The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A246543 a(n) = 2/n^3*( sum_{k=0}^{n-1} (-1)^k*(3*k^2+3*k+1)*C(n-1,k)^3*C(n+k,k)^3 ), where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!). 6
 2, -47, 1142, 3793, -4094806, 371557891, -13021558306, -1374157073639, 281067953420114, -22220280272696387, -51611579093593498, 257837341935815261683, -35155217354672369625958, 1761633462267526777842223, 202464167122130621896038062 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Conjecture: Let n be any positive integer. For m = 0, 2, 4, ..., we have sum_{k=0}^{n-1} (3k^2+3k+1)*(C(n-1,k)*C(n+k,k))^m == 0 (mod n^3); for m = 1, 3, 5, ... we have 2*sum_{k=0}^{n-1}(-1)^k*(3k^2+3k+1)*(C(n-1,k)*C(n+k,k))^m == 0 (mod n^3). The Zeilberger algorithm could yield a complicated fifth-order recurrence for a(n). The author proved the conjecture in the latest version of arXiv:1408.5381. - Zhi-Wei Sun, Sep 14 2014 LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..70 Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, arXiv:1408.5381, 2014. EXAMPLE a(2) = -47 since 2/2^3*( sum_{k=0,1} (-1)^k*(3k^2+3k+1)*C(1,k)^3*C(2+k,k)^3 ) = 1/4*(1-7*3^3) = -47. MATHEMATICA a[n_] := Sum[(3 k^2 + 3 k + 1) (-1)^k (Binomial[n - 1, k] Binomial[n + k, k])^3, {k, 0, n - 1}] 2/n^3 Table[a[n], {n, 1, 14}] CROSSREFS Cf. A246512, A246542. Sequence in context: A120050 A112783 A277655 * A119776 A087265 A079307 Adjacent sequences:  A246540 A246541 A246542 * A246544 A246545 A246546 KEYWORD sign AUTHOR Zhi-Wei Sun, Aug 29 2014 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified May 16 17:22 EDT 2021. Contains 343949 sequences. (Running on oeis4.)