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
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Sep 07 2014
STATUS
approved