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 A090791 Minimal numbers n such that numerator(Bernoulli(2*n)/(2*n)) is different from numerator(Bernoulli(2*n)/(2*n*(2*n-r))) for some integer r. 1

%I

%S 52,80,95,134,114,141,213,187,274,338,312,312,292

%N Minimal numbers n such that numerator(Bernoulli(2*n)/(2*n)) is different from numerator(Bernoulli(2*n)/(2*n*(2*n-r))) for some integer r.

%C These values of n 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, n=pk+r/2 = 37*k + 30/2 = 37k+15 = 52 = 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. A090790, A090495, A090496.

%K nonn,more

%O 1,1

%A _Cino Hilliard_, Feb 16 2004

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Last modified May 26 17:16 EDT 2022. Contains 354092 sequences. (Running on oeis4.)