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A291008
p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 7*S^2.
3
0, 7, 14, 70, 224, 868, 3080, 11368, 41216, 150640, 548576, 2000992, 7293440, 26592832, 96946304, 353449600, 1288577024, 4697851648, 17127165440, 62441440768, 227645874176, 829940392960, 3025756030976, 11031154419712, 40216845025280, 146620616568832
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 A291000 for a guide to related sequences.
FORMULA
G.f.: 7*x/(1 - 2*x - 6*x^2).
a(n) = 2*a(n-1) + 6*a(n-2) for n >= 3.
a(n) = 7*A083099(n).
a(n) = (sqrt(7)*((1+sqrt(7))^n - (1-sqrt(7))^n)) / 2. - Colin Barker, Aug 23 2017
a(n) = 7*i^(n-1)*6^((n-1)/2)*ChebyshevU(n-1, -i/sqrt(6)). - G. C. Greubel, Jun 01 2023
MATHEMATICA
z = 60; s = x/(1 - x); p = 1 - s^7;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291008 *)
LinearRecurrence[{2, 6}, {0, 7}, 40] (* G. C. Greubel, Jun 01 2023 *)
PROG
(Magma) [n le 2 select 7*(n-1) else 2*Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jun 01 2023
(SageMath)
A291008=BinaryRecurrenceSequence(2, 6, 0, 7)
[A291008(n) for n in range(41)] # G. C. Greubel, Jun 01 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 22 2017
STATUS
approved