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a(n) = denominator of 3*Sum_{j=0..n+1} 1/(2*j+1).
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%I #21 Sep 06 2024 22:22:13

%S 1,5,35,105,1155,15015,15015,255255,4849845,4849845,111546435,

%T 557732175,1673196525,48522699225,1504203675975,1504203675975,

%U 1504203675975,55655536011075,55655536011075,2281876976454075,98120709987525225

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

%H G. C. Greubel, <a href="/A111877/b111877.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = denominator of (3/2)*(digamma(n+5/2) + 2*log(2) + euler_gamma).

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

%F 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 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 (Magma) [Denominator((2*HarmonicNumber(2*n+4) - HarmonicNumber(n+2)))/3: n in [0..40]]; // _G. C. Greubel_, Jul 24 2023

%o (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 Cf. A001620, A025547, A350669 (numerators).

%K easy,nonn

%O 0,2

%A _Paul Barry_, Aug 19 2005

%E Name edited by _G. C. Greubel_, Jul 24 2023