|
|
A292327
|
|
p-INVERT of the Fibonacci sequence (A000045), where p(S) = (1 - S)^2.
|
|
2
|
|
|
2, 5, 14, 38, 102, 271, 714, 1868, 4858, 12569, 32374, 83058, 212350, 541219, 1375570, 3487384, 8821170, 22266413, 56098206, 141087934, 354268502, 888238903, 2223968666, 5561234916, 13889778218, 34652529473, 86361653126, 215021205770, 534861620718
(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
|
|
|
FORMULA
|
G.f.: -(2 + x) *(-1 + 2 *x))/(-1 + 2 x + x^2)^2.
a(n) = 4*a(n-1) - 2*a(n-2) - 4*a(n-3) - a(n-4) for n >= 5.
|
|
MATHEMATICA
|
z = 60; s = x/(1 - x - x^2); p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292327 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|