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 A176285 a(1) = 1, and then 4*(2*n + 1)^2*a(n+1) + n^2*a(n) = (205*n^2 + 160*n + 32)*binomial(2n-1,n)^3 (n = 1, 2, 3, ...). 2
 1, 11, 316, 12011, 522376, 24593348, 1219951188, 62798884331, 3323228619736, 179665076698136, 9880531254032176, 550994628527745476, 31084678988906064016, 1770908612898043660556, 101738260887234550287316 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS On Apr 04 2010, Zhi-Wei Sun introduced this sequence and conjectured that each term a(n) is a positive integer. He also guessed that a(n) is odd if and only if n is a power of two. It is easy to see that 8*n^2*binomial(2*n,n)^2*a(n) equals Sum_{k=0..n-1}(205*k^2 + 160*k + 32)*(-1)^{n - 1 - k}*binomial(2*k,k)^5. Sun also conjectured that for any prime p > 5 we have Sum_{k=0..p-1}(205*k^2 + 160*k + 32)(-1)^k*binomial(2*k,k)^5 = 32*p^2 + 64*p^3*Sum_{k=1..p-1}1/k (mod p^7) and Sum_{k=0..(p-1)/2}(205*k^2 + 160*k + 32)(-1)^k*binomial(2*k,k)^5 = 32*p^2 + 896/3*p^5*B_{p-3} (mod p^6), where B_0, B_1, B_2, ... are Bernoulli numbers. It is also remarkable that Sum_{n>0}(-1)^n(205*n^2 - 160*n + 32)/(n^5*binomial(2*n,n)^5) = -2*zeta(3) as proved by T. Amdeberhan and D. Zeilberger via the WZ method. LINKS T. Amdeberhan and D. Zeilberger, Hypergeometric series acceleration via the WZ method, Electron. J. Combin. 4(1997), no.2, #R3. Kasper Andersen, Re: A somewhat surprising conjecture J. Guillera, Hypergeometric identities for 10 extended Ramanujan-type series, arXiv:1104.0396 [math.NT], 2011; Ramanujan J. 15(2008), 219-234. Z. W. Sun, Open Conjectures on Congruences, preprint, arXiv:0911.5665 [math.NT], 2009-2011. Zhi-Wei Sun, A somewhat surprising conjecture Zhi-Wei Sun, Re: A somewhat surprising conjecture FORMULA a(n) = (Sum_{k=0..n-1} (205*k^2 + 160*k + 32)(-1)^{n - 1 - k}*binomial(2*k,k)^5)/(8*n^2*binomial(2*n,n)^2). EXAMPLE For n = 2 we have a(2) = 11 since 4*(2*1 + 1)^2*a(2) = -1^2*a(1) + (205*1^2 + 160*1 + 32)*binomial(2*1 - 1,1)^3 = 396. MATHEMATICA u[n_]:=u[n]=((205(n-1)^2+160(n-1)+32)Binomial[2n-3, n-1]^3-(n-1)^2*u[n-1])/(4(2n-1)^2) u=1 Table[u[n], {n, 1, 50}] CROSSREFS Cf. A173774, A000984, A001700. Sequence in context: A266029 A106827 A276977 * A266484 A219979 A115609 Adjacent sequences:  A176282 A176283 A176284 * A176286 A176287 A176288 KEYWORD nonn AUTHOR Zhi-Wei Sun, Apr 14 2010 STATUS approved

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Last modified July 5 00:40 EDT 2020. Contains 335457 sequences. (Running on oeis4.)