|
| |
|
|
A292325
|
|
p-INVERT of (1,0,0,0,1,0,0,0,0,0,...), where p(S) = (1 - S)^2.
|
|
1
|
|
|
|
2, 3, 4, 5, 8, 13, 20, 29, 40, 56, 80, 115, 164, 230, 320, 445, 620, 864, 1200, 1660, 2290, 3155, 4344, 5975, 8206, 11252, 15408, 21078, 28810, 39344, 53680, 73173, 99662, 135640, 184480, 250740, 340578, 462316, 627200, 850420, 1152480, 1561043, 2113420
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,1
|
|
|
COMMENTS
|
Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
|
|
|
LINKS
|
Index entries for linear recurrences with constant coefficients, signature (2, -1, 0, 0, 2, -2, 0, 0, 0, -1)
|
|
|
FORMULA
|
G.f.: -(((-1 + x) (1 + x^4) (2 + x + x^2 + x^3 + x^4))/((1 - x + x^2)^2 (-1 + x^2 + x^3)^2)).
a(n) = 2*a(n-1) - a(n-2) + 2*a(n-5) - 2*a(n-6) - a(n-10) for n >= 11.
|
|
|
MATHEMATICA
|
z = 60; s = x + x^5; p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292325 *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
