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%I #59 Jul 10 2022 16:12:41
%S 1,1,-1,1,-2,1,1,-6,9,-4,1,-33,171,-247,108,1,-289,8619,-44023,63340,
%T -27648,1,-3413,911744,-26978398,137635215,-197965148,86400000,1,
%U -50070,160195328,-42565306462,1258841772303,-6421706556188,9236348345088,-4031078400000
%N Triangle read by rows. Row n gives the coefficients of Product_{k=0..n} (x - k^k) expanded in decreasing powers of x, with row 0 = {1}.
%C Let M be an n X n matrix filled by binomial(i*j, i) with rows and columns j = 1..n, k = 1..n; then its determinant equals unsigned T(n, n). Can we find a general formula for T(n+m, n) based on determinants of matrices and binomials?
%F T(n, 0) = 1.
%F T(n, 1) = -A062970(n).
%F T(n, 2) = Sum_{m=0..n-1} A062970(m)*m^m.
%F T(n, k) = Sum_{m=0..n-1} -T(m, k-1)*m^m.
%F T(n, n) = (-1)^n*A002109(n).
%e The triangle begins:
%e 1;
%e 1, -1;
%e 1, -2, 1;
%e 1, -6, 9, -4;
%e 1, -33, 171, -247, 108;
%e 1, -289, 8619, -44023, 63340, -27648;
%e 1, -3413, 911744, -26978398, 137635215, -197965148, 86400000;
%e ...
%e Row 4: x^4-33*x^3+171*x^2-247*x+108 = (x-1)*(x-1^1)*(x-2^2)*(x-3^3).
%o (PARI) T(n, k) = polcoeff(prod(m=0, n-1, (x-m^m)), n-k);
%Y Cf. A002109, A062970.
%Y Cf. A008276 (The Stirling numbers of the first kind in reverse order).
%Y Cf. A039758 (Coefficients for polynomials with roots in odd numbers).
%Y Cf. A355540 (Coefficients for polynomials with roots in factorials).
%K sign,tabl
%O 0,5
%A _Thomas Scheuerle_, Jul 07 2022
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