Irregular table of numbers m such that the sum (k+1)^(2n) + (k+2)^(2n) +…+ (k+m)^(2n) may be prime.
2, 3, 6, 2, 3, 5, 6, 10, 15, 30, 3, 6, 7, 14, 21, 42, 2, 3, 5, 6, 10, 15, 30, 3, 6, 11, 22, 33, 66, 3, 5, 6, 7, 10, 13, 14, 15, 21, 26, 30, 35, 39, 42, 65, 70, 78, 91, 105, 130, 182, 195, 210, 273, 390, 455, 546, 910, 1365, 2730, 3, 6, 2, 3, 5, 6, 10, 15, 17
1
That is, numbers m such that the sum of m consective 2n-powers can be prime. It appears that the n-th row has most of the divisors of the denominator of the Bernoulli number B(2n). Why? Note that the n-th row begins with 2 only if n is a power of 2.
T. D. Noe, Plot of 15 rows
T. D. Noe, Table of 15 rows
Eric W. Weisstein, MathWorld: Bernoulli Number
Cf. A002445, A163251, A165347, S000786-S000789.
nonn,tabf,hard,nice
T. D. Noe, Dec 07 2015