OFFSET
1,1
COMMENTS
The identity (100*n+1)^2-(100*n^2+2*n)*10^2 = 1 can be written as a(n)^2-A158127(n)*10^2 = 1. - Vincenzo Librandi, Feb 11 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
E. J. Barbeau, Polynomial Excursions, Chapter 10: Diophantine equations (2010), pages 84-85 (row 15 in the first table at p. 85, case d(t) = t*(10^2*t+2)).
Index entries for linear recurrences with constant coefficients, signature (2,-1).
FORMULA
G.f.: x*(101-x)/(1-x)^2. - Vincenzo Librandi, Feb 11 2012
a(n) = 2*a(n-1)-a(n-2). - Vincenzo Librandi, Feb 11 2012
MATHEMATICA
LinearRecurrence[{2, -1}, {101, 201}, 50] (* Vincenzo Librandi, Feb 11 2012 *)
100*Range[50]+1 (* Harvey P. Dale, Apr 12 2024 *)
PROG
(Magma) I:=[101, 201]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; \\ Vincenzo Librandi, Feb 11 2012
(PARI) for(n=1, 40, print1(100*n + 1", ")); \\ Vincenzo Librandi, Feb 11 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 13 2009
STATUS
approved