A369435 Square array A(n, k) = n! * [t^n] (exp(t)/(1+k-k*exp(t))) for n >= 0 and k >= 0, read by antidiagonals upwards.
1, 1, 1, 1, 2, 1, 1, 6, 3, 1, 1, 26, 15, 4, 1, 1, 150, 111, 28, 5, 1, 1, 1082, 1095, 292, 45, 6, 1, 1, 9366, 13503, 4060, 605, 66, 7, 1, 1, 94586, 199815, 70564, 10845, 1086, 91, 8, 1, 1, 1091670, 3449631, 1471708, 243005, 23826, 1771, 120, 9, 1, 1, 14174522, 68062695, 35810212, 6534045, 653406, 45955, 2696, 153, 10, 1
Offset: 0
Examples
Array A(n, k) starts: n\k : 0 1 2 3 4 5 6 7 8 ================================================================================ 0 : 1 1 1 1 1 1 1 1 1 1 : 1 2 3 4 5 6 7 8 9 2 : 1 6 15 28 45 66 91 120 153 3 : 1 26 111 292 605 1086 1771 2696 3897 4 : 1 150 1095 4060 10845 23826 45955 80760 132345 5 : 1 1082 13503 70564 243005 653406 1490587 3024008 5618169 6 : 1 9366 199815 1471708 6534045 21502866 58018051 135878520 286195833 7 : 1 94586 3449631 35810212 8 : 1 1091670 68062695 9 : 1 14174522 . Triangle T(n, k) starts: [0] 1; [1] 1, 1; [2] 1, 2, 1; [3] 1, 6, 3, 1; [4] 1, 26, 15, 4, 1; [5] 1, 150, 111, 28, 5, 1; [6] 1, 1082, 1095, 292, 45, 6, 1; [7] 1, 9366, 13503, 4060, 605, 66, 7, 1; [8] 1, 94586, 199815, 70564, 10845, 1086, 91, 8, 1; [9] 1, 1091670, 3449631, 1471708, 243005, 23826, 1771, 120, 9, 1;
Crossrefs
Programs
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Maple
egf := exp(t) / (1 + x*(1 - exp(t))): sert := series(egf, t, 12): col := k -> local j; seq(subs(x=k, j!*coeff(sert, t, j)), j=0..9): T := (n, k) -> col(k)[n - k + 1]: # Triangle for n from 0 to 9 do seq(T(n, k), k=0..n) od; # Peter Luschny, Jan 24 2024 with(combinat): # WP Worpitzky polynomials, WC coefficients of WP. WC := (n, k) -> local j; add(eulerian1(n, j)*binomial(n-j, n-k), j=0..n): WP := n -> local j; add(WC(n, j) * x^j, j=0..n): A369435row := (n, k) -> subs(x = k, WP(n)): seq(lprint(seq(A369435row(n, k), k = 0..7)), n = 0..7); # Peter Luschny, Apr 26 2024
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PARI
{A(n, k) = n! * polcoeff(exp(x+x*O(x^n)) / (1+k-k*exp(x+x*O(x^n))), n)}
Formula
A(n, k) = Sum_{i=0..n} A163626(n, i) * (-k)^i for n >= 0 and k >= 0.
A(n, k) = Sum_{i=0..n} A028246(n+1, i+1) * k^i for n >= 0 and k >= 0.
E.g.f. of column k >= 0: exp(t) / (1 + k - k * exp(t)).
Conjecture: A(n, n) = (n + 1) * A321189(n) for n >= 0. [This is true. - Peter Luschny, Apr 26 2024]
A(n, n) = A372312(n). - Peter Luschny, Apr 26 2024
Comments