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A111877 a(n) = denominator of 3*Sum_{j=0..n+1} 1/(2*j+1).

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%I A111877 #21 Sep 06 2024 22:22:13
%S A111877 1,5,35,105,1155,15015,15015,255255,4849845,4849845,111546435,
%T A111877 557732175,1673196525,48522699225,1504203675975,1504203675975,
%U A111877 1504203675975,55655536011075,55655536011075,2281876976454075,98120709987525225
%N A111877 a(n) = denominator of 3*Sum_{j=0..n+1} 1/(2*j+1).
%H A111877 G. C. Greubel, <a href="/A111877/b111877.txt">Table of n, a(n) for n = 0..1000</a>
%F A111877 a(n) = denominator of (3/2)*(digamma(n+5/2) + 2*log(2) + euler_gamma).
%F A111877 a(n) = denominator of ( 3*Sum_{j=0..n+1} 1/(2*j+1) ).
%F A111877 a(n) = (1/3) * denominator of ( 2*H_{2*n+4} - H_{n+2} ), where H_{n} is the n-th Harmonic number. - _G. C. Greubel_, Jul 24 2023
%t A111877 f[x_]:= 2*x+1; a[1]= f[1]; a[n_]:= LCM[f[n], a[n-1]]; Array[a, 21]/3 (* _Robert G. Wilson v_, Jan 04 2013 *)
%o A111877 (Magma) [Denominator((2*HarmonicNumber(2*n+4) - HarmonicNumber(n+2)))/3: n in [0..40]]; // _G. C. Greubel_, Jul 24 2023
%o A111877 (SageMath) [denominator(2*harmonic_number(2*n+4,1) - harmonic_number(n+2,1))/3 for n in range(41)] # _G. C. Greubel_, Jul 24 2023
%Y A111877 Cf. A001620, A025547, A350669 (numerators).
%K A111877 easy,nonn
%O A111877 0,2
%A A111877 _Paul Barry_, Aug 19 2005
%E A111877 Name edited by _G. C. Greubel_, Jul 24 2023