OFFSET
0,1
COMMENTS
The identity (1250*n^2 - 1800*n + 649)^2 - (25*n^2 - 36*n + 13)*(250*n - 180)^2 = 1 can be written as A154358(n)^2 - a(n)*A154360(n)^2 = 1. See also the third comment in A154357.
Numbers of the form (3n-2)^2 + (4n-3)^2. - Bruno Berselli, Dec 12 2011
From Klaus Purath, May 06 2025: (Start)
25*a(n)-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*(5*y)^2 = -1 for any integer n where a(1-n) = A154357(n). The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*5^2 - 1), x(0) = 1, x(1) = 4*D*5^2 - 1, y(0) = 1, y(1) = 4*D*5^2 - 3. The two recurrences are of the form (4*D*5^2 - 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) = A007533(n-1), n>0. - R. J. Mathar, Jan 14 2009
G.f.: (13 - 37*x + 74*x^2) / (1-x)^3. - R. J. Mathar, Jan 05 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Feb 21 2012
E.g.f.: (13 - 11*x + 25*x^2) * exp(x). - G. C. Greubel, Sep 14 2016
From Klaus Purath, May 06 2025: (Start)
a(n) = (3*n-2)^2 + (4*n-3)^2.
25*a(n) - 1 = (25*n - 18)^2. (End)
MATHEMATICA
Table[25n^2-36n+13, {n, 0, 40}] (* Harvey P. Dale, Apr 02 2011 *)
LinearRecurrence[{3, -3, 1}, {13, 2, 41}, 50] (* Vincenzo Librandi, Feb 21 2012 *)
PROG
(PARI) for(n=0, 40, print1(25*n^2 - 36*n + 13", ")); \\ Vincenzo Librandi, Feb 21 2012
(Magma) [25*n^2-36*n+13: n in [0..40]]; // Bruno Berselli, Sep 15 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Jan 07 2009
EXTENSIONS
Offset corrected from R. J. Mathar, Jan 05 2011
First comment rewritten by Bruno Berselli, Dec 12 2011
STATUS
approved
