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.
52, 80, 95, 134, 114, 141, 213, 187, 274, 338, 312, 312, 292
Offset: 1
Examples
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.
Programs
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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 */
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.
Comments