OFFSET
2,1
COMMENTS
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.
PROG
(PARI) \ prestore some ireg primes in iprime[] or use slower PARI BIF prime() 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(r", ")) ) ) }
CROSSREFS
KEYWORD
nonn
AUTHOR
Cino Hilliard, Feb 16 2004
STATUS
approved