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)!).
-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
Keywords
Examples
a(2) = 7 since sum_{k=0,1,2}C(2,k)*C(2+k,k)/(2k-1) = -1 + 6 + 6/3 = 7.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 0..200
- 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.
Programs
-
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
-
Mathematica
a[n_]:=Sum[Binomial[n,k]Binomial[n+k,k]/(2k-1),{k,0,n}]; Table[a[n],{n, 0, 25}]
-
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
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
Comments