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 A243101 a(n) = (sum_{k=0}^{n-1}(4*k^3-1)*C(n-1,k)*C(n+k,k))/n^2, where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!). 3
 -1, 2, 37, 324, 2403, 16582, 109961, 711176, 4521607, 28394442, 176648877, 1090974156, 6698429931, 40930511502, 249105886737, 1510954029072, 9138180665871, 55128765109906, 331851698655797, 1993747216968788 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Conjecture: Let n be any positive integer. (i) For any m = 0,1,2,..., we have sum_{k=0}^{n-1}(-1)^k*(4*k^3-1)*C(n-1,k)^m*C(-n-1,k)^m == 0 (mod n^2). Also, we may replace 4*k^3-1  by 2*k^2+2*k+1 or 2*k^3+k^2+k. (ii) For any s,t = 0,1,2,..., both sum_{k=0}^{n-1}(4*k^3-1)*C(n-1,k)^s*C(-n-1,k)^t and sum_{k=0}^{n-1}(-1)^k*(4*k^3-1)*C(n-1,k)^s*C(-n-1,k)^t are multiples of n. The author proved the conjecture in the latest version of arXiv:1408.5381, thus a(n) is always an integer. - Zhi-Wei Sun, Sep 01 2014 LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..200 Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, arXiv:1408.5381, 2014. FORMULA Recurrence (obtained via the Zeilberger algorithm): n*(2*n+3)*(n^2+3*n+1)*a(n) - 2*(2*n^2+4*n+1)*(3*n^2+6*n+2)*a(n+1) + (n+2)*(2*n+1)*(n^2+n-1)*a(n+2) = 0. a(n) ~ (3+2*sqrt(2))^n * sqrt(n) / (2^(5/4) * sqrt(Pi)). - Vaclav Kotesovec, Sep 02 2014 EXAMPLE a(2) = 2 since (sum_{k=0,1}(4*k^3-1)*C(1,k)*C(2+k,k))/2^2 = (-1 + 3*3)/4 = 2. MAPLE A243101:=n->add((4*k^3-1)*binomial(n-1, k)*binomial(n+k, k), k=0..n-1)/n^2: seq(A243101(n), n=1..20); # Wesley Ivan Hurt, Sep 01 2014 MATHEMATICA a[n_]:=Sum[(4k^3-1)*Binomial[n-1, k]Binomial[n+k, k], {k, 0, n-1}]/n^2 Table[a[n], {n, 1, 20}] CROSSREFS Cf. A246542, A246543. Sequence in context: A053788 A078976 A200911 * A001569 A092853 A297796 Adjacent sequences:  A243098 A243099 A243100 * A243102 A243103 A243104 KEYWORD sign AUTHOR Zhi-Wei Sun, Aug 29 2014 STATUS approved

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Last modified June 12 19:06 EDT 2021. Contains 344959 sequences. (Running on oeis4.)