OFFSET
1,1
COMMENTS
The identity (2048*n^2+128*n+1)^2 - (16*n^2+n)*(512*n+16)^2 = 1 can be written as A157476(n)^2 - a(n)*A157475(n)^2 = 1 (see also second comment in A157476).
Sequence found by reading the line from 17, in the direction 17, 66,... in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Nov 02 2012
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
a(n) = A173511(2*n). - Reinhard Zumkeller, Feb 20 2010
a(1)=17, a(2)=66, a(3)=147, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Nov 08 2011
G.f.: x*(17 + 14*x + 3*x^2 - 3*x^3 + x^4)/(1-x)^3. - Vincenzo Librandi, Jan 01 2015
MATHEMATICA
Table[16n^2+n, {n, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {17, 66, 147}, 50] (* Harvey P. Dale, Nov 08 2011 *)
CoefficientList[Series[(17 + 14 x + 3 x^2 - 3 x^3 + x^4) / (1-x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Jan 01 2015 *)
PROG
(PARI) a(n)=16*n^2+n \\ Charles R Greathouse IV, Feb 09 2012
(Magma) [16*n^2 + n: n in [1..40]]; // Vincenzo Librandi, Jan 01 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 01 2009
EXTENSIONS
Comment rewritten by Bruno Berselli, Aug 22 2011
STATUS
approved