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A291005
p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 2 S - 2 S^3.
2
2, 6, 20, 68, 230, 774, 2600, 8732, 29330, 98526, 330980, 1111868, 3735110, 12547374, 42150440, 141596132, 475664450, 1597901526, 5367837140, 18032197268, 60575633990, 203491974774, 683591422280, 2296391457932, 7714277207570, 25914602943726, 87055031555300
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.: 2*(1 - 2*x + 2*x^2)/(1 - 5*x + 7*x^2 - 5*x^3).
a(n) = 5*a(n-1) - 7*a(n-2) + 5*a(n-3) for n >= 4.
a(n) = 2*A291337(n) for n >= 0.
MATHEMATICA
z = 60; s = 1 - 2 s - 2 s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291005 *)
u / 2 (* A291337 *)
LinearRecurrence[{5, -7, 5}, {2, 6, 20}, 30] (* Vincenzo Librandi, Aug 27 2017 *)
PROG
(Magma) I:=[2, 6, 20]; [n le 3 select I[n] else 5*Self(n-1)-7*Self(n-2)+5*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Aug 27 2017
(SageMath)
def A291005_list(prec):
P.<x> = PowerSeriesRing(ZZ, prec)
return P( 2*(1-2*x+2*x^2)/(1-5*x+7*x^2-5*x^3) ).list()
A291005_list(30) # G. C. Greubel, Jun 01 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 23 2017
STATUS
approved