Bernoulli numbers and arithmetic progressions

[Hans Havermann]

T. D. Noe:

> For irregular prime p, I think that p first appears as a quotient in
> A090496 for
>
> n = p*k - (p-1)/2,
>
> where k is least integer such that k > 0 and p | numerator(B(2k)).

I was curious to see if this formulation could predict subsequent 
appearances of the irregular primes. First-appearance numbers given by

ip={A000928}; Table[p = ip[[i]]; k = 1; 
While[Mod[Numerator[BernoulliB[2k]], p] > 0, k++]; p*k-(p-1)/2, {i, 1, 
65}]

are: {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}

The second time Mod[Numerator[BernoulliB[2k]], p] == 0: {1240, 2980, 
4121, 8434, 6438, 9891, 20637, 8557, 36698, ...}

Looks good. It even correctly predicts the 8557 for irregular prime 157 
which, as Cino Hilliard pointed out, violates Roland Bacher's initial 
conjecture that "all indices yielding a given prime p form an 
arithmetic progression of step ((p-1)/2)*p". Continuing in this vein, 
the third appearances should be: {1351, 3452, 4456, ...}

Well no! We know that the first irregular prime, 37, appears at 574, 
1240, 1906, 2572, 3238, 3904, 4570, 5236, 5902, 6568, 7234, 7900, 8566, 
9232, 9898, ... clearly not at 1351. What are the values of p*k-(p-1)/2 
where Mod[Numerator[BernoulliB[2k]], p] == 0 for p=37?

{574, 1240, 1351, 1906, 2572, 2720, 3238, 3904, 4089, 4570, 5236, 5458, 
5902, 6568, 6827, 7234, 7900, 8196, 8566, 9232, 9565, 9898, ...}

Every third term in this calculated sequence, it seems, is a number 
that does not figure in the *actual* 37-appearance sequence. So, 
perhaps the *fourth* time Mod[Numerator[BernoulliB[2k]], p] == 0 for 
the irregular primes yields only the actual *third* appearance: {1906, 
4691, 6332, ...}. Again, looks good. Can we demonstrate constant 
periodicity then for each irregular prime, based on these sequences?

The second-appearance numbers minus the first-appearance numbers yield: 
{666, 1711, 2211, 5050, 5253, 8515, 11026, 3768, 27028, 32896, 34453, 
36585, 39903, 42778, 46971, 48205, 60031, 20121, 14023, 75466, 80200, 
83436, 88410, 93528, 106030, 106953, 23350, 10802, 136503, 146070, 
59076, 154846, 166176, 587, 175528, 183921, 187578, 47509, 191271, 
46063, 1941, 212878, 216811, 31631, 228826, 232903, 64954, 263901, 
281625, 286146, 289180, 298378, 317206, 120541, 328455, 336610, 341551, 
351541, 384126, 387640, 392941, 139350, 453628, 470935, 562330}

This should equal the third-appearance numbers minus the 
second-appearance numbers, but they do not! The first failure is at 
index-8, irregular prime 157, which we already noted failed Roland 
Bacher's arithmetic-progression conjecture. Subsequent failures are at 
index-18 & index-19. and surprisingly common after index-26.
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