OFFSET
1,1
COMMENTS
Let S_(2*n+1)(m) denote difference between multiples of 2*n+1 in interval [0,m), m>=1, with even and odd digit sums in base 2*n. As is shown in the Shevelev and Moses link, a recursion for S_(2*n+1)(m) is connected with the periodicity of a special digit function, the smallest period of which is a(n).
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..100
Vladimir Shevelev and Peter J. C. Moses, A family of digit functions with large periods, arXiv:1209.5705 [math.NT], 2012.
MATHEMATICA
Table[2^(2*n + 1)*(2*n + 1)*n^(2*n), {n, 15}] (* Wesley Ivan Hurt, Apr 28 2020 *)
PROG
(Maxima) A217971(n):=2^(2*n+1)*(2*n+1)*n^(2*n)$ makelist(A217971(n), n, 1, 10); /* Martin Ettl, Nov 15 2012 */
(PARI) a(n) = {2^(2*n+1) * (2*n+1)*n^(2*n)} \\ Andrew Howroyd, Apr 28 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev and Peter J. C. Moses, Oct 16 2012
EXTENSIONS
Terms a(11) and beyond from Andrew Howroyd, Apr 28 2020
STATUS
approved