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A092291
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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.
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3
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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MATHEMATICA
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(* 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 *)
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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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
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STATUS
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approved
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