A098443 Expansion of 1/sqrt(1-8*x-4*x^2).
1, 4, 26, 184, 1366, 10424, 80996, 637424, 5064166, 40528984, 326251276, 2638751504, 21426682876, 174563719984, 1426219233416, 11681133293024, 95877105146246, 788433553532824, 6494463369141116, 53576199709855184
Offset: 0
Examples
G.f. = 1 + 4*x + 26*x^2 + 184*x^3 + 1366*x^4 + 10424*x^5 + 80996*x^6 + ...
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
- Hacène Belbachir, Abdelghani Mehdaoui, László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
- Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.
Programs
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Mathematica
CoefficientList[Series[1/Sqrt[1 - 8*x - 4*x^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 15 2012, updated Mar 21 2024 *) -
PARI
x='x+O('x^66); Vec(1/sqrt(1-8*x-4*x^2)) \\ Joerg Arndt, May 11 2013
Formula
E.g.f.: exp(4*x) * BesselI(0, 2*sqrt(5)*x).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k) * binomial(2(n-k), n) * 2^(n-2k).
D-finite with recurrence: n*a(n) = 4*(2*n-1)*a(n-1) + 4*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 15 2012
a(n) ~ sqrt(50+20*sqrt(5))*(4+2*sqrt(5))^n/(10*sqrt(Pi*n)). Equivalently, a(n) ~ 2^(n-1/2) * phi^(3*n + 3/2) / (5^(1/4) * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Oct 15 2012, updated Mar 21 2024
G.f.: 1/(1 - 2*x*(2+x)*Q(0)), where Q(k)= 1 + (4*k+1)*x*(2+x)/(k+1 - x*(2+x)*(2*k+2)*(4*k+3)/(2*x*(2+x)*(4*k+3) + (2*k+3)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 15 2013
G.f.: Q(0), where Q(k) = 1 + 2*x*(x+2)*(4*k+1)/( 2*k+1 - x*(x+2)*(2*k+1)*(4*k+3)/(x*(x+2)*(4*k+3) + (k+1)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 16 2013
From Peter Bala, Mar 16 2024: (Start)
a(n) = (-2*i)^n * P(n, 2*i), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial.
Sum_{n >= 1} (-1)^(n+1)*4^n/(n*a(n-1)*a(n)) = 2*arctan(1/2) = 2*A073000. (End)
From Seiichi Manyama, Aug 29 2025: (Start)
a(n) = Sum_{k=0..n} (2-i)^k * (2+i)^(n-k) * binomial(n,k)^2, where i is the imaginary unit.
a(n) = Sum_{k=0..floor(n/2)} 5^k * 4^(n-2*k) * binomial(n,2*k) * binomial(2*k,k).
a(n) = [x^n] (1+4*x+5*x^2)^n. (End)
Comments