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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
52, 80, 95, 134, 114, 141, 213, 187, 274, 338, 312, 312, 292 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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

LINKS

Table of n, a(n) for n=1..13.

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, n=pk+r/2 = 37*k + 30/2 = 37k+15 = 52 = 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

Cf. A090790, A090495, A090496.

Sequence in context: A118148 A111173 A090793 * A234099 A026067 A039475

Adjacent sequences:  A090788 A090789 A090790 * A090792 A090793 A090794

KEYWORD

nonn,more

AUTHOR

Cino Hilliard, Feb 16 2004

STATUS

approved

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Last modified January 22 20:00 EST 2022. Contains 350504 sequences. (Running on oeis4.)