OFFSET
1,1
COMMENTS
Follows the integer values from 1 on the quadratic equation 5*x^2 + 4*n - 15, this is the case x=n.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
WolframAlpha, Table of 5n^2+4n-15
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
From Bruno Berselli, Mar 27 2014: (Start)
G.f.: -x*(6 - 31*x + 15*x^2)/(1 - x)^3.
a(n+1) - a(n) = A017377(n).
a(n) - a(-n) = A008590(n). (End)
EXAMPLE
For n=3, a(3) = 5*3^2 + 4*3 - 15 = 42; for n=6, a(6) = 5*6^2 + 4*6 - 15 = 189.
MATHEMATICA
Table[5 n^2 + 4 n - 15, {n, 50}]
CoefficientList[Series[(6 - 31 x + 15 x^2)/(x - 1)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Mar 29 2014 *)
PROG
(Magma) [5*n^2+4*n-15: n in [1..50]];
(PARI) a(n)=5*n^2+4*n-15 \\ Charles R Greathouse IV, Jun 17 2017
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Katherine Guo, Mar 26 2014
STATUS
approved