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A245769
a(n) = Sum_{k=0..n} C(n, k)*C(n+k, k)/(2k-1), where C(n, k) denotes the binomial coefficient n!/(k!*(n-k)!).
8
-1, 1, 7, 25, 87, 329, 1359, 6001, 27759, 132689, 649815, 3242377, 16421831, 84196761, 436129183, 2278835681, 11996748255, 63568974241, 338777252263, 1814623238137, 9763917858359, 52750451120361, 286036294786287, 1556185889290065, 8492182185653327, 46471113779766769
OFFSET
0,3
COMMENTS
Note that a(n) is always an integer since C(n,k)*C(n+k,k) = C(n+k,2k)*C(2k,k), and C(2k,k)/(2k-1) = 2*Catalan(k-1) for k > 0.
Conjecture: The sequence a(n+1)/a(n) (n = 3,4,...) is strictly increasing to the limit 3+2*sqrt(2), and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 5,6,...) is strictly decreasing to the limit 1.
LINKS
Victor J. W. Guo, and Ji-Cai Liu, Proof of a conjecture of Z.-W. Sun on the divisibility of a triple sum, Journal of Number Theory, Volume 156, November 2015, Pages 154-160. Rn is a(n).
Zhi-Wei Sun, A new kind of numbers and their arithmetic properties, arXiv:1408.5381 [math.NT], 2017.
FORMULA
Recurrence (obtained via the Zeilberger algorithm): (n+1)*a(n) - (7*n+15)*a(n+1) + (7*n+13)*a(n+2) - (n+3)*a(n+3) = 0.
a(n) ~ A006318(n)/2 as n tends to the infinity, thus a(n)^(1/n) has the limit 3+2*sqrt(2).
0 = +a(n)*(+a(n+1) -15*a(n+2) +13*a(n+3) -3*a(n+4)) +a(n+1)*(+a(n+1) +50*a(n+2) -34*a(n+3) +13*a(n+4)) +a(n+2)*(-63*a(n+2) +50*a(n+3) -15*a(n+4)) +a(n+3)*(+a(n+3) +a(n+4)) for all n in Z. - Michael Somos, Aug 24 2014
EXAMPLE
a(2) = 7 since sum_{k=0,1,2}C(2,k)*C(2+k,k)/(2k-1) = -1 + 6 + 6/3 = 7.
MATHEMATICA
a[n_]:=Sum[Binomial[n, k]Binomial[n+k, k]/(2k-1), {k, 0, n}]; Table[a[n], {n, 0, 25}]
PROG
(PARI) for(n=0, 25, print1(sum(k=0, n, binomial(n, k)*binomial(n+k, k)/(2*k -1)), ", ")) \\ G. C. Greubel, Aug 05 2018
(Magma) [(&+[Binomial(n, k)*Binomial(n+k, k)/(2*k-1): k in [0..n]]): n in [0..25]]; // G. C. Greubel, Aug 05 2018
CROSSREFS
Sequence in context: A155233 A183914 A279216 * A146933 A155258 A261722
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Aug 24 2014
STATUS
approved