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%I #31 Aug 18 2017 10:27:42
%S 1,35,2431,6678671,2756205443,2781907990776503,3749562977351496827,
%T 34227405074603836560875299,10994118229823875586790445536799,
%U 3383080509296917481189798760796480670771162183
%N a(n) is the factor P(n) having prime factors between n^2 and 2*n^2 in A285388(n) = R(n)P(n) for n > 1, a(1)=1.
%C All A290564(n) prime factors in the interval occur with multiplicity 1. Thus a(n) is squarefree.
%F a(n) = A285388(n)/A290583(n).
%e a(3) = 2431: A285388(3) = 36465 = (R(3) = 15)*(P(3) = 11*13*17 = 2431);
%e a(4) = 6678671: A285388(4) = 300540195 = (R(4) = 45)*(P(4) = 17*19*23*29*31 = 6678671).
%t 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 *)
%o (PARI) a285388(n) = my(m=n*binomial(2*n^2, n^2)); m>>valuation(m, 2);
%o 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
%Y Cf. A285388, A290583 (R()), A290564 (number of primes in interval).
%K nonn
%O 1,2
%A _Ralf Steiner_, Aug 07 2017
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