A077420 Bisection of Chebyshev sequence T(n,3) (odd part) with Diophantine property.
1, 33, 1121, 38081, 1293633, 43945441, 1492851361, 50713000833, 1722749176961, 58522759015841, 1988051057361633, 67535213191279681, 2294209197446147521, 77935577499977736033, 2647515425801796877601
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Z. Cerin and G. M. Gianella, On sums of squares of Pell-Lucas Numbers, INTEGERS 6 (2006) #A15
- Tanya Khovanova, Recursive Sequences
- S. Vidhyalakshmi, V. Krithika, and K. Agalya, On The Negative Pell Equation y^2 = 72x^2 - 8, International Journal of Emerging Technologies in Engineering Research (IJETER) Volume 4, Issue 2, February (2016).
- Index entries for sequences related to Chebyshev polynomials.
- Index entries for linear recurrences with constant coefficients, signature (34,-1).
Crossrefs
Programs
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Magma
I:=[1,33]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011
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Mathematica
LinearRecurrence[{34,-1},{1,33},20] (* Vincenzo Librandi, Nov 22 2011 *) a[c_, n_] := Module[{}, p := Length[ContinuedFraction[ Sqrt[ c]][[2]]]; d := Denominator[Convergents[Sqrt[c], n p]]; t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}]; Return[t]; ] (* Complement of A041027 *) a[18, 20] (* Gerry Martens, Jun 07 2015 *)
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Maxima
makelist(expand(((1+sqrt(2))^(4*n+2)+(1-sqrt(2))^(4*n+2))/6),n,0,14); /* _Bruno Berselli, Nov 22 2011 */
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PARI
Vec((1-x)/(1-34*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Nov 22 2011
Formula
a(n) = 34*a(n-1) - a(n-2), a(-1)=1, a(0)=1.
a(n) = T(2*n+1, 3)/3 = S(n, 34) - S(n-1, 34), with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x)=0, S(n, 34)= A029547(n), T(n, 3)=A001541(n).
G.f.: (1-x)/(1-34*x+x^2).
a(n) = sqrt(8*A046176(n+1)^2 + 1)/3.
a(n) = (k^n)+(k^(-n))-a(n-1) = A003499(2*n)-a(n-1), where k = (sqrt(2)+1)^4 = 17+12*sqrt(2) and a(0)=1. - Charles L. Hohn, Apr 05 2011
a(n) = a(-n-1) = A029547(n)-A029547(n-1) = ((1+sqrt(2))^(4n+2)+(1-sqrt(2))^(4n+2))/6. - Bruno Berselli, Nov 22 2011
Comments