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Numerators of b(n) = b(n-1)/2 + 1/(2*n), b(0)=0.
4

%I #37 Nov 14 2022 09:55:11

%S 0,1,1,5,1,4,13,151,16,83,73,1433,647,15341,28211,10447,608,19345,

%T 18181,651745,771079,731957,2786599,122289917,14614772,140001721,

%U 134354573,774885169,745984697,41711914513,80530073893,4825521853483

%N Numerators of b(n) = b(n-1)/2 + 1/(2*n), b(0)=0.

%C See the denominators in A241519.

%C b(n) = 0, 1/2, 1/2, 5/12, 1/3, 4/15, 13/60, 151/840, 16/105, 83/630, 73/630, ... (Ta0(n) in A241269) is an autosequence of the first kind.

%H Ralf Stephan, <a href="/A242376/b242376.txt">Table of n, a(n) for n = 0..999</a>

%F 0 = b(n)*(+b(n+1) - 4*b(n+2) + 4*b(n+3)) + b(n+1)*(-2*b(n+1) + 9*b(n+2) - 10*b(n+3)) + b(n+2)*(-2*b(n+2) + 4*b(n+3)) if n>=0. - _Michael Somos_, May 26 2014

%F b(n) = -Re(Phi(2, 1, n + 1)). - _Eric W. Weisstein_, Dec 11 2017

%F G.f. for b(n): -log(1-x)/(2*(1-x/2)). - _Vladimir Kruchinin_, Nov 14 2022

%e 0, 1/2, 1/2, 5/12, 1/3, 4/15, 13/60, 151/840, 16/105, 83/630, 73/630, ...

%t Table[-Re[LerchPhi[2, 1, n + 1]], {n, 0, 20}] // Numerator (* _Eric W. Weisstein_, Dec 11 2017 *)

%t -Re[LerchPhi[2, 1, Range[20]]] // Numerator (* _Eric W. Weisstein_, Dec 11 2017 *)

%t RecurrenceTable[{b[n] == b[n - 1]/2 + 1/(2 n), b[0] == 0}, b[n], {n, 20}] // Numerator (* _Eric W. Weisstein_, Dec 11 2017 *)

%o (Sage)

%o def a():

%o b = n = 0

%o while True:

%o yield numerator(b)

%o n = n + 1

%o b = (b/2 + 1/(2*n)) # _Ralf Stephan_, May 18 2014

%Y Cf. A241519 (denominators).

%K nonn,frac

%O 0,4

%A _Paul Curtz_, May 12 2014

%E a(14)-a(25) from _Jean-François Alcover_, May 12 2014

%E Corrected a(22) and a(24), more terms from _Ralf Stephan_, May 18 2014