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A097731 Chebyshev U(n,x) polynomial evaluated at x=99 gives 2*7^2+1. 6
1, 198, 39203, 7761996, 1536836005, 304285766994, 60247045028807, 11928610629936792, 2361804657682456009, 467625393610496352990, 92587466130220595436011, 18331850668390067399977188, 3629613844875103124600047213, 718645209434602028603409370986 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Used to form integer solutions of Pell equation a^2 - 50*b^2 =-1. See A097732 with A097733.
LINKS
R. Flórez, R. A. Higuita, A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
Tanya Khovanova, Recursive Sequences
FORMULA
a(n) = 2*99*a(n-1) - a(n-2), n>=1, a(0)=1, a(-1):=0.
a(n) = S(n, 2*99)= U(n, 99), Chebyshev's polynomials of the second kind. See A049310.
G.f.: 1/(1-198*x+x^2).
a(n) = sum((-1)^k*binomial(n-k, k)*198^(n-2*k), k=0..floor(n/2)), n>=0.
a(n) = ((99+70*sqrt(2))^(n+1) - (99-70*sqrt(2))^(n+1))/(140*sqrt(2)), n>=0.
a(n) = Pell(6*n + 6)/Pell(6). Sum_{n >= 0} 1/( 14*a(n) + 1/(14*a(n)) ) = 1/14. - Peter Bala, Mar 25 2015
a(n) = A002965(12*(n+1))/70. - Gerry Martens, Jul 14 2023
MAPLE
with(combinat): seq(fibonacci(6*n+6, 2)/70, n=0..12); # Zerinvary Lajos, Apr 21 2008
MATHEMATICA
LinearRecurrence[{198, -1}, {1, 198}, 12] (* Ray Chandler, Aug 11 2015 *)
CROSSREFS
Cf. A002965.
Sequence in context: A348602 A270849 A324454 * A006243 A171391 A151738
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Aug 31 2004
STATUS
approved

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Last modified March 28 14:38 EDT 2024. Contains 371254 sequences. (Running on oeis4.)