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%I #21 Oct 17 2018 08:56:14
%S 2,10,530,4734,474986,5153122,676701794,1232820800342,15623119507746,
%T 34472401720246110,6163354867874693078,83483882991733501114,
%U 15658391111267929558466,42132263940113324754864134
%N a(n) = (C(2p,p)-2) / p^3, where p = prime(n).
%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 R. P. Stanley, <a href="http://ocw.mit.edu/courses/mathematics/18-s66-the-art-of-counting-spring-2003/">MIT Course OCW 18.S66, The Art of Counting, Spring 2003</a>. (Problem 18)
%F a(n) = A060842(n) / A000040(n).
%F a(n) = 2 * A034602(n).
%e a(6)=4734 since 13 is the sixth prime and (C(26,13)-2)/13^3 = (10400600-2)/2197 = 4734.
%t Table[(Binomial[2p,p]-2)/p^3,{p,Prime[Range[3,20]]}] (* _Harvey P. Dale_, Oct 23 2017 *)
%Y Cf. A268512, A268589, A268590.
%K nonn
%O 3,1
%A _Henry Bottomley_, Oct 02 2003
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