OFFSET
1,2
COMMENTS
In the latest version of arXiv:1408:5381, the author proved that a(n) is always an integer. Notice that a(65) is relatively prime to 65. - Zhi-Wei Sun, Sep 14 2014
Conjecture: The sequence a(n+1)/a(n) (n = 1,2,3,...) is strictly increasing to the limit 17+12*sqrt(2), and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 1,2,3,...) is strictly decreasing to the limit 1.
Note that sum_{k=0}^{n-1}(2k+1)*A(k) = n^5*a(n) for all n > 0, where A(n) = sum_{k=0..n}C(n,k)^2*C(n+k,k)^2*(6k^3+9k^2+5k+1) for n = 0,1,2,....
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], 2014-2018.
Zuo-Ru Zhang, Proof of two conjectures of Z.-W. Sun on combinatorial sequences, arXiv:2112.12427 [math.CO], 2021.
FORMULA
Recurrence (obtained via the Zeilberger algorithm):
n^3*(n + 1)*(2n + 5)*(3n^2 + 12n + 11)*(6n^2 + 24n + 25)*a(n) - (n + 1)*(2n + 5)*(630n^7 + 6552n^6 + 28137n^5 + 64134n^4 + 82777n^3 + 59512n^2 + 21646n + 3076)*a(n+1) + (n + 2)*(2n + 1)*(630n^7 + 6678n^6 + 29271n^5 + 68751n^4 + 93469n^3 + 73445n^2 + 30640n + 5072)*a(n+2) - (n + 2)*(n + 3)^3*(2n + 1)*(3n^2 + 6n + 2)*(6n^2 + 12n + 7)*a(n+3) = 0.
EXAMPLE
a(2) = 8 since sum_{k=0,1} (3k^2+3k+1)C(1,k)^2*C(2+k,k)^2 = 1 + 7*3^2 = 64 = 2^3*8.
MATHEMATICA
a[n_]:=Sum[(3k^2+3k+1)*(Binomial[n-1, k]Binomial[n+k, k])^2, {k, 0, n-1}]/(n^3)
Table[a[n], {n, 1, 20}]
PROG
(PARI) a(n) = sum(k=0, n-1, (3*k^2+3*k+1)*binomial(n-1, k)^2*binomial(n+k, k)^2) /n^3; \\ Michel Marcus, Dec 24 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 28 2014
STATUS
approved