|
|
A243101
|
|
a(n) = (sum_{k=0}^{n-1}(4*k^3-1)*C(n-1,k)*C(n+k,k))/n^2, where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).
|
|
3
|
|
|
-1, 2, 37, 324, 2403, 16582, 109961, 711176, 4521607, 28394442, 176648877, 1090974156, 6698429931, 40930511502, 249105886737, 1510954029072, 9138180665871, 55128765109906, 331851698655797, 1993747216968788
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Conjecture: Let n be any positive integer.
(i) For any m = 0,1,2,..., we have sum_{k=0}^{n-1}(-1)^k*(4*k^3-1)*C(n-1,k)^m*C(-n-1,k)^m == 0 (mod n^2). Also, we may replace 4*k^3-1 by 2*k^2+2*k+1 or 2*k^3+k^2+k.
(ii) For any s,t = 0,1,2,..., both sum_{k=0}^{n-1}(4*k^3-1)*C(n-1,k)^s*C(-n-1,k)^t and sum_{k=0}^{n-1}(-1)^k*(4*k^3-1)*C(n-1,k)^s*C(-n-1,k)^t are multiples of n.
The author proved the conjecture in the latest version of arXiv:1408.5381, thus a(n) is always an integer. - Zhi-Wei Sun, Sep 01 2014
|
|
LINKS
|
|
|
FORMULA
|
Recurrence (obtained via the Zeilberger algorithm):
n*(2*n+3)*(n^2+3*n+1)*a(n) - 2*(2*n^2+4*n+1)*(3*n^2+6*n+2)*a(n+1) + (n+2)*(2*n+1)*(n^2+n-1)*a(n+2) = 0.
a(n) ~ (3+2*sqrt(2))^n * sqrt(n) / (2^(5/4) * sqrt(Pi)). - Vaclav Kotesovec, Sep 02 2014
|
|
EXAMPLE
|
a(2) = 2 since (sum_{k=0,1}(4*k^3-1)*C(1,k)*C(2+k,k))/2^2 = (-1 + 3*3)/4 = 2.
|
|
MAPLE
|
|
|
MATHEMATICA
|
a[n_]:=Sum[(4k^3-1)*Binomial[n-1, k]Binomial[n+k, k], {k, 0, n-1}]/n^2
Table[a[n], {n, 1, 20}]
|
|
CROSSREFS
|
|
|
KEYWORD
|
sign
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|