OFFSET
0,1
COMMENTS
The identity (57122*n^2 +47320*n +9801)^2 - (169*n^2 +140*n +29)*(4394*n +1820)^2 = 1 can be written as A156735(n)^2 - a(n)*A156636(n)^2 = 1.
The continued fraction expansion of sqrt(a(n)) is [13n+5; {2, 1, 1, 2, 26n+10}]. - Magus K. Chu, Sep 15 2022
From Klaus Purath, Apr 06 2025: (Start)
a(n)*13^2-1 is a square, and a(n) is the sum of two squares (see FORMULA). There are no squares in this sequence. The odd prime factors of these terms are always of the form 4*k + 1.
All a(n) = D satisfy the Pell equation (k*x)^2 - D*(13*y)^2 = -1 for any integer n where a(1-n) = A156639(n). The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*13^2 - 1), x(0) = 1, x(1) = 4*D*13^2 - 1, y(0) = 1, y(1) = 4*D*13^2 - 3. The two recurrences are of the form (4*D*13^2 - 2, -1).
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3) for n>2.
G.f.: (29 + 251*x + 58*x^2)/(1-x)^3. - Vincenzo Librandi, May 03 2014
E.g.f.: (29 +309*x +169*x^2)*exp(x). - G. C. Greubel, Feb 28 2021
From Klaus Purath, Apr 06 2025: (Start)
a(n) = (5*n + 2)^2 + (12*n + 5)^2 for any integer n.
169*a(n) - 1 = (169*n + 70)^2 for any integer n. (End)
MAPLE
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {29, 338, 985}, 50]
CoefficientList[Series[(29 +251x +58x^2)/(1-x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, May 03 2014 *)
PROG
(Magma) I:=[29, 338, 985]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]];
(PARI) a(n)=169*n^2+140*n+29 \\ Charles R Greathouse IV, Dec 23 2011
(SageMath) [169*n^2 + 140*n + 29 for n in (0..50)] # G. C. Greubel, Feb 28 2021
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Feb 15 2009
EXTENSIONS
Edited by Charles R Greathouse IV, Jul 25 2010
STATUS
approved