OFFSET
1,2
COMMENTS
All A290564(n) prime factors in the interval occur with multiplicity 1. Thus a(n) is squarefree.
EXAMPLE
MATHEMATICA
Table[First@ Apply[Times, Map[Power, #]] &@ Select[FactorInteger@ Numerator[Sum[Binomial[2 k, k]/4^k, {k, 0, n^2 - 1}]/n], Function[p, # <= First@ p < 2 #] &[n^2]], {n, 10}] (* Michael De Vlieger, Aug 10 2017 *)
PROG
(PARI) a285388(n) = my(m=n*binomial(2*n^2, n^2)); m>>valuation(m, 2);
a(n) = if (n==1, 1, my(f=factor(a285388(n))); for (k=1, #f~, if ((n^2 > f[k, 1]) || (f[k, 1] > 2*n^2), f[k, 1] = 1)); factorback(f)); \\ Michel Marcus, Aug 07 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Ralf Steiner, Aug 07 2017
STATUS
approved