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%I #44 Dec 23 2013 06:23:01
%S 2,4,6,8,12,14,18,24,26,34,36,38,40,42,54,62,68,70,72,74,76,78,86,88,
%T 94,98,100,102,108,110,114,118,120,122,124,126,130,134,142,146,152,
%U 158,162,182,186,188,190,194,196,202,204,206,208,210,214,216,218,220,222,228,230,232,234
%N n >= 2 such that the denominator/6 of Bernoulli(n) is congruent to {1, 5, 7, 13 or 19} modulo 30.
%C Conjecture: for these and only these n, the absolute value of the numerator of Bernoulli(n) is congruent 1 modulo 6. If my conjecture is true, then you can obtain the residue modulo 6 of the abs. value of Bernoulli numerators by calculating their denominators/6 modulo 30. Program uses the von Staudt-Clausen Theorem. None of these n are in the complementary sequence, A233579 (n such that the denominator/6 of Bernoulli(n) is congruent to {11, 17, 23, 25 or 29} modulo 30. I have checked and verified that, up to n = 50446, the union of A233578 and A233579 is all even numbers >= 2.
%H Michael G. Kaarhus, <a href="/A233578/b233578.txt">Table of n, a(n) for n = 1..10000</a>
%H M. G. Kaarhus, <a href="http://www.christaboveme.com/bern/bern-2-types.pdf">Splitting the Bernoulli Numbers</a>
%e 100 is in this sequence, because the denominator of Bernoulli(100) = 33330, and 33330/6 = 5555, and 5555 is congruent to 5 modulo 30. As for the conjecture, the abs. val. of the numerator of Bernoulli(100) is congruent to 1 modulo 6.
%o (Maxima) float(true)$ load(basic)$ i:[1]$ n:2$ for r:1 thru 10000 step 0 do (for p:3 while p-1<=n step 0 do (p:next_prime(p), if mod(n, p-1)=0 then push(p,i)), d:(product(i[k],k,1,length(i))), x:mod(d,30), if (x=1 or x=5 or x=7 or x=13 or x=19) then (print(r, ", ",n), r:r+1), i:[1], n:n+2)$
%Y Cf. A233579, subsequence of A005843.
%K nonn
%O 1,1
%A _Michael G. Kaarhus_, Dec 13 2013
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