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A097310
Chebyshev T-polynomials T(n,14) with Diophantine property.
5
1, 14, 391, 10934, 305761, 8550374, 239104711, 6686381534, 186979578241, 5228741809214, 146217791079751, 4088869408423814, 114342125644787041, 3197490648645613334, 89415396036432386311, 2500433598371461203374, 69922725358364481308161, 1955335876435834015425134
OFFSET
0,2
COMMENTS
a(n)^2 - 195 b(n)^2 = +1 with b(n):=A097311(n) gives all nonnegative solutions of this Pell equation.
a(195+390k)-1 and a(195+390k)+1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006
Except for the first term, positive values of x (or y) satisfying x^2 - 28xy + y^2 + 195 = 0. - Colin Barker, Feb 23 2014
FORMULA
a(n) = 28*a(n-1) - a(n-2), a(-1):= 14, a(0)=1.
a(n) = T(n, 14)= (S(n, 28)-S(n-2, 28))/2 = S(n, 28)-14*S(n-1, 28) with T(n, x), resp. S(n, x), Chebyshev's polynomials of the first, resp.second, kind. See A053120 and A049310. S(n, 28)=A097311(n).
a(n) = (ap^n + am^n)/2 with ap := 14+sqrt(195) and am := 14-sqrt(195).
a(n) = sum(((-1)^k)*(n/(2*(n-k)))*binomial(n-k, k)*(2*14)^(n-2*k), k=0..floor(n/2)), n>=1.
G.f.: (1-14*x)/(1-28*x+x^2).
a(n) = sqrt(1 + 195*A097311(n)^2), n>=0.
MATHEMATICA
LinearRecurrence[{28, -1}, {1, 14}, 20] (* Harvey P. Dale, Jan 29 2014 *)
CoefficientList[Series[(1 - 14 x)/(1 - 28 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 24 2014 *)
PROG
(Sage) [lucas_number2(n, 28, 1)/2 for n in range(0, 16)] - Zerinvary Lajos, Jun 27 2008
(PARI) Vec((1-14*x)/(1-28*x+x^2) + O(x^100)) \\ Colin Barker, Feb 23 2014
CROSSREFS
Sequence in context: A000473 A233094 A211421 * A223003 A228185 A041367
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Aug 31 2004
EXTENSIONS
More terms from Colin Barker, Feb 23 2014
STATUS
approved