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%I #12 Aug 11 2015 01:21:44
%S 1,23,1,1,23,1,1,1,19,1,1,1,23,23,1,1,17,13,1,23,7,1,23,23,1,23,11,1,
%T 23,7,1,1,1,1,19,7,1,1,7,17,1,1,1,23,19,19,1,7,1,23,7,1,1,19,1,7,23,1,
%U 1,7,1,23,29,1,23,13,1,23,7,1,1,19,1,1,19,1,23
%N Solutions to numerator(Bernoulli(k)) == denominator/6 (Bernoulli(k)) (mod 30).
%C Conjecture: the residues mod 30 of the numerator and the denominator/6 of Bernoulli(20(n-1) + 2) are equal. Conjecture: the only solutions to the above equation are {1, 7, 11, 13, 17, 19, 23 or 29}. Observation: the differences between these solutions are (6, 4, 2, 4, 2, 4, 6), a sequence with bilateral symmetry. Program checks all nonzero Bernoulli numbers except B(1), but if the above conjecture is true, then it needs check only every 20th Bernoulli Number starting with B(2).
%H Michael G. Kaarhus, <a href="/A234788/b234788.txt">Table of n, a(n) for n = 1..300</a>
%F This formula is conjectural, but the program verified it for each of the first 300 numbers in this sequence: to obtain k from the n of this sequence, k = 20(n-1) + 2.
%e 13 is in this sequence because both the numerator and the denominator/6 of a Bernoulli Number are congruent to 13 mod 30. Using my conjectural formula, you can find which Bernoulli Number: 13 is the 18th number in this sequence. k = 20(18-1) + 2. k = 342. So, both the numerator and the denominator/6 of Bernoulli(342) are congruent to 13 mod 30.
%o (Maxima) k:-2$ for n:1 thru 300 step 0 do (k:k+2, b:bern(k), f:mod(num(b), 30), a:mod(denom(b)/6, 30), if f=a then (print(n, ", ", a), if 20*(n-1)+2#k then (print("Exception at k=", k, " n=", n), n:4000), n:n+1))$
%Y Similar to A233578 and A233579.
%K nonn
%O 1,2
%A _Michael G. Kaarhus_, Dec 30 2013
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