OFFSET
0,3
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).
Note that in A290890, s = (1,2,3,4,...); i.e., A000027(n+1) for n>=0, whereas in A290990, s = (0,1,2,3,4,...); i.e., A000027(n) for n>=0.
Guide to p-INVERT sequences using s = (1,2,3,4,5,...) = A000027:
p(S) t(1,2,3,4,5,...)
1 - S A001906
1 - S^3 A290891
1 - S^4 A290892
1 - S^5 A290893
1 - S^6 A290894
1 - S^7 A290895
1 - S^8 A290896
1 - S - S^2 A289780
1 - S - S^3 A290897
1 - S - S^4 A290898
1 - S^2 - S^4 A290899
1 - S^2 - S^3 A290900
1 - S^3 - S^4 A290901
(1 - S)^2 A290917
(1 - S)^3 A290918
(1 - S)^4 A290919
(1 - S)^5 A290920
(1 - S)^6 A290921
1 - S - 2*S^2 A290922
1 - 3*S - 2*S^2 A290925
(1 - S^2)^2 A290926
(1 - S^2)^3 A290927
(1 - S^3)^2 A290928
(1 - S)(1 - S^2) A290929
(1 - S^2)(1 - S^4) A290930
1 - 3 S + S^2 A291025
1 - 4 S + S^2 A291026
1 - 5 S + S^2 A291027
1 - 6 S + S^2 A291028
1 - S - S^2 - S^3 A291029
1 - S - S^2 - S^3 - S^4 A201030
1 - 3 S + 2 S^3 A291031
1 - S - S^2 - S^3 + S^4 A291032
1 - 6 S A291033
1 - 7 S A291034
1 - 8 S A291181
1 - 3 S + 2 S^3 A291031
1 - 3 S + 2 S^2 A291182
1 - 4 S + 2 S^3 A291183
1 - 4 S + 3 S^3 A291184
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, -5, 4, -1)
FORMULA
G.f.: x/(1 - 4 x + 5 x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) - 5*a(n-2) + 4*a(n-3) - a(n-4).
EXAMPLE
(See the examples at A289780.)
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 15 2017
STATUS
approved