%I #12 Dec 29 2018 07:53:52
%S 1,1,-3,-21,75,1475,-5005,-221389,593523,57764619,-89101881,
%T -23273632371,953636541,13409519997705,23908442020749,
%U -10469975115603501,-40844292735050541,10646036726696597027,66995992524016223543,-13672657170891872702719,-122282221141986787179519,21647316686778755963070321,256325163531592225309743129,-41426918732532942751217361155,-620418821801458605268716606275,94275566307675915918535250768725
%N a(n) = [x^n] Product_{k=1..n} (k + x - k*x^2), for n >= 0.
%C a(n+1) = -2*(n+1) * A322227(n) + a(n), for n >= 1.
%C a(n+1) = -n*(n+1)^2 * A322226(n) + a(n), for n >= 1.
%H Paul D. Hanna, <a href="/A322228/b322228.txt">Table of n, a(n) for n = 0..300</a>
%e The irregular triangle A322225 formed from coefficients of x^k in Product_{m=1..n} (m + x - m*x^2), for n >= 0, k = 0..2*n, begins
%e 1;
%e 1, 1, -1;
%e 2, 3, -3, -3, 2;
%e 6, 11, -12, -21, 12, 11, -6;
%e 24, 50, -61, -140, 75, 140, -61, -50, 24;
%e 120, 274, -375, -1011, 540, 1475, -540, -1011, 375, 274, -120;
%e 720, 1764, -2696, -8085, 4479, 15456, -5005, -15456, 4479, 8085, -2696, -1764, 720;
%e 5040, 13068, -22148, -71639, 42140, 169266, -50932, -221389, 50932, 169266, -42140, -71639, 22148, 13068, -5040; ...
%e in which the central terms equal this sequence.
%e RELATED SEQUENCES.
%e Note that the terms in the secondary diagonal A322227 in the above triangle
%e [1, 3, -12, -140, 540, 15456, -50932, -3176172, 7343325, 1053842295, ...]
%e may be divided by triangular numbers to obtain A322226:
%e [1, 1, -2, -14, 36, 736, -1819, -88227, 163185, 19160769, -15294993, ...].
%t a[n_] := SeriesCoefficient[Product[k + x - k x^2, {k, 1, n}], {x, 0, n}];
%t Array[a, 26, 0] (* _Jean-François Alcover_, Dec 29 2018 *)
%o (PARI) {T(n, k) = polcoeff( prod(m=1, n, m + x - m*x^2) +x*O(x^k), k)}
%o /* Print the irregular triangle */
%o for(n=0, 10, for(k=0, 2*n, print1( T(n, k), ", ")); print(""))
%o /* Print this sequence */
%o for(n=0, 30, print1( T(n, n), ", "))
%Y Cf. A322235, A322236, A322237.
%Y Cf. A322238 (variant).
%K sign
%O 0,3
%A _Paul D. Hanna_, Dec 15 2018