|
Search: id:A147559
|
|
|
| A147559 |
|
Result of using the perfect squares 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(3x^3)... |
|
+0
5
|
|
| 1, 4, 5, 11, -6, -22, -4, 155, 16, -182, -158, 376, 56, -1456, 680, 23155, -4966, -28674, 6132, 117946, 15792, -415426, -162814, 512550, 333904, -4231332, 235968, 15171332, -5259270, -68578566, 15199212, 736983115, -4403208, -1097465342
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
EXAMPLE
|
From the perfect squares, construct the series 1+x+4x^2+9x^3+16x^4+25x^5+... 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)=4. Then divide this quotient by (1+a(2)x^2), i.e. here (1+4x^2), to get (1+a(3)x^3+...), giving a(3)=5.
|
|
CROSSREFS
|
Cf. A000290
Sequence in context: A052508 A074098 A126069 this_sequence A007429 A064945 A069820
Adjacent sequences: A147556 A147557 A147558 this_sequence A147560 A147561 A147562
|
|
KEYWORD
|
sign
|
|
AUTHOR
|
N. Fernandez (primeness(AT)borve.org), Nov 07 2008
|
|
EXTENSIONS
|
Terms from a(11) on corrected by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 11 2008
|
|
|
Search completed in 0.002 seconds
|
