OFFSET
1,2
COMMENTS
Originally defined as the first column of A027448, but now contains numerator in reduced form (cf. A329122). - Sean A. Irvine, Nov 05 2019
LINKS
Donal F. Connon, Some series and integrals involving the Riemann zeta function, binomial coefficients and the harmonic numbers. Vol. II(a), arXiv:0710.4023 [math.HO], 2007-2008.
Jerry Metzger and Thomas Richards, A Prisoner Problem Variation, Journal of Integer Sequences, Vol. 18 (2015), Article 15.2.7.
FORMULA
Numerators of sequence a(1,n) in (a(i,j))^4 where a(i,j) = 1/i if j <= i, 0 if j > i.
Numerators of (H(n,1)^3 + 3*H(n,1)*H(n,2) + 2*H(n,3))/(6*n) = ((gamma+Psi(n+1))^3 + 3*(gamma+Psi(n+1))*(1/6*Pi^2 - Psi(1, n+1)) + 2*Zeta(3) + Psi(2, n+1))/(6*n), where H(n, m) = Sum_{i=1..n} 1/i^m are generalized harmonic numbers. - Vladeta Jovovic, Aug 10 2002
From Groux Roland, Feb 11 2011: (Start)
For n>=1, (H(n,1)^3 + 3*H(n,1)*H(n,2) + 2*H(n,3))/(6*n) = -(1/6)*Integral_{x=0..1} x^(n-1)*log^3(1-x) dx = (1/n)*Sum_{j=1..n} (H(n,1) - H(j-1,1))*H(j,1)/j.
For every k>=1 the first column of (a(i,j))^k is the binomial transform of (-1)^n/(n+1)^k.
Proof: the sequence S(n,k) = ((-1)^k/k!)*Integral_{x=0..1} x^(n-1)*log^k(1-x) dx gives the binomial transform of (-1)^n/(n+1)^(k+1) and can be evaluated by parts with S(n,k) = (1/n)*Sum_{j=1..k} S(j,k-1) according to the generation of the first column of (a(i,j))^k.
(End)
MATHEMATICA
a[n_] := -1/6 Integrate[x^(n-1) Log[1-x]^3, {x, 0, 1}] // Numerator;
Table[a[n], {n, 1, 16}] (* Jean-François Alcover, Aug 06 2018 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
Corrected by Vladeta Jovovic, Aug 10 2002
Title changed by Sean A. Irvine, Nov 05 2019
STATUS
approved