A189286 a(n):=(Sum_{k=0}^n C(6k,3k)C(3k,k)C(6(n-k),3(n-k))C(3(n-k),n-k))/((2n-1)Binomial[3n,n])

-1, 40, 696, 23408, 969496, 44602560, 2187147600, 111957721920, 5911097451480, 319469892415808, 17584481176101952, 982222958294603040, 55530668360895219728, 3171318959654377396864, 182670436050532943578560, 10599737781026193970325760, 619014530633087163062727000, 36353266320338484003053582400, 2145559172529803104937217263040, 127190916635938933740168015020160

Offset

0,2

Comments

On Apr 19 2011, Zhi-Wei Sun conjectured that a(n) is an integer for every n=0,1,2,.... He proved that a(p-1)=[(p+1)/6] (mod p) for any prime p, and also made the following conjecture:

(i) a(n)^{1/n} tends to 64 as n tends to the infinity.

(ii) For any positive integer n, we have a(n)=0 (mod 8), and a(n)/8 is odd if and only if n is a power of two.

Links

Table of n, a(n) for n=0..19.

Formula

Recursion: (n+2)^2*(3n+2)(3n+4)(3n+5)a(n+2)

=16(2n+1)(2n+3)(3n+2)(18n^2+54n+41)a(n+1) - 9216(n+1)^2(4n^2-1)(3n+5)a(n).

Example

For n=1 we have a(1)=(C(6,3)C(3,1)+C(6,3)C(3,1))/C(3,1)=120/3=40.

Mathematica

S[n_]:=Sum[Binomial[6k, 3k]Binomial[3k, k]Binomial[3(n-k), n-k]Binomial[6(n-k), 3(n-k)], {k, 0, n}]/((2n-1)Binomial[3n, n])
Table[S[n], {n, 0, 19}]

Crossrefs

Sequence in context: A247410 A183464 A359932 * A160445 A010840 A233725

Adjacent sequences: A189283 A189284 A189285 * A189287 A189288 A189289

Keyword

sign

Author

Zhi-Wei Sun, Apr 19 2011

Status

approved