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A099930
a(n) = Pell(n)*Pell(n-1)*Pell(n-2) / 10.
3
1, 12, 174, 2436, 34307, 482664, 6791772, 95567064, 1344731653, 18921807828, 266250046986, 3746422451772, 52716164405255, 741772724044560, 10437534301224120, 146867252940711408, 2066579075472320521
OFFSET
3,2
LINKS
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006.
N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.
FORMULA
G.f.: x^3 / ((1+2*x-x^2)*(1-14*x-x^2)).
a(n) = 12*a(n-1) + 30*a(n-2) - 12*a(n-3) - a(n-4); a(3)=1, a(4)=12, a(5)=174, a(6)=2436. - Harvey P. Dale, Feb 26 2012
From Peter Bala, Mar 30 2015: (Start)
The following remarks assume an offset of 0.
The o.g.f. A(x) = 1/( (1 + 2*x - x^2)*(1 - 14*x - x^2) ). Hence A(x) (mod 4) = 1/(1 + 2*x - x^2)^2 (mod 4). It follows by Theorem 1 of Heninger et al. that sqrt(A(x)) = 1 + 6*x + 69*x^2 + 804*x^3 + ... has integral coefficients.
Sum_{n >= 0} a(n)*x^n = exp( Sum_{n >= 1} Pell(4*n)/Pell(n)*x^n/n ). Cf. A001656, A084175. (End)
a(n+1) = (1/2)*Sum_{k=1..n-1} ( A014445(k)*A110272(n-k) ) for n > 1. - Michael A. Allen, Jan 25 2023
MATHEMATICA
Drop[CoefficientList[Series[x^3/((1+2x-x^2)(1-14x-x^2)), {x, 0, 20}], x], 3] (* or *) LinearRecurrence[{12, 30, -12, -1}, {1, 12, 174, 2436}, 20] (* Harvey P. Dale, Feb 26 2012 *)
CROSSREFS
Cf. A000129. Third column of triangle A099927. Cf. A001656, A084175.
Sequence in context: A239335 A120662 A230815 * A307549 A052208 A234531
KEYWORD
nonn,easy
AUTHOR
Ralf Stephan, Nov 03 2004
STATUS
approved