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A291010
p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - 2*S)*(1 - 3*S).
2
5, 24, 108, 468, 1980, 8244, 33948, 138708, 563580, 2280564, 9200988, 37040148, 148869180, 597602484, 2396787228, 9606280788, 38482518780, 154102262004, 616925608668, 2469252116628, 9881657512380, 39540577187124, 158204150161308, 632942124883668
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 A291000 for a guide to related sequences.
FORMULA
G.f.: (5 - 11*x)/(1 - 7*x + 12*x^2).
a(n) = 7*a(n-1) - 12*a(n-2) for n >= 3.
a(n) = 9*4^n - 4*3^n. - Colin Barker, Aug 23 2017
E.g.f.: 9*exp(4*x) - 4*exp(3*x). - G. C. Greubel, Jun 04 2023
MATHEMATICA
z = 60; s = x/(1-x); p = (1-2s)(1-3s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* this sequence *)
LinearRecurrence[{7, -12}, {5, 24}, 40] (* G. C. Greubel, Jun 04 2023 *)
PROG
(PARI) Vec((5-11*x)/((1-3*x)*(1-4*x)) + O(x^30)) \\ Colin Barker, Aug 23 2017
(Magma) [36*(4^(n-1)-3^(n-2)): n in [0..40]]; // G. C. Greubel, Jun 04 2023
(SageMath)
A291010=BinaryRecurrenceSequence(7, -12, 5, 24)
[A291010(n) for n in range(41)] # G. C. Greubel, Jun 04 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 23 2017
STATUS
approved