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%I #16 Jun 26 2017 23:04:10
%S 30,42,56,66,22,20,128,60,82,162,98,82,18
%N Numbers r arising in A090791.
%C These values of r correspond to the first 13 irregular primes produced by a/b.
%F 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.
%e 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.
%o (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));
%o b = numerator(a/(n2-r)); v=a/b; if(a <> b && v==p,print(k","n","v);break) ) ) } /* A001067 */
%o (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 */
%Y Cf. A090495, A090496, A090791.
%K nonn,more
%O 1,1
%A _Cino Hilliard_, Feb 16 2004
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