1,1

These values of r correspond to the first 13 irregular primes produced by a/b.

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.

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.

(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)); \ A001067 b = numerator(a/(n2-r)); v=a/b; if(a <> b && v==p, print(k", "n", "v); break) ) ) } \ compute irregular primes irprime from - Michael Somos Feb 04 2004 irprime(n) = { local(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) }

Cf. A090495, A090496, A090791.

Sequence in context: A127663 A008885 A097036 this_sequence A090800 A114816 A000977

Adjacent sequences: A090787 A090788 A090789 this_sequence A090791 A090792 A090793

nonn

Cino Hilliard (hillcino368(AT)gmail.com), Feb 16 2004