A307419 Triangle of harmonic numbers T(n, k) = [t^n] Gamma(n+k+t)/Gamma(k+t) for n >= 0 and 0 <= k <= n, read by rows.
1, 0, 1, 0, 3, 1, 0, 11, 9, 1, 0, 50, 71, 18, 1, 0, 274, 580, 245, 30, 1, 0, 1764, 5104, 3135, 625, 45, 1, 0, 13068, 48860, 40369, 11515, 1330, 63, 1, 0, 109584, 509004, 537628, 203889, 33320, 2506, 84, 1, 0, 1026576, 5753736, 7494416, 3602088, 775929, 81900, 4326, 108, 1
Offset: 0
Examples
Triangle starts: 0: [1] 1: [0, 1] 2: [0, 3, 1] 3: [0, 11, 9, 1] 4: [0, 50, 71, 18, 1] 5: [0, 274, 580, 245, 30, 1] 6: [0, 1764, 5104, 3135, 625, 45, 1] 7: [0, 13068, 48860, 40369, 11515, 1330, 63, 1] 8: [0, 109584, 509004, 537628, 203889, 33320, 2506, 84, 1] 9: [0, 1026576, 5753736, 7494416, 3602088, 775929, 81900, 4326, 108, 1] Col: A000254, A001706, A001713, A001719, ...
Crossrefs
Programs
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Maple
# Note that for n > 16 Maple fails (at least in some versions) to compute the # terms properly. Inserting 'simplify' or numerical evaluation might help. A307419Row := proc(n) local ogf, ser; ogf := (n, k) -> GAMMA(n+k+x)/GAMMA(k+x); ser := (n, k) -> series(ogf(n,k),x,k+2); seq(coeff(ser(n,k),x,k),k=0..n) end: seq(A307419Row(n), n=0..9); # Alternatively by the egf for column k: A307419Col := proc(n, len) local f, egf, ser; f := (n,x) -> (log(1-x)/(x-1))^n/n!; egf := (n,x) -> diff(f(n, x), [x$n]); ser := n -> series(egf(n, x), x, len); seq(k!*coeff(ser(n), x, k), k=0..len-1) end: seq(print(A307419Col(k, 10)), k=0..9); # Peter Luschny, Apr 12 2019 T := (n, k) -> add((-1)^(n-j)*binomial(j, k)*Stirling1(n, j)*k^(j-k), j = k..n): seq(seq(T(n,k), k = 0..n), n = 0..9); # Peter Luschny, Jun 09 2022
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Mathematica
f[n_, x_] := f[n, x] = D[(Log[1 - x]/(x - 1))^n/n!, {x, n}]; T[n_, k_] := (n - k)! SeriesCoefficient[f[k, x], {x, 0, n - k}]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 13 2019 *) -
Maxima
T(n,k):=n!*sum((binomial(k+i-1,i)*abs(stirling1(n-i,k)))/(n-i)!,i,0,n-k);
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Maxima
taylor((1-t)^(-x/(1-t)),t,0,7,x,0,7);
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Maxima
T(n,k):=coeff(taylor(gamma(n+k+t)/gamma(k+t),t,0,10),t,k);
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PARI
T(n, k) = n!*sum(i=0, n-k, abs(stirling(n-i, k, 1))*binomial(i+k-1, i)/(n-i)!); \\ Michel Marcus, Apr 13 2019
Formula
E.g.f.: A(t,x) = (1-t)^(-x/(1-t)).
T(n, k) = n!*Sum_{L1+L2+...+Lk=n} H(L1)H(L2)...H(Lk) with Li > 0, where H(n) are the harmonic numbers A001008.
T(n, k) = n!*Sum_{i=0..n-k} abs(Stirling1(n-i, k))/(n-i)!*binomial(i+k-1, i).
T(n, k) = k! [x^k] (d^n/dx^n) ((log(1-x)/(x-1))^n/n!), the e.g.f. for column k where Col(k) = [T(n+k, k) for n = 0, 1, 2, ...]. - Peter Luschny, Apr 12 2019
T(n, k) = Sum_{j=k..n} (-1)^(n-j)*binomial(j, k)*Stirling1(n, j)*k^(j-k). - Peter Luschny, Jun 09 2022