|
|
A292534
|
|
p-INVERT of the squares (A000290), where p(S) = 1 + S - S^2.
|
|
1
|
|
|
-1, -2, 4, 21, 30, 11, 80, 622, 2055, 4584, 10711, 34354, 115480, 341213, 934750, 2640483, 7874188, 23564242, 68738591, 198108496, 575654335, 1688669686, 4951141372, 14443935957, 42064267934, 122731975243, 358682023576, 1047906654118, 3058580566407
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
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).
See A292479 for a guide to related sequences.
|
|
LINKS
|
|
|
FORMULA
|
G.f.: ((1 + x) (-1 + 4 x - 2 x^2 + x^3))/(1 - 5 x + 12 x^2 - 22 x^3 + 16 x^4 - 7 x^5 + x^6).
a(n) = 5*a(n-1) - 12*a(n-2) + 22*a(n-3) - 16*a(n-4) + 7*a(n-5) - a(n-6) for n >= 7.
|
|
MATHEMATICA
|
z = 60; s = x (x + 1)/(1 - x)^3; p = 1 + s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292534 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
easy,sign
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|