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A349226 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}.

%I A349226 #59 Jul 10 2022 16:12:41
%S A349226 1,1,-1,1,-2,1,1,-6,9,-4,1,-33,171,-247,108,1,-289,8619,-44023,63340,
%T A349226 -27648,1,-3413,911744,-26978398,137635215,-197965148,86400000,1,
%U A349226 -50070,160195328,-42565306462,1258841772303,-6421706556188,9236348345088,-4031078400000
%N A349226 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 A349226 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 A349226 T(n, 0) = 1.
%F A349226 T(n, 1) = -A062970(n).
%F A349226 T(n, 2) = Sum_{m=0..n-1} A062970(m)*m^m.
%F A349226 T(n, k) = Sum_{m=0..n-1} -T(m, k-1)*m^m.
%F A349226 T(n, n) = (-1)^n*A002109(n).
%e A349226 The triangle begins:
%e A349226   1;
%e A349226   1,    -1;
%e A349226   1,    -2,      1;
%e A349226   1,    -6,      9,        -4;
%e A349226   1,   -33,    171,      -247,       108;
%e A349226   1,  -289,   8619,    -44023,     63340,     -27648;
%e A349226   1, -3413, 911744, -26978398, 137635215, -197965148, 86400000;
%e A349226   ...
%e A349226 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 A349226 (PARI) T(n, k) = polcoeff(prod(m=0, n-1, (x-m^m)), n-k);
%Y A349226 Cf. A002109, A062970.
%Y A349226 Cf. A008276 (The Stirling numbers of the first kind in reverse order).
%Y A349226 Cf. A039758 (Coefficients for polynomials with roots in odd numbers).
%Y A349226 Cf. A355540 (Coefficients for polynomials with roots in factorials).
%K A349226 sign,tabl
%O A349226 0,5
%A A349226 _Thomas Scheuerle_, Jul 07 2022