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 A246875 a(n) = (Sum_{k=0..n-1} C(n-1,k)^2*C(-n-1,k)^2/C(k+2,2))/n. 1
 1, 2, 13, 134, 1783, 27950, 491335, 9401390, 192033565, 4131488426, 92723165533, 2155279960586, 51602299168639, 1267128734047142, 31803430252162579, 813628992468938750, 21168533016938471665, 559044288633621863810, 14962460440143262653685, 405299365266569619086462 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS The author proved in arXiv:1408.5381 that a(n) is always an integer. Note that Sum_{k=0..n-1} C(n-1,k)*C(-n-1,k)/C(k+2,2) = 0 for n > 1. Conjecture: The sequence a(n+1)/a(n) (n > 0) is strictly increasing to the limit 17+12*sqrt(2), and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n > 1) is strictly decreasing to the limit 1. 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 [math.NT], 2014. FORMULA Recurrence (obtained via the Zeilberger algorithm): -n*(n-1)^2*(2*n+3)*a(n) + 4*(17*n^4+68*n^3+92*n^2+48*n+9)*a(n+1) - (n+2)*(n+3)^2*(2*n+1)*a(n+2) = 0. a(n) ~ (17+12*sqrt(2))^n / (2^(5/4) * Pi^(3/2) * n^(9/2)). - Vaclav Kotesovec, Sep 07 2014 a(n) = 4F3(1-n,1-n,1+n,1+n;1,1,3;1)/n. - Benedict W. J. Irwin, Apr 04 2017 EXAMPLE a(2) = 2 since (Sum_{k=0..1} C(2-1,k)^2*C(-2-1,k)^2/C(2+k,2))/2 = (1 + (-3)^2/3)/2 = 2. MATHEMATICA a[n_]:=Sum[(Binomial[n-1, k]*Binomial[-n-1, k])^2/Binomial[k+2, 2], {k, 0, n-1}]/n Table[a[n], {n, 1, 20}] Table[HypergeometricPFQ[{1-n, 1-n, 1+n, 1+n}, {1, 1, 3}, 1]/n, {n, 1, 10}] (* Benedict W. J. Irwin, Apr 04 2017 *) PROG (PARI) a(n) = sum(k=0, n - 1, (binomial(n - 1, k) * binomial(-n - 1, k))^2/binomial(k + 2, 2))/n; \\ Indranil Ghosh, Apr 04 2017 CROSSREFS Cf. A243101, A246460, A246511, A246543, A246567. Sequence in context: A239748 A065132 A047856 * A242004 A187021 A152059 Adjacent sequences:  A246872 A246873 A246874 * A246876 A246877 A246878 KEYWORD nonn AUTHOR Zhi-Wei Sun, Sep 07 2014 STATUS approved

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Last modified May 22 20:52 EDT 2022. Contains 353959 sequences. (Running on oeis4.)