%I #16 Dec 13 2014 08:06:14
%S 37,59,67,101,103,131,149,157,157,233,257,263,271,283,293,307,311,347,
%T 353,353,379,379,389,401,409,421,433,461,463,467,467,491,491,491,523,
%U 541,547,547,557,577,587,587,593,607,613,617,617,617,619,631,631,647
%N Irregular primes in the ratio numerator(Bernoulli(2*n)/(2*n)) / numerator(Bernoulli(2*n)/(2*n*(2*n-r))) when these numerators are different and n is a minimum for some integer r. Duplication indicates irregularity index > 1.
%C Only even values of r need to be tested.
%C See Table A.3, "Calculated irregular pairs of order 10 of primes below 1000," in B. C. Kellner.
%H Robert G. Wilson v, <a href="/A090798/b090798.txt">Table of n, a(n) for n = 1..2000</a>
%H Bernd C. Kellner, <a href="http://dx.doi.org/10.1090/S0025-5718-06-01887-4"> On irregular prime power divisors of the Bernoulli numbers</a>, Math. Comp. 76 (2007) 405-441.
%F Given a = numerator(Bernoulli(2*n)/(2*n)) and b = numerator(a/(2*n-r)) for integer r positive or negative, then n>0 n = p + r/2 For every irregular prime p there is an r such that n is minimum.
%t f[p_] := Block[{c = 0, k = 1}, While[ 2k <= p - 3, If[ Mod[ Numerator@ BernoulliB[ 2k], p] == 0, c++]; k++]; c]; p = 5; lst = {}; While[p < 1001, AppendTo[lst, Table[p, {f@ p}]]; p = NextPrime@ p]; Flatten@ lst
%o (PARI) \ prestore some ireg primes in iprime[] bernmin(m) = { for(x=1,m, p=iprime[x]; forstep(r=2,p,2, n=r/2+p; n2=n+n; a = numerator(bernfrac(n2)/(n2)); \ A001067 b = numerator(a/(n2-r)); \ if(a <> b,print(r","n","a/b)) if(a <> b,print1(a/b",")) ) ) }
%Y Cf. A090495 A090496.
%K nonn
%O 1,1
%A _Cino Hilliard_, Feb 16 2004