|
|
A124161
|
|
a(n) = n*(n-1)*(n^3 + 21*n^2 - 4*n + 96)/120.
|
|
1
|
|
|
0, 0, 3, 15, 48, 121, 261, 504, 896, 1494, 2367, 3597, 5280, 7527, 10465, 14238, 19008, 24956, 32283, 41211, 51984, 64869, 80157, 98164, 119232, 143730, 172055, 204633, 241920, 284403, 332601, 387066, 448384, 517176, 594099, 679847, 775152, 880785, 997557
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
Arises in studying the Goldbach conjecture.
|
|
LINKS
|
|
|
FORMULA
|
G.f.: x^2*(3-3*x+3*x^2-2*x^3)/(1-x)^6. - Matthew House, Jan 16 2017
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>5. - Colin Barker, Jan 16 2017
|
|
MATHEMATICA
|
Table[n(n-1)(n^3+21n^2-4n+96)/120, {n, 0, 50}] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 0, 3, 15, 48, 121}, 50] (* Harvey P. Dale, Nov 26 2022 *)
|
|
PROG
|
(PARI) concat(vector(2), Vec(x^2*(3-3*x+3*x^2-2*x^3) / (1-x)^6 + O(x^40))) \\ Colin Barker, Jan 16 2017
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|