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%I #34 Jan 29 2019 02:17:33
%S 2,17,19,31,47,59,61,71,101,103,107,109,137,149,151,157,167,181,197,
%T 211,223,227,229,241,257,263,269,271,283,311,313,317,331,337,347,349,
%U 353,367,379,383,389,397,401,421,439,449,457,461,463,467,479,503,521
%N Primes p such that B(4*p) has denominator 30 where B(2n) are the Bernoulli numbers.
%C Complement of A087634, primes p such that phi(k) = 4p has a solution, where phi is Euler's totient function.
%C The sequences a(n), A005384 and A023212 form a partition of the set of primes > 3: Using von Staudt-Clausen formula the divisors of 4p increased by 1 are {2,3,5,p+1,2p+1,4p+1}, p+1 is clearly an even number, and if 2p+1 and 4p+1 are not prime, then denom(B(4p))=30. - _Enrique Pérez Herrero_, Aug 15 2011
%C Also 2 with the primes p such that both 2*p+1 and 4*p+1 are composite: A210684. For these numbers k > 2 the equation: phi(n)=k*tau(n), where phi is A000010 and tau is A000005, has no solutions: A112954(a(n))=0. - _Enrique Pérez Herrero_, May 12 2012
%H E. Pérez Herrero, <a href="/A043297/b043297.txt">Table of n, a(n) for n=1..30000</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Von_Staudt%E2%80%93Clausen_theorem">Von Staudt-Clausen theorem</a>
%t Select[Prime[Range[100]], Denominator[BernoulliB[4# ]]==30&] (* _T. D. Noe_, Feb 19 2004 *)
%t Select[Prime[Range[100]],!PrimeQ[4#+1]&&!PrimeQ[2#+1]||(#==2)&] (* _Enrique Pérez Herrero_, Aug 16 2011 *)
%Y Cf. A051225, A053176, A087634.
%Y Cf. A005384, A023212.
%Y Cf. A210684.
%K easy,nonn
%O 1,1
%A _Benoit Cloitre_, Mar 24 2002
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