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Revision History for A309132 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

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a(n) is the denominator of F(n) = A027641(n-1)/n + A027642(n-1)/n^2.
(history; published version)
#110 by Charles R Greathouse IV at Thu Sep 08 08:46:21 EDT 2022
PROG

(MAGMAMagma) [Denominator(Numerator(Bernoulli(n-1))/n + Denominator(Bernoulli(n-1))/n^2): n in [1..70]]; // Vincenzo Librandi, Jul 14 2019

Discussion
Thu Sep 08
08:46
OEIS Server: https://oeis.org/edit/global/2944
#109 by Harvey P. Dale at Sat Mar 21 15:43:40 EDT 2020
STATUS

editing

approved

#108 by Harvey P. Dale at Sat Mar 21 15:43:36 EDT 2020
MATHEMATICA

Table[Denominator[Numerator[BernoulliB[n - 1]] / n + Denominator[ BernoulliB[ n - 1]] / n^2], {n, 70}] (* Vincenzo Librandi, Jul 14 2019 *)

STATUS

approved

editing

#107 by Peter Luschny at Sun Aug 18 04:01:20 EDT 2019
STATUS

proposed

approved

#106 by Jonathan Sondow at Fri Aug 16 19:26:54 EDT 2019
STATUS

editing

proposed

#105 by Jonathan Sondow at Fri Aug 16 19:26:42 EDT 2019
COMMENTS

The values of F(n) when n is prime are A327033. - Jonathan Sondow, Aug 16 2019

STATUS

approved

editing

#104 by Bruno Berselli at Mon Jul 22 06:33:25 EDT 2019
STATUS

reviewed

approved

#103 by Peter Luschny at Mon Jul 22 04:17:31 EDT 2019
STATUS

proposed

reviewed

#102 by Jonathan Sondow at Sun Jul 21 11:58:52 EDT 2019
STATUS

editing

proposed

#101 by Jonathan Sondow at Sun Jul 21 11:49:17 EDT 2019
COMMENTS

Theorem 2. If n is a prime or a Carmichael number, then a(n) = A326690(n) = denominator of (Sum_{prime p | n} 1/p - 1/n). The proof is a generalization of that of Theorem 1. (Note that Theorem 2 implies Theorem 1, since if n is prime, then (Sum_{prime p | n} 1/p - 1/n) = 1/n - 1/n = 0, /1, so a(p) = A326690(n) = 1.) For n a prime or a Carmichael number, an application of Theorem 2 is to compute computing a(n) without calculating Bernoulli numbers(n-1) which may be huge; see A309268 and A326690. - Jonathan Sondow, Jul 19 2019

STATUS

proposed

editing