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%I #37 Dec 23 2013 06:20:35
%S 10,16,20,22,28,30,32,44,46,48,50,52,56,58,60,64,66,80,82,84,90,92,96,
%T 104,106,112,116,128,132,136,138,140,144,148,150,154,156,160,164,166,
%U 168,170,172,174,176,178,180,184,192,198,200,212,224,226,238,240,242,246,252,260,262,268
%N Numbers n such that the denominator/6 of Bernoulli(n) is congruent to {11, 17, 23, 25 or 29} modulo 30.
%C Conjecture: for these and only these n, the absolute value of the numerator of Bernoulli(n) is congruent 5 modulo 6. If this is true, then you can obtain the residue modulo 6 of the absolute value of Bernoulli numerators by calculating their denominators/6 modulo 30. The program uses the von Staudt-Clausen Theorem. None of these n are in the complementary sequence, A233578 (n >= 2 such that the denominator/6 of Bernoulli_n is congruent to {1, 5, 7, 13 or 19} 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="/A233579/b233579.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 112 is in this sequence, because the denominator of Bernoulli(112) = 1671270, and 1671270/6 = 278545, and 278545 is congruent to 25 modulo 30. As for the conjecture, the absolute value of the numerator of Bernoulli(112) is congruent to 5 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=11 or x=17 or x=23 or x=25 or x=29) then (print(r, ", ",n), r:r+1), i:[1], n:n+2)$
%Y Cf. A233578, subsequence of A005843.
%K nonn
%O 1,1
%A _Michael G. Kaarhus_, Dec 13 2013
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