OFFSET
0,2
COMMENTS
Multiply the square of each term by 5 to form a bisection of A243945.
Limit_{n->oo} a(n+1)/a(n) = 9 + 4*sqrt(5).
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..500
FORMULA
a(n)^2 = (1/5) * Sum_{k=0..2*n+1} C(2*k, k)^2 * C(2*n+k+1, 2*n-k+1).
a(n) ~ (9+4*sqrt(5))^(n+1) / (2*5^(1/4)*sqrt(2*Pi*n) * sqrt(5+2*sqrt(5))). - Vaclav Kotesovec, Aug 18 2014. Equivalently, a(n) ~ phi^(6*n + 9/2) / (2^(3/2) * sqrt(5*Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021
From Peter Bala, Mar 15 2018: (Start)
a(n) = (1/sqrt(5))*P(2*n+1,sqrt(5)), where P(n,x) denotes the n-th Legendre polynomial. See A008316.
a(n) = Sum_{k = 0..n} (1/2)*C(2*k+1,k)*C(n,k)*C(2*n+2*k+2,2*n+1)/C(n+k+1,n). In general, (1/sqrt(1 + 4*x))*P(2*n+1,sqrt(1+4*x)) = (1/(2*C(2*n+1,n))) * Sum_{k = 0..n} C(n,k)*C(n+k+1,k)*C(2*n+2*k+2,n+k+1)*x^k.
a(n) = (1/sqrt(5))*Sum_{k = 0..2*n+1} C(2*n+1,k)^2 * phi^(2*n-2*k+1), where phi = (sqrt(5) + 1)/2.
a(n) = (1/sqrt(5))*Sum_{k = 0..2*n+1} C(2*n+1,k)*C(2*n+1+k,k) * Phi^k, where Phi = (sqrt(5) - 1)/2. (End)
a(n) = hypergeom([-n, n + 3/2], [1], -4). - Peter Luschny, Mar 16 2018
From Peter Bala, Mar 17 2018: (Start)
a(n) = Sum_{k = 0..n} C(2*n+1,2*k)*C(2*k,k)*5^(n-k).
D-finite with recurrence: n*(4*n-3)*(2*n+1)*a(n) = (4*n-1)*(36*n^2-18*n-7)*a(n-1) - (n-1)*(2*n-1)*(4*n+1)*a(n-2). (End)
EXAMPLE
G.f.: A(x) = 1 + 11*x + 155*x^2 + 2365*x^3 + 37555*x^4 + 610897*x^5 + ...
where
A(x)^2 = (1+x - sqrt(1-18*x+x^2)) / (10*x*(1-18*x+x^2)).
MAPLE
seq(add(1/2*binomial(2*k+1, k)*binomial(n, k)*binomial(2*n+2*k+2, 2*n+1)/binomial(n+k+1, n), k = 0..n), n = 0..20); # Peter Bala, Mar 15 2018
MATHEMATICA
CoefficientList[Series[Sqrt[((1+x-Sqrt[1-18x+x^2]))/(10x(1-18x+x^2))], {x, 0, 20}], x] (* Harvey P. Dale, Jul 31 2016 *)
a[n_] := Hypergeometric2F1[-n, n + 3/2, 1, -4];
Table[a[n], {n, 0, 18}] (* Peter Luschny, Mar 16 2018 *)
PROG
(PARI) /* From definition: */
{a(n)=polcoeff( sqrt( (1+x - sqrt(1-18*x+x^2 +x^2*O(x^n))) / (10*x*(1-18*x+x^2 +x*O(x^n))) ), n)}
for(n=0, 20, print1(a(n), ", "))
(PARI) /* From a(n) = sqrt( A243945(2*n+1)/5 ): */
{a(n)=sqrtint( (1/5)*sum(k=0, 2*n+1, binomial(2*k, k)^2*binomial(2*n+k+1, 2*n-k+1)) )}
for(n=0, 20, print1(a(n), ", "))
(Python)
from math import comb
def A243947(n): return sum(5**(n-k)*comb(m:=k<<1, k)*comb((n<<1)+1, m) for k in range(n+1)) # Chai Wah Wu, Mar 23 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul D. Hanna, Aug 17 2014
STATUS
approved