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%I #8 Mar 09 2022 01:35:34
%S 1,1,0,3,1,0,22,0,9,1,0,121,0,235,0,27,1,0,620,0,3446,0,1996,0,81,1,0,
%T 3119,0,40314,0,63854,0,15349,0,243,1,0,15618,0,422087,0,1434812,0,
%U 963327,0,112546,0,729,1,0,78117,0,4157997,0,26672209,0,37898739,0,12960063,0,806047,0,2187
%N Irregular triangle of coefficients of p(n, x) = (1 - x^2)^(n+1)*Sum_{j >= 0} (4*j+ 1)^n*x^(2*j), read by rows.
%C Define the series q(x, n) = (1 - x^2)^(n+1)*Sum_{j >= 1} (4*k+1)^n*x^(2*k) then the sum r(x, n) = p(x, n) + q(x, n) is symmetrical and gives r(x, n) =(x+1)^(2*n+1)*A060187(x, n).
%H G. C. Greubel, <a href="/A158782/b158782.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+1)*Sum_{j >= 0} (4*j+ 1)^n*x^(2*j).
%e The irregular triangle begins as:
%e 1;
%e 1, 0, 3;
%e 1, 0, 22, 0, 9;
%e 1, 0, 121, 0, 235, 0, 27;
%e 1, 0, 620, 0, 3446, 0, 1996, 0, 81;
%e 1, 0, 3119, 0, 40314, 0, 63854, 0, 15349, 0, 243;
%e 1, 0, 15618, 0, 422087, 0, 1434812, 0, 963327, 0, 112546, 0, 729;
%t p[n_, x_]= (1-x^2)^(n+1)*Sum[(4*k+1)^n*x^(2*k), {k,0,Infinity}];
%t Table[FullSimplify[p[n,x]], {n,0,12}];
%t Table[CoefficientList[p[n,x], x], {n, 0, 12}]//Flatten (* modified by _G. C. Greubel_, Mar 08 2022 *)
%o (Sage)
%o def p(n,x): return (1-x^2)^(n+1)*sum( (4*j+1)^n*x^(2*j) for j in (0..n+1) )
%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)]) # _G. C. Greubel_, Mar 08 2022
%Y Cf. A060187.
%K nonn,tabf
%O 0,4
%A _Roger L. Bagula_, Mar 26 2009
%E Edited by _G. C. Greubel_, Mar 08 2022
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