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%I #29 Dec 12 2017 15:29:53
%S 1,2,2,12,3,15,60,840,105,630,630,13860,6930,180180,360360,144144,
%T 9009,306306,306306,11639628,14549535,14549535,58198140,2677114440,
%U 334639305,3346393050
%N Denominators of b(n) = b(n-1)/2 + 1/(2*n), b(0)=0.
%C Generally, 2*b(n) = b(n-1) + f(n). See, for f(n)=n, A000337(n)/2^n.
%C a(0)=1. b(n) is mentioned in A241269.
%C Difference table of b(n):
%C 0, 1/2, 1/2, 5/12, 1/3, 4/15, ...
%C 1/2, 0, -1/12, -1/12, -1/15, -1/20, ...
%C -1/2, -1/12, 0, 1/60, 1/60, 11/840, ...
%C 5/12, 1/12, 1/60, 0, -1/280, -1/280, ...
%C etc.
%C b(n) is mentioned in A241269 as an autosequence of the first kind.
%C The denominators of the first two upper diagonals are the positive Apéry numbers, A005430(n+1). Compare to the array in A003506.
%C Numerators: 0, 1, 1, 5, 1, 4, 13, 151, 16, 83, 73, 1433, 647, 15341, ... .
%F b(n) = -Re(Phi(2, 1, n + 1)) where Phi denotes the Lerch transcendent. - _Eric W. Weisstein_, Dec 11 2017
%e 0, 1/2, 1/2, 5/12, 1/3, 4/15, 13/60, 151/840, 16/105, 83/630, 73/630, ...
%e b(1) = (0+1)/2, hence a(1)=2.
%e b(2) = (1/2+1/2)/2 = 1/2, hence a(2)=2.
%e b(3) = (1/2+1/3)/2 = 5/12, hence a(3)=12.
%t b[0] = 0; b[n_] := b[n] = 1/2*(b[n-1] + 1/n); Table[b[n] // Denominator, {n, 0, 25}] (* _Jean-François Alcover_, Apr 25 2014 *)
%t Table[-Re[LerchPhi[2, 1, n + 1]], {n, 0, 20}] // Denominator (* _Eric W. Weisstein_, Dec 11 2017 *)
%t -Re[LerchPhi[2, 1, Range[20]]] // Denominator (* _Eric W. Weisstein_, Dec 11 2017 *)
%t RecurrenceTable[{b[n] == b[n - 1]/2 + 1/(2 n), b[0] == 0}, b[n], {n, 20}] // Denominator (* _Eric W. Weisstein_, Dec 11 2017 *)
%Y Cf. A086466.
%Y Cf. A242376 (numerators).
%K nonn
%O 0,2
%A _Paul Curtz_, Apr 24 2014
%E Extension, after a(13), from _Jean-François Alcover_, Apr 24 2014