A158557 a(n) = 225*n^2 + 15.
15, 240, 915, 2040, 3615, 5640, 8115, 11040, 14415, 18240, 22515, 27240, 32415, 38040, 44115, 50640, 57615, 65040, 72915, 81240, 90015, 99240, 108915, 119040, 129615, 140640, 152115, 164040, 176415, 189240, 202515, 216240, 230415, 245040, 260115, 275640, 291615
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..10000
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Programs
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Magma
I:=[15, 240, 915]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 14 2012
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Mathematica
LinearRecurrence[{3, -3, 1}, {15, 240, 915}, 50] (* Vincenzo Librandi, Feb 14 2012 *) 225*Range[0,40]^2+15 (* Harvey P. Dale, Apr 06 2019 *)
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PARI
for(n=0, 22, print1(225*n^2 + 15", ")); \\ Vincenzo Librandi, Feb 14 2012
Formula
G.f.: 15*(1 + 13*x + 16*x^2)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 09 2023: (Start)
Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(15))*Pi/sqrt(15) + 1)/30.
Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(15))*Pi/sqrt(15) + 1)/30. (End)
E.g.f.: 15*exp(x)*(1 + 15*x + 15*x^2). - Elmo R. Oliveira, Jan 15 2025
Extensions
Comment rewritten, a(0) added by R. J. Mathar, Oct 16 2009
Comments