A090798 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.

37, 59, 67, 101, 103, 131, 149, 157, 157, 233, 257, 263, 271, 283, 293, 307, 311, 347, 353, 353, 379, 379, 389, 401, 409, 421, 433, 461, 463, 467, 467, 491, 491, 491, 523, 541, 547, 547, 557, 577, 587, 587, 593, 607, 613, 617, 617, 617, 619, 631, 631, 647

Offset: 1

Author

Cino Hilliard, Feb 16 2004

Keywords

• nonn

Comments

Only even values of r need to be tested.

See Table A.3, "Calculated irregular pairs of order 10 of primes below 1000," in B. C. Kellner.

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..2000
• Bernd C. Kellner, On irregular prime power divisors of the Bernoulli numbers, Math. Comp. 76 (2007) 405-441.

Crossrefs

Cf. A090495 A090496.

Programs

Mathematica

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

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",")) ) ) }

Formula

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.