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A176285
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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, ...).
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2
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1, 11, 316, 12011, 522376, 24593348, 1219951188, 62798884331, 3323228619736, 179665076698136, 9880531254032176, 550994628527745476, 31084678988906064016, 1770908612898043660556, 101738260887234550287316
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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FORMULA
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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).
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EXAMPLE
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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.
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MATHEMATICA
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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]=1 Table[u[n], {n, 1, 50}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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