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%I #28 Jun 07 2020 01:24:13
%S 5,25,85,185,235,295,305,335,355,365,395,425,505,535,635,685,695,745,
%T 815,835,925,985,995,1115,1135,1145,1285,1315,1345,1385,1415,1445,
%U 1475,1525,1535,1555,1565,1585,1655,1675,1735,1765
%N Numbers m such that the Bernoulli number B_{2*m} has denominator 66.
%C From the von Staudt-Clausen theorem, denominator(B_{2*m}) = product of primes p such that (p-1)|2*m.
%D B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 75.
%H T. D. Noe, <a href="/A051229/b051229.txt">Table of n, a(n) for n = 1..1000</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Von_Staudt%E2%80%93Clausen_theorem">Von Staudt-Clausen theorem</a>.
%H <a href="/index/Be#Bernoulli">Index entries for sequences related to Bernoulli numbers</a>.
%F a(n) = 5*A119456(n). - _G. C. Greubel_, Jun 06 2020
%e The numbers m = 5, 25 belong to the list because B_10 = 5/66 and B_50 = 495057205241079648212477525/66. - _Petros Hadjicostas_, Jun 06 2020
%t Select[Range[2000],Denominator[BernoulliB[2 #]]==66&] (* _Harvey P. Dale_, Mar 11 2012 *)
%o (PARI) is(n)=denominator(bernfrac(2*n))==66 \\ _Charles R Greathouse IV_, Feb 06 2017
%o (Sage) [n for n in (1..2000) if denominator(bernoulli(2*n))==66 ] # _G. C. Greubel_, Jun 06 2020
%Y Cf. A045979, A051222, A051225, A051226, A051227, A051228.
%Y Equals A051230/2.
%K nonn,nice,easy
%O 1,1
%A _N. J. A. Sloane_
%E More terms from _Michael Somos_
%E Name edited by _Petros Hadjicostas_, Jun 06 2020
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