|
| |
|
|
A291031
|
|
p-INVERT of the positive integers, where p(S) = 1 - 3*S + 2*S^3.
|
|
2
|
|
|
|
3, 15, 70, 321, 1461, 6624, 29967, 135399, 611318, 2758881, 12447753, 56154744, 253306119, 1142572767, 5153589754, 23244956169, 104843981505, 472885383744, 2132882300571, 9620044596687, 43389716584682, 195702453488433, 882684641446989, 3981207177094608
(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).
See A290890 for a guide to related sequences.
|
|
|
LINKS
|
|
|
|
FORMULA
|
G.f.: (3 - 12 x + 16 x^2 - 12 x^3 + 3 x^4)/(1 - 9 x + 27 x^2 - 36 x^3 + 27 x^4 - 9 x^5 + x^6).
a(n) = 9*a(n-1) - 27*a(n-2) + 36*a(n-3) - 27*a(n-4) + 90*a(n-5) - a(n-6).
|
|
|
MATHEMATICA
|
z = 60; s = x/(1 - x)^2; p = 1 - 3 s + 2 s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291031 *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
