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 A246511 a(n) = (sum_{k=0}^{n-1} (-1)^k*(2k+1)*C(n-1,k)^2*C(n+k,k)^2)/n, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!). 5
 1, -13, 103, 219, -26139, 503957, -4066061, -54914149, 2550230113, -43157232273, 192777017511, 10118180981037, -318814450789587, 4344955121014089, 6807591584551563, -1781238363905009253, 42912636577174295769, -425791821468024981709, -5452095049517604924017, 305524943325956601071159 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Zhi-Wei Sun proved that a(n) is always an integer, and that sum_{k=0}^{n-1}(2k+1)*A(k) = n^3*a(n), where A(k) = sum_{j=0..k} (-1)^j*(2j+1)^2*C(k,j)^2*C(k+j,j)^2. The Zeilberger algorithm could yield a complicated fourth-order recurrence for this sequence. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..100 Zhi-Wei Sun, Two new kinds of numbers and their arithmetic properties, arXiv:1408.5381, 2014. FORMULA a(n) = hypergeom([3/2, 1-n, 1-n, n+1, n+1], [1/2, 1, 1, 1], -1)/n. - Robert Israel, Aug 28 2014 Recurrence: (n-1)^2*n^3*(2*n-7)*(2*n-5)*(40*n^6 - 600*n^5 + 3612*n^4 - 11120*n^3 + 18354*n^2 - 15270*n + 4949)*a(n) = -2*(n-1)^2*(2*n-7)*(1120*n^10 - 21280*n^9 + 173136*n^8 - 789528*n^7 + 2217244*n^6 - 3965700*n^5 + 4511984*n^4 - 3162198*n^3 + 1267357*n^2 - 247675*n + 14910)*a(n-1) - 2*(n-2)*(2*n-7)*(2*n-1)*(9080*n^10 - 181600*n^9 + 1569004*n^8 - 7670464*n^7 + 23311258*n^6 - 45445432*n^5 + 56332869*n^4 - 42029480*n^3 + 16243359*n^2 - 1773884*n - 347928)*a(n-2) - 2*(n-3)^2*(2*n-1)*(1120*n^10 - 23520*n^9 + 213456*n^8 - 1095144*n^7 + 3485308*n^6 - 7092252*n^5 + 9139424*n^4 - 7057450*n^3 + 2811541*n^2 - 317773*n - 61278)*a(n-3) - (n-4)^3*(n-3)^2*(2*n-3)*(2*n-1)*(40*n^6 - 360*n^5 + 1212*n^4 - 1872*n^3 + 1266*n^2 - 234*n - 35)*a(n-4). - Vaclav Kotesovec, Sep 07 2014 EXAMPLE a(2) = -13 since sum_{k=0,1}(-1)^k*(2k+1)C(1,k)^2*C(2+k,k)^2 = 1 - 3*3^2 = 2*(-13). MAPLE a:= n -> add((-1)^k*(2*k+1)*binomial(n-1, k)^2*binomial(n+k, k)^2, k=0..n-1)/n: seq(a(n), n=1..40); # Robert Israel, Aug 28 2014 MATHEMATICA a[n_]:=Sum[(-1)^k*(2k+1)*Binomial[n-1, k]^2*Binomial[n+k, k]^2, {k, 0, n-1}]/n Table[a[n], {n, 1, 20}] CROSSREFS Cf. A246065, A246138, A246459, A246460, A246461, A246462. Sequence in context: A289420 A050670 A142318 * A085957 A119594 A066034 Adjacent sequences:  A246508 A246509 A246510 * A246512 A246513 A246514 KEYWORD sign AUTHOR Zhi-Wei Sun, Aug 27 2014 STATUS approved

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