|
|
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
|
|
|
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
|
|
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
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|