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a(n) = (4/n^2)*( Sum_{k=0..n-1} k*A246459(k) ).
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%I #36 Nov 12 2023 12:08:02

%S 0,7,52,378,2832,21785,171036,1364391,11023264,89985681,740894700,

%T 6144227430,51267563280,430045297695,3623966778180,30662599042530,

%U 260367332354496,2217928838577641,18947382204700044,162281586037920126

%N a(n) = (4/n^2)*( Sum_{k=0..n-1} k*A246459(k) ).

%C Conjecture: a(n) is always an integer.

%C Note: the formula for a(n) in terms of A005802 proves that a(n) is an integer, divisible by n-1. - _Mark van Hoeij_, Nov 06 2023

%H Zhi-Wei Sun, <a href="/A246513/b246513.txt">Table of n, a(n) for n = 1..150</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1408.5381">Two new kinds of numbers and their arithmetic properties</a>, arXiv:1408.5381 [math.NT], 2014-2018.

%F Recurrence: (n-2)*n^2*(2*n-7)*(4*n-5)*a(n) = (n-1)*(80*n^4 - 532*n^3 + 1126*n^2 - 893*n + 195)*a(n-1) - 9*(n-2)^2*(n-1)*(2*n-5)*(4*n-1)*a(n-2). - _Vaclav Kotesovec_, Aug 28 2014

%F a(n) ~ 3^(2*n+1/2) / (2*Pi*n). - _Vaclav Kotesovec_, Aug 28 2014

%F a(n) = (n-1) * ((n+1)^2 * A005802(n-1) - (n-1)*n * A005802(n-2)). - _Mark van Hoeij_, Nov 06 2023

%e a(2) = 7 since (4/2^2)*( Sum_{k=0..1} k*A246459(k) ) = A246459(1) = 7.

%p h := n -> hypergeom([1/2, 1 - n, -n], [2, 2], 4):

%p a := n -> (n - 1) * ((n + 1)^2 * h(n) / n - n * h(n - 1)):

%p seq(simplify(a(n)), n = 1..20); # _Peter Luschny_, Nov 06 2023

%p ogf := (((-54*x^4+18*x^3+30*x^2+6*x)*hypergeom([4/3, 4/3],[2],-27*x*(x-1)^2/(9*x-1)^2)+(-1701*x^3+783*x^2-111*x+5)*hypergeom([1/3, 1/3],[1],-27*x*(x-1)^2/(9*x-1)^2))/(1-9*x)^(8/3) - 5)/6;

%p series(ogf, x=0, 25); # _Mark van Hoeij_, Nov 12 2023

%t s[n_] := Sum[Binomial[n, k]^2 Binomial[2 k, k] (2 k + 1), {k, 0, n}]

%t a[n_] := Sum[k s[k], {k, 0, n-1}] 4/n^2

%t Table[a[n], {n, 1, 20}]

%Y Cf. A246459, A246460.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Aug 28 2014