|
| |
|
|
A129863
|
|
Sums of three consecutive pentagonal numbers.
|
|
4
| |
|
|
6, 18, 39, 69, 108, 156, 213, 279, 354, 438, 531, 633, 744, 864, 993, 1131, 1278, 1434, 1599, 1773, 1956, 2148, 2349, 2559, 2778, 3006, 3243, 3489, 3744, 4008, 4281, 4563, 4854, 5154, 5463, 5781, 6108, 6444, 6789, 7143, 7506, 7878, 8259, 8649, 9048, 9456
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,1
|
|
|
COMMENTS
| Arises in pentagonal number analogue to A129803 Triangular numbers which are the sum of three consecutive triangular numbers. What are the pentagonal numbers which are the sum of three consecutive pentagonal numbers?
|
|
|
FORMULA
| a(n) = P(n) + P(n+1) + P(n+2) where P(n) = A000326(n) = n(3n-1)/2.
a(n) = P(n) + P(n+1) + P(n+2) where P(n) = A000326(n) = n(3n-1)/2. For n = 0, 1, 2, ... a(n+1) = (9/2)*(n^2) + (15/2)*n + 6.
a(n) = (3n^2+5n+4)*(3/2) - Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), May 27 2007
G.f.: 3*x*(2+x^2)/(1-x)^3. - Colin Barker, Feb 13 2012
|
|
|
EXAMPLE
| a(1) = 6 = A000326(0) + A000326(1) + A000326(2) = 0 + 1 + 5.
a(2) = 18 = A000326(1) + A000326(2) + A000326(3) = 1 + 5 + 12.
|
|
|
MATHEMATICA
| Table[(3/2)*(4 + 5*n + 3*n^2), {n, 0, 100}] - Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), May 27 2007
|
|
|
CROSSREFS
| Cf. A000326, A007667, A034961, A129803.
Sequence in context: A132432 A005899 A180118 * A035489 A122061 A002411
Adjacent sequences: A129860 A129861 A129862 * A129864 A129865 A129866
|
|
|
KEYWORD
| easy,nonn,changed
|
|
|
AUTHOR
| Jonathan Vos Post (jvospost3(AT)gmail.com), May 23 2007, May 24 2007
|
| |
|
|