A295132 -id:A295132 - cp's OEIS frontend

1, 3, 19, 127, 921, 6921, 53523, 422199, 3382417, 27429043, 224636259, 1854761437, 15419579761, 128941830993, 1083686483259, 9147887134119, 77520233226537, 659167237928691, 5622149927918763, 48083938099637247

Offset: 1

Zhi-Wei Sun, Nov 20 2017

nonn

From Mark van Hoeij, Nov 10 2022: (Start)
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.
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.
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)

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..200
Heba Bou KaedBey, Mark van Hoeij, and Man Cheung Tsui, Solving Third Order Linear Difference Equations in Terms of Second Order Equations, arXiv:2402.11121 [math.AC], 2024. See p. 6.

Cf. A000984, A295113, A295132, A002426.

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

f[n_,k_]:=f[n,k]=Binomial[n-1,k]Binomial[n+k,k]Binomial[2k,k](k+2)(-3)^(n-1-k);
s[n_]:=a[n]=Sum[f[n,k],{k,0,n-1}]/(2n);
Table[s[n],{n,1,20}]

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).
From Mark van Hoeij, Nov 10 2022: (Start)
G.f.: (hypergeom([1/2, 1/2], [1], 16*x/(1 + 3*x)^2) - 1)/4.
a(n) = (A002426(n)^2 + 3 * A002426(n-1)^2)/4. (End)
G.f.: EllipticK((4*sqrt(x))/(3*x + 1))/(2*Pi) - (1/4). - Peter Luschny, Nov 10 2022

Name simplified based on the proof of Mark van Hoeij by Peter Luschny, Nov 10 2022

cp's OEIS Frontend

A295371 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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions