|
Search: id:A147654
|
|
|
| A147654 |
|
Result of using the positive integers 1,2,3,... as coefficients in an infinite polynomial series in x and then expressing this series as (1+a(1)x)(1+a(2)x^2)(1+a(3)x^3) |
|
+0
5
|
|
| 1, 2, 1, 3, 0, -2, 0, 9, 0, -6, 0, 4, 0, -18, 0, 93, 0, -54, 0, 72, 0, -186, 0, 232
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
EXAMPLE
|
From the positive integers 1,2,3,..., construct the series 1+x+2x^2+3x^3+4x^4+... a(1) is always the coefficient of x, here 1. Divide by (1+a(1)x), i.e. here (1+x), to get the quotient (1+a(2)x^2+...), which here gives a(2)=2. Then divide this quotient by (1+a(2)x^2), i.e. here (1+2x^2), to get (1+a(3)x^3+...), giving a(3)=1.
|
|
CROSSREFS
|
Cf. A147541
Sequence in context: A137639 A113288 A035215 this_sequence A071467 A125073 A071461
Adjacent sequences: A147651 A147652 A147653 this_sequence A147655 A147656 A147657
|
|
KEYWORD
|
sign
|
|
AUTHOR
|
N. Fernandez (primeness(AT)borve.org), Nov 09 2008
|
|
|
Search completed in 0.002 seconds
|
