A092291 Let p = n-th irregular prime, A000928(n). Then a(n) = smallest value of m such that numerator(Bernoulli(2*m)/(2*m)) / numerator(Bernoulli(2*m)/(2*m*(2*m-1))) equals p.

574, 1269, 1910, 3384, 1185, 1376, 9611, 4789, 9670, 20946, 13019, 11247, 2689, 22708, 13355, 45251, 48407, 32653, 18761, 38706, 76391, 25563, 50310, 79023, 44948, 29864, 21716, 71441, 104339, 22993, 73572, 61549, 14714, 26122, 6227, 179369, 159687, 5862, 132157, 24925, 76023, 15346, 73479, 136956, 212240, 10587, 3801, 137040, 108520, 194171, 98550, 282532, 87272, 133081, 220187, 305002, 41764, 27268, 380180, 70921, 184940, 241076, 73858, 80108, 250927

Offset: 1

Author

N. J. A. Sloane, based on a suggestion of Roland Bacher, Feb 05 2004

Keywords

• nonn
• nice

Comments

It was conjectured that a(n) = (1 + A000928(n) * (A035112(n) - 1))/2. However, Bernd Kellner's insightful paper shows that this formula first fails for the irregular prime 6449. - T. D. Noe, Feb 10 2004

Links

• Bernd Kellner, A conjecture about numerators of Bernoulli numbers

Crossrefs

Term in A090495 corresponding to first occurrence of p in A090496.

Programs

Mathematica

(* This program is not convenient for a large number of terms *) irregularPrimeQ[p_] := Module[{k = 1}, While[2*k <= p-3 && Mod[ Numerator[ BernoulliB[2*k]], p] != 0, k++]; 2*k <= p-3]; irregularPrime[1] = 37; irregularPrime[n_] := irregularPrime[n] = Module[{p}, For[p = NextPrime[ irregularPrime[n-1]], True, p = NextPrime[p], If[ irregularPrimeQ[p], Return[p]]]]; a[n_] := a[n] = For[m = 1, True, m++, If[ Numerator[BernoulliB[2*m]/(2*m)] / Numerator[ BernoulliB[2*m]/(2*m*(2*m-1))] == irregularPrime[n], Return[m]]]; Table[ Print[a[n]]; a[n], {n, 1, 15}] (* Jean-François Alcover, Sep 27 2013 *)

Extensions

Initial terms were computed by Roland Bacher, Feb 04 2004; further terms from Hans Havermann, Feb 05 2004 and T. D. Noe, Feb 06 2004

Offset modified by Jean-François Alcover, Sep 27 2013