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 A246460 a(n) = (sum_{k=0}^{n-1} (2k+1)*C(n-1,k)^2*C(n+k,k)^2)/n^2, where C(n,k) denotes the binomial coefficient n!/(k!(n-k)!). 10
 1, 7, 77, 1211, 23009, 489035, 11203765, 270937315, 6825612185, 177559028087, 4739821161173, 129244697791951, 3587524535220001, 101099089948850323, 2886373390151379397, 83343790441133767475, 2430567530705659113545, 71508611747063572974095, 2120357936904537499679125, 63315310358625743871987019 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Conjecture: a(n) is always an integer. The author proved this in the latest version of arXiv:1408.5381. - Zhi-Wei Sun, Sep 01 2014 LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..100 Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, arXiv:1408.5381, 2014. FORMULA Recurrence (obtained via the Zeilberger algorithm): -n^3*(2*n + 5)*(3*n^2 + 12*n + 11)*a(n) + (2*n + 5)*(105*n^5 + 675*n^4 + 1579*n^3 + 1663*n^2 + 768*n+126)*a(n+1) - (2*n + 1)*(105*n^5 + 900*n^4 + 2929*n^3 + 4448*n^2 + 3048*n + 684)*a(n+2) + (n + 3)^3*(2*n + 1)*(3*n^2 + 6*n + 2)*a(n+3) = 0. a(n) ~ 2^(1/4) * (17+12*sqrt(2))^n / (8 * Pi^(3/2) * n^(5/2)). - Vaclav Kotesovec, Aug 27 2014 EXAMPLE a(2) = 7 since sum_{k=0,1} (2k+1)C(1,k)^2*C(2+k,k)^2 = 1 + 3*3^2 = 28 = 2^2*7. MAPLE A246460:=n->add((2*k+1)*binomial(n-1, k)^2*binomial(n+k, k)^2/n^2, k=0..n-1): seq(A246460(n), n=1..20); # Wesley Ivan Hurt, Aug 26 2014 MATHEMATICA a[n_]:=Sum[(2k+1)*Binomial[n-1, k]^2*Binomial[n+k, k]^2, {k, 0, n-1}]/n^2 Table[a[n], {n, 1, 20}] CROSSREFS Cf. A245769, A246065, A246138, A246459. Sequence in context: A107866 A034176 A001765 * A222145 A240404 A093980 Adjacent sequences:  A246457 A246458 A246459 * A246461 A246462 A246463 KEYWORD nonn AUTHOR Zhi-Wei Sun, Aug 26 2014 EXTENSIONS Typo in cross-reference corrected by Vaclav Kotesovec, Aug 27 2014 STATUS approved

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Last modified January 19 12:48 EST 2020. Contains 331049 sequences. (Running on oeis4.)