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%I #42 Nov 15 2022 09:26:46
%S 1,7,97,1791,38241,892039,22092673,571387903,15271248769,418796912007,
%T 11725812711009,333962374092543,9648543623050593,282164539499639559,
%U 8338391167566634497,248661515283002490879,7474768663941435203073
%N a(n) = (1/n^2) * Sum_{k = 0..n-1} (2*k+1)*(D_k)^2, where D_0, D_1, ... are central Delannoy numbers given by A001850.
%C On Jun 14 2010, _Zhi-Wei Sun_ conjectured that a(n) = (1/n^2) * Sum_{k = 0..n-1} (2*k+1)*(D_k)^2 is always an integer and that p^2*a(p) = p^2 - 4*p^3*q_p(2) - 2*p^4*q_p(2)^2 (mod p^5) for any prime p > 3, where q_p(2) denotes the Fermat quotient (2^(p-1) - 1)/p (see Sun, Remark 4.3, p. 26, 2014). He also conjectured that Sum_{k = 0..n-1} (2*k+1)*(-1)^k*(D_k)^2 == 0 (mod n*D_n/(3,D_n)) for all n = 1,2,3,....
%C The fact that a(n) is an integer follows directly from the formulas for a(n) in the formula section below. - _Mark van Hoeij_, Nov 13 2022
%H G. C. Greubel, <a href="/A178808/b178808.txt">Table of n, a(n) for n = 1..500</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1006.2776">Arithmetic properties of Apery numbers and central Delannoy numbers</a>, arXiv:1006.2776 [math.NT], 2011.
%H Zhi-Wei Sun, <a href="https://arxiv.org/abs/1008.3887">Congruences involving generalized central trinomial coefficients</a>, Sci. China Math. 57 (2014), no. 7, 1375-1400; arXiv:1008.3887 [math.NT], 2010-2014.
%F a(n) ~ (1 + sqrt(2))^(4*n) / (16*Pi*n^2). - _Vaclav Kotesovec_, Jan 24 2019
%F G.f.: Integral(hypergeom([1/2, 1/2], [2], -32*x/(1 - 34*x + x^2))/((1 - x)*(1 - 34*x + x^2)^(1/2))). - _Mark van Hoeij_, Nov 10 2022
%F a(n) = (6*A001850(n)*A001850(n-1) - A001850(n)^2 - A001850(n-1)^2)/8. - _Mark van Hoeij_, Nov 12 2022
%F a(n) = (3*f(n)*f(n-1) - g(n))/4, where g(n) = hypergeom([n, -n, 1/2], [1, 1], -8) and f(n) = hypergeom([-n, -n], [1], 2). This formula also gives an integer value for n = 0. - _Peter Luschny_, Nov 13 2022
%e For n = 3 we have a(3) = (D_0^2 + 3*D_1^2 + 5*D_2^2)/3^2 = (1 + 3*3^2 + 5*13^2)/3^2 = 97.
%p A001850 := n -> LegendreP(n, 3); seq((6*A001850(n)*A001850(n-1)-A001850(n)^2-A001850(n-1)^2)/8, n=1..20); # _Mark van Hoeij_, Nov 12 2022
%p # Alternative:
%p g := n -> hypergeom([n, -n, 1/2], [1, 1], -8): # A358388
%p f := n -> hypergeom([-n, -n], [1], 2): # A001850
%p a := n -> (3*f(n)*f(n-1) - g(n)) / 4:
%p seq(simplify(a(n)), n = 1..17); # _Peter Luschny_, Nov 13 2022
%t DD[n_]:=Sum[Binomial[n+k,2k]Binomial[2k,k],{k,0,n}]; SS[n_]:= Sum[(2k+1)*DD[k]^2,{k,0,n-1}]/n^2; Table[SS[n],{n,1,25}]
%t Table[Sum[(2k+1)*JacobiP[k,0,0,3]^2, {k, 0, n-1}]/n^2, {n, 1, 30}] (* _G. C. Greubel_, Jan 23 2019 *)
%o (Python) # prepends a(0) = 0
%o def A178808List(size: int) -> list[int]:
%o A358387 = A358387gen()
%o A358388 = A358388gen()
%o return [(next(A358387) - next(A358388)) // 4 for n in range(size)]
%o print(A178808List(18)) # _Peter Luschny_, Nov 15 2022
%Y Cf. A001850, A178790, A178791, A173774, A358388.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Jun 16 2010
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