 3853, 35032, 125746, 188908, 125746, 35032, 3853, 114, 1;

%t p[n_, x_]:= p[n, x]= (1/x^2)*(1-x)^(2*n+4)*Sum[k^(n+1)*x^k, {k, 0, Infinity}]^2;

%t Table[CoefficientList[p[n, x], x], {n,0,12}]//Flatten (* modified by _G. C. Greubel_, Mar 09 2022 *)

%o (Sage)

%o 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

%o def T(n,k): return ( p(n,x) ).series(x, 2*n+1).list()[k]

%o flatten([[T(n,k) for k in (0..2*n)] for n in (0..12)])

%Y Cf. A165883, A165890, A165891.

%K nonn,tabf

%O 0,3

%A _Roger L. Bagula_, Sep 29 2009

%E Edited by _G. C. Greubel_, Mar 09 2022