A077242 Combined Diophantine Chebyshev sequences A077240 and A077239.

5, 7, 23, 37, 133, 215, 775, 1253, 4517, 7303, 26327, 42565, 153445, 248087, 894343, 1445957, 5212613, 8427655, 30381335, 49119973, 177075397, 286292183, 1032071047, 1668633125, 6015350885, 9725506567, 35060034263, 56684406277, 204344854693, 330380931095

Offset: 0

Author

Wolfdieter Lang, Nov 08 2002

Keywords

• nonn
• easy

Comments

a(n)^2 - 8*b(n)^2 = 17, with the companion sequence b(n)= A077241(n).

Because there is only one primitive Pythagorean triangle with sum of the legs L = 17 (see also A120681), namely (5,12,13), all positive solutions (x(n), y(n)) = (a(n), 2*A077241(n)) of the (generalized) Pell equation x^2 - 2*y^2 = +17 satisfy x(n) < 2*y(n), for n >= 1, only 5 = x(0) > 2*y(0) = 4. The proof runs along the same line as the one given in a comment on the L=7 case in A077443. - Wolfdieter Lang, Feb 05 2015

Examples

			23 = a(2) = sqrt(8*A077241(2)^2 + 17) = sqrt(8*8^2 + 17)= sqrt(529) = 23.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Index entries for sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Programs

Magma

I:=[5,7,23,37]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Feb 18 2014

Mathematica

A077239 = Table[2*ChebyshevT[n+1, 3] + ChebyshevT[n, 3], {n, 0, 12}]; A077240 = Table[ChebyshevT[n+1, 3] + 2*ChebyshevT[n, 3], {n, 0, 12}]; Riffle[A077240, A077239] (* Jean-François Alcover, Dec 19 2013 *)
CoefficientList[Series[(1 - x) (5 + 12 x + 5 x^2)/(1 - 6 x^2 + x^4), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 18 2014 *)

PARI

Vec((1-x)*(5+12*x+5*x^2)/(1-6*x^2+x^4) + O(x^50)) \\ Colin Barker, Mar 27 2016

Formula

a(2*k) = A077240(k) and a(2*k+1) = A077239(k), k>=0.

G.f.: (1-x)*(5+12*x+5*x^2)/(1-6*x^2+x^4).

a(n) = 6*a(n-2)-a(n-4) for n>3. - Vincenzo Librandi, Feb 18 2014

a(n) = ((6-5*sqrt(2))*(1-sqrt(2))^n - (-1-sqrt(2))^n*(-4+sqrt(2)) + 4*(-1+sqrt(2))^n + sqrt(2)*(-1+sqrt(2))^n + 6*(1+sqrt(2))^n + 5*sqrt(2)*(1+sqrt(2))^n)/4. - Colin Barker, Mar 27 2016