A102902 a(n) = 9*a(n-1) - 16*a(n-2), with a(0) = 1, a(1) = 9.
1, 9, 65, 441, 2929, 19305, 126881, 833049, 5467345, 35877321, 235418369, 1544728185, 10135859761, 66507086889, 436390025825, 2863396842201, 18788331166609, 123280631024265, 808912380552641, 5307721328585529
Offset: 0
Links
- Indranil Ghosh, Table of n, a(n) for n = 0..1221
- R. Flórez, R. A. Higuita, and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5, Journal of Integer Sequences, Vol. 17 (2014).
- Index entries for linear recurrences with constant coefficients, signature (9,-16).
Programs
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Magma
[4^n*Evaluate(ChebyshevSecond(n+1), 9/8): n in [0..30]]; // G. C. Greubel, Dec 09 2022
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Mathematica
LinearRecurrence[{9,-16},{1,9},20] (* Harvey P. Dale, Jul 28 2016 *) -
SageMath
[lucas_number1(n,9,16) for n in range(1, 21)] # Zerinvary Lajos, Apr 23 2009
Formula
G.f.: 1/(1-9*x+16*x^2).
a(n) = Sum_{k=0..n} binomial(2*n-k+1, k)*4^k.
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-16)^k*9^(n-2*k).
a(n) = 4^n * ChebyshevU(n, 9/8). - G. C. Greubel, Dec 09 2022
From Peter Bala, Jul 23 2025: (Start)
a(n) := ((9 + sqrt(17))^(n+1) - (9 - sqrt(17))^(n+1))/(2^(n+1)*sqrt(17)).
The following products telescope:
Product_{k >= 1} 1 + 4^k/a(k) = (1 + sqrt(17))/2.
Product_{k >= 1} 1 - 4^k/a(k) = (1 + sqrt(17))/18.
Product_{k >= 1} 1 + (-4)^k/a(k) = (17 + sqrt(17))/34.
Product_{k >= 1} 1 - (-4)^k/a(k) = (17 + sqrt(17))/18. (End)