OFFSET
1,1
COMMENTS
These values of r correspond to the first 13 irregular primes produced by 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*k+(p+r)/2 if r is odd and n = p*k+r/2 if r is even where k = 1, 2.. For every irregular prime p there is an r such that n is minimum.
EXAMPLE
Given a, b as defined above and p=37, r=30, 52 = pk+r/2 = 37*1 + 30/2 is the smallest number that for a<>b a/b = 37.
PROG
(PARI) bern3(m, r) = { for(i=m, m, p=irprime(i); /* use the Somos script below to get irregular prime */ for(k=1, p, if(r%2, n=p*k+(p+r)/2, n=p*k+r/2); n2=n+n; a = numerator(bernfrac(n2)/(n2));
b = numerator(a/(n2-r)); v=a/b; if(a <> b && v==p, print(k", "n", "v); break) ) ) } /* A001067 */
(PARI) irprime(n) = { my(p); if(n<1, 0, p=irprime(n-1) + (n==1); while(p = nextprime(p+2), forstep(i=2, p-3, 2, if( numerator(bernfrac(i))%p == 0, break(2)))); p) }; /* compute irregular primes irprime from - Michael Somos, Feb 04 2004 */
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Cino Hilliard, Feb 16 2004
STATUS
approved