OFFSET
1,1
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 A156721(n)^2 - a(n)*A156627(n)^2 = 1.
The continued fraction expansion of sqrt(a(n)) is [13n-6; {1, 1, 1, 1, 1, 1, 26n-12}]. - Magus K. Chu, Sep 06 2022
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
G.f.: x*(58 + 251*x + 29*x^2)/(1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {58, 425, 1130}, 40]
Table[169n^2-140n+29, {n, 40}] (* Harvey P. Dale, Mar 24 2023 *)
PROG
(Magma) I:=[58, 425, 1130]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..40]];
(PARI) a(n)=169*n^2-140*n+29 \\ Charles R Greathouse IV, Dec 23 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Feb 15 2009
EXTENSIONS
Edited by Charles R Greathouse IV, Jul 25 2010
STATUS
approved