A243943 a(n) = A006442(n)^2.

1, 25, 1369, 93025, 6974881, 553425625, 45558768025, 3848757330625, 331434586569025, 28966516730025625, 2561512789823546329, 228690489716580520225, 20579914168308199841761, 1864413002713001259355225, 169871744046114667846619929, 15554069096581207471331850625

Offset: 0

Author

Paul D. Hanna, Aug 17 2014

Keywords

• nonn

Comments

In general, we have the binomial identity:

if b(n) = Sum_{k=0..n} t^k * C(2*k, k) * C(n+k, n-k),

then b(n)^2 = Sum_{k=0..n} (t^2+t)^k * C(2*k, k)^2 * C(n+k, n-k),

where the g.f. of b(n) is 1/sqrt(1 - (4*t+2)*x + x^2),

and the g.f. of b(n)^2 is 1 / AGM(1-x, sqrt((1+x)^2 - (4*t+2)^2*x)), where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) is the arithmetic-geometric mean.

Note that the g.f. of A006442 is 1/sqrt(1 - 10*x + x^2).

Limit_{n -> oo} a(n+1)/a(n) = (5 + 2*sqrt(6))^2 = 49 + 20*sqrt(6).

Examples

			G.f.: A(x) = 1 + 9*x + 169*x^2 + 3969*x^3 + 103041*x^4 + 2832489*x^5 +...
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..500

Crossrefs

Sequences of the form LegendreP(n, 2*m+1)^2: A000012 (m=0), A243949 (m=1), this sequence (m=2), A243944 (m=3), A243007 (m=4).

Cf. A006442.

Programs

Magma

[Evaluate(LegendrePolynomial(n), 5)^2 : n in [0..40]]; // G. C. Greubel, May 17 2023

Mathematica

Table[Sum[6^k * Binomial[2*k, k]^2 * Binomial[n+k, n-k], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Sep 28 2019 *)
LegendreP[Range[0,40], 5]^2 (* G. C. Greubel, May 17 2023 *)

PARI

{a(n) = sum(k=0, n, 6^k * binomial(2*k, k)^2 * binomial(n+k, n-k) )}
for(n=0, 20, print1(a(n), ", "))

PARI

{a(n) = sum(k=0, n, 2^k * binomial(2*k, k) * binomial(n+k, n-k) )^2}
for(n=0, 20, print1(a(n), ", "))

PARI

/* Using AGM: */
{a(n)=polcoeff( 1 / agm(1-x, sqrt((1+x)^2 - 10^2*x +x*O(x^n))), n)}
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 30 2014

SageMath

[gen_legendre_P(n,0,5)^2 for n in range(41)] # G. C. Greubel, May 17 2023

Formula

G.f.: 1 / AGM(1-x, sqrt(1-98*x+x^2)). - Paul D. Hanna, Aug 30 2014

a(n) = Sum_{k=0..n} 6^k * C(2*k, k)^2 * C(n+k, n-k).

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

a(n) ~ (5+2*sqrt(6))^(2*n+1) / (4*Pi*sqrt(6)*n). - Vaclav Kotesovec, Sep 28 2019