OFFSET
1,1
COMMENTS
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.
LINKS
Michael G. Kaarhus, Table of n, a(n) for n = 1..10000
M. G. Kaarhus, Splitting the Bernoulli Numbers
EXAMPLE
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.
PROG
(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)$
CROSSREFS
KEYWORD
nonn
AUTHOR
Michael G. Kaarhus, Dec 13 2013
STATUS
approved