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a(n) = (1/(2n))*Sum_{k=0..n-1} C(n-1, k)*C(n+k, k)*C(2k, k)*(k+2)*(-3)^(n-1-k).
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%I #48 Feb 22 2024 09:41:33

%S 1,3,19,127,921,6921,53523,422199,3382417,27429043,224636259,

%T 1854761437,15419579761,128941830993,1083686483259,9147887134119,

%U 77520233226537,659167237928691,5622149927918763,48083938099637247

%N a(n) = (1/(2n))*Sum_{k=0..n-1} C(n-1, k)*C(n+k, k)*C(2k, k)*(k+2)*(-3)^(n-1-k).

%C From _Mark van Hoeij_, Nov 10 2022: (Start)

%C The fact that a(n) is an integer follows from the formula for the generating function. It is not difficult to show that f(x) := (hypergeom([1/2, 1/2], [1], 16*x) - 1)/4 is an element of Z[[x]]. Substituting x -> x/(1 + 3*x)^2 then shows that the given g.f. is in Z[[x]] as well. The fact that this formula is indeed the g.f. follows from the recurrence.

%C One can also show that a(n) is odd, as follows. Reducing f(x) in Z[[x]] modulo 2 gives: x + x^2 + x^4 + x^8 + x^16 + ... Again substitute x -> x/(1 + 3*x)^2, which modulo 2 is: x + x^3 + x^5 + x^7 + ... Then use the fact that (a+b)^(2^i) is congruent to a^(2^i) + b^(2^i) modulo 2 to see that f(x/(1 + 3*x)^2) is congruent to x + x^2 + x^3 + x^4 + ... modulo 2, so every a(n) is congruent to 1 modulo 2.

%C The formula a(n) = (A002426(n)^2 + 3*A002426(n-1)^2)/4 gives a second proof that a(n) is an odd integer. The numbers A002426(n) are odd, and so their squares are congruent to 1 modulo 8. Hence A002426(n)^2 + 3*A002426(n-1)^2 is congruent to 1 + 3 * 1 modulo 8. Since a(n) is that number divided by 4, it follows that a(n) is an odd integer. (End)

%H Zhi-Wei Sun, <a href="/A295371/b295371.txt">Table of n, a(n) for n = 1..200</a>

%H Heba Bou KaedBey, Mark van Hoeij, and Man Cheung Tsui, <a href="https://arxiv.org/abs/2402.11121">Solving Third Order Linear Difference Equations in Terms of Second Order Equations</a>, arXiv:2402.11121 [math.AC], 2024. See p. 6.

%F Via the Zeilberger algorithm we find that the sequence the following recurrence: (2n + 1)*(n + 3)^2*a(n + 3) = (2n + 1)*(7n^2 + 38n + 52)*a(n + 2) + 3*(2n + 5)*(7n^2 + 4n + 1)*a(n + 1) - 27*(2n + 5)*n^2*a(n).

%F From _Mark van Hoeij_, Nov 10 2022: (Start)

%F G.f.: (hypergeom([1/2, 1/2], [1], 16*x/(1 + 3*x)^2) - 1)/4.

%F a(n) = (A002426(n)^2 + 3 * A002426(n-1)^2)/4. (End)

%F G.f.: EllipticK((4*sqrt(x))/(3*x + 1))/(2*Pi) - (1/4). - _Peter Luschny_, Nov 10 2022

%e a(3) = 19 since (1/6)*Sum_{k=0,1,2} binomial(2,k)*binomial(3+k,k)*binomial(2k,k)*(k+2)*(-3)^(2-k) = (2*(-3)^2 + 2*4*2*3*(-3) + 10*6*4)/6 = 19.

%p ogf := EllipticK((4*sqrt(x))/(3*x + 1))/(2*Pi) - (1/4); ser := series(ogf, x, 22): seq(coeff(ser, x, n), n = 1..20); # _Peter Luschny_, Nov 10 2022

%t f[n_,k_]:=f[n,k]=Binomial[n-1,k]Binomial[n+k,k]Binomial[2k,k](k+2)(-3)^(n-1-k);

%t s[n_]:=a[n]=Sum[f[n,k],{k,0,n-1}]/(2n);

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

%Y Cf. A000984, A295113, A295132, A002426.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Nov 20 2017

%E Name simplified based on the proof of _Mark van Hoeij_ by _Peter Luschny_, Nov 10 2022