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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.
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
1,1
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
Bernd C. Kellner, On irregular prime power divisors of the Bernoulli numbers, Math. Comp. 76 (2007) 405-441.
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.
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
PROG
(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", ")) ) ) }
CROSSREFS
KEYWORD
nonn
AUTHOR
Cino Hilliard, Feb 16 2004
STATUS
approved