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%I #28 Oct 17 2018 08:59:00
%S 6,3,2,3,6,13,30,75,200,555,1590,4693,14202,43863,137882,440235,
%T 1424958,4668304,15459366,51692379,174362770,592815459,2030105382,
%U 6998177293,24270836436,84646997613,296744311172,1045283877639,3698462401026,13140509079977,46869358523238,167781751129899
%N a(n) = (3/n^3) * Sum_{d|n} (-1)^(n+d)*moebius(n/d)*binomial(2*d,d).
%H G. C. Greubel, <a href="/A254593/b254593.txt">Table of n, a(n) for n = 1..1650</a>
%H R. R. Aidagulov, M. A. Alekseyev. On p-adic approximation of sums of binomial coefficients. Journal of Mathematical Sciences 233:5 (2018), 626-634. doi:<a href="http://doi.org/10.1007/s10958-018-3948-0">10.1007/s10958-018-3948-0</a> arXiv:<a href="http://arxiv.org/abs/1602.02632">1602.02632</a>
%H M. Kontsevich, R. Stanley, O. Gorodetsky, et al. <a href="http://mathoverflow.net/q/195339">A congruence involving binomial coefficients</a>, Mathoverflow, 2015.
%F a(n) = 6*A131868(n)/n.
%F For n == 0, 1, or 3 (mod 4), a(n) = A268592(n)/2; for n == 2 (mod 4), a(n) = A268592(n)/2 + A268592(n/2)/8.
%t a[n_] := 3/n^3 DivisorSum[n, (-1)^(n+#) MoebiusMu[n/#] Binomial[2#, #]&]; Array[a, 40] (* _Jean-François Alcover_, Dec 18 2015 *)
%o (PARI) { a(n) = sumdiv(n,d,(-1)^(n+d)*moebius(n/d)*binomial(2*d,d))*3/n^3 }
%K nonn
%O 1,1
%A _Max Alekseyev_, Feb 01 2015
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