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A260667
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a(n) = (1/n^2) * Sum_{k=0..n-1} (2k+1)*S(k,n)^2, where S(k,x) denotes the polynomial Sum_{j=0..k} binomial(k,j)*binomial(x,j)*binomial(x+j,j).
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34
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1, 37, 1737, 102501, 6979833, 523680739, 42129659113, 3572184623653, 315561396741609, 28807571694394593, 2701627814373536601, 259121323945378645947, 25330657454041707496017, 2516984276442279642274311, 253667099464270541534450025, 25884030861250181046253181349, 2670255662315910532447096232073
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OFFSET
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1,2
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COMMENTS
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Conjecture: For k = 0,1,2,... define S(k,x):= Sum_{j=0..k} binomial(k,j)*binomial(x,j)*binomial(x+j,j).
(i) For any integer n > 0, the polynomial (1/n^2) * Sum_{k=0..n-1}(2k+1)*S(k,x)^2 is integer-valued (and hence a(n) is always integral).
(ii) Let r be 0 or 1, and let x be any integer. Then, for any positive integers m and n, we have the congruence
Sum_{k=0..n-1} (-1)^(k*r)*(2k+1)*S(k,x)^(2m) == 0 (mod n).
(iii) For any odd prime p, we have Sum_{k=0..p-1} S(k,-1/2)^2 == (-1/p)(1-7*p^3*B_{p-3}) (mod p^4), where (a/p) is the Legendre symbol, and B_0,B_1,B_2,... are Bernoulli numbers. Also, for any prime p > 3 we have Sum_{k=0..p-1} S(k,-1/3)^2 == p - (14/3)*(p/3)*p^3*B_{p-2}(1/3) (mod p^4), where B_n(x) denotes the Bernoulli polynomial of degree n; Sum_{k=0..p-1} S(k,-1/4)^2 == (2/p)*p - 26*(-2/p)*p^3*E_{p-3} (mod p^4), where E_0,E_1,E_2,... are Euler numbers; Sum_{k=0..p-1} S(k,-1/6)^2 == (3/p)*p - (155/12)*(-1/p)*p^3*B_{p-2}(1/3) (mod p^4).
Our conjecture is motivated by a conjecture of Kimoto and Wakayama which states that Sum_{k=0..p-1} S(k,-1/2)^2 == (-1/p) (mod p^3) for any odd prime p. The Kimoto-Wakayama conjecture was confirmed by Long, Osburn and Swisher in 2014.
For more related conjectures, see Sun's paper arXiv.1512.00712. - Zhi-Wei Sun, Dec 03 2015
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LINKS
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FORMULA
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a(n) ~ phi^(10*n + 3) / (10 * Pi^2 * n^3), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Nov 06 2021
Conjecture: a(p-1) == 1 (mod p^3) for all primes p >= 5. - Peter Bala, Aug 15 2022
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EXAMPLE
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a(2) = 37 since (1/2^2) * Sum_{k=0..1} (2k+1)*S(k,2)^2 = (S(0,2)^2 + 3*S(1,2)^2)/4 = (1^2 + 3*7^2)/4 = 148/4 = 37.
G.f. = x + 37*x^2 + 1737*x^3 + 102501*x^4 + 6979833*x^5 + 523680739*x^6 + ...
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MAPLE
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c := n -> binomial(2*n, n)/(n + 1):
h := n -> simplify(hypergeom([-n, -n, -n], [1, -2*n], 1)):
b := n -> c(n)^2*((n+11)*(2+4*n)^2*h(n+1)^2-2*(n+1)*(11*n+16)*(1+2*n)*h(n)*h(n+1)-h(n)^2*(n+1)^3)/(25*(n+2)):
a := n -> b(n-1): seq(a(n), n = 1..17); # Peter Luschny, Nov 11 2022
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MATHEMATICA
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S[k_, x_]:=S[k, x]=Sum[Binomial[k, j]Binomial[x, j]Binomial[x+j, j], {j, 0, k}]
a[n_]:=a[n]=Sum[(2k+1)*S[k, n]^2, {k, 0, n-1}]/n^2
Do[Print[n, " ", a[n]], {n, 1, 17}]
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CROSSREFS
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The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692, A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
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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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