|
|
A239796
|
|
a(n) = 7*n^2 + 2*n - 15.
|
|
1
|
|
|
-6, 17, 54, 105, 170, 249, 342, 449, 570, 705, 854, 1017, 1194, 1385, 1590, 1809, 2042, 2289, 2550, 2825, 3114, 3417, 3734, 4065, 4410, 4769, 5142, 5529, 5930, 6345, 6774, 7217, 7674, 8145, 8630, 9129, 9642, 10169, 10710, 11265, 11834, 12417, 13014, 13625, 14250, 14889, 15542, 16209, 16890
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Follows the integer values from 1 on the parabola: 7*n^2 + 2*n - 15.
The first in the family of parabolas of the form: prime(k+3)*n^2 + prime(k)*n - prime(k+1)*prime(k+2), where k >= 1 (k=1 gives a(n)). - Wesley Ivan Hurt, Mar 26 2014
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
For n=3, a(3) = 7*3^2 + 2*3 - 15 = 54; for n=6, a(6) = 7*6^2 + 2*6 - 15 = 249.
|
|
MAPLE
|
|
|
MATHEMATICA
|
CoefficientList[Series[(6 - 35 x + 15 x^2)/(x - 1)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Mar 29 2014 *)
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
sign,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|