

A234788


Solutions to numerator(Bernoulli(k)) == denominator/6 (Bernoulli(k)) (mod 30).


1



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, 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, 1, 7, 1, 23, 29, 1, 23, 13, 1, 23, 7, 1, 1, 19, 1, 1, 19, 1, 23
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OFFSET

1,2


COMMENTS

Conjecture: the residues mod 30 of the numerator and the denominator/6 of Bernoulli(20(n1) + 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).


LINKS

Michael G. Kaarhus, Table of n, a(n) for n = 1..300


FORMULA

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(n1) + 2.


EXAMPLE

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(181) + 2. k = 342. So, both the numerator and the denominator/6 of Bernoulli(342) are congruent to 13 mod 30.


PROG

(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*(n1)+2#k then (print("Exception at k=", k, " n=", n), n:4000), n:n+1))$


CROSSREFS

Similar to A233578 and A233579.
Sequence in context: A348750 A040530 A040529 * A144445 A174729 A022186
Adjacent sequences: A234785 A234786 A234787 * A234789 A234790 A234791


KEYWORD

nonn


AUTHOR

Michael G. Kaarhus, Dec 30 2013


STATUS

approved



