OFFSET
0,1
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
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).
FORMULA
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
EXAMPLE
23 = a(2) = sqrt(8*A077241(2)^2 + 17) = sqrt(8*8^2 + 17)= sqrt(529) = 23.
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 *)
PROG
(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
(PARI) Vec((1-x)*(5+12*x+5*x^2)/(1-6*x^2+x^4) + O(x^50)) \\ Colin Barker, Mar 27 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Nov 08 2002
STATUS
approved