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A292403
p-INVERT of (1,0,0,0,0,1,0,0,0,0,0,0,...), where p(S) = 1 - S^2.
2
0, 1, 0, 1, 0, 1, 2, 1, 4, 1, 6, 2, 8, 7, 10, 16, 12, 29, 18, 46, 36, 67, 74, 93, 140, 136, 242, 224, 388, 401, 592, 727, 900, 1278, 1422, 2147, 2364, 3467, 4060, 5491, 7004, 8736, 11890, 14191, 19724, 23589, 32128, 39744, 51964, 66991, 84406, 111930, 138588
OFFSET
0,7
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).
LINKS
Index entries for linear recurrences with constant coefficients, signature (0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1)
FORMULA
G.f.: -((x (1 + x)^2 (1 - x + x^2 - x^3 + x^4)^2)/((-1 + x + x^6) (1 + x + x^6))).
a(n) = a(n-2) + 2*a(n-7) + a(n-12) for n >= 13.
MATHEMATICA
z = 60; s = x + x^4; p = 1 - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292403 *)
LinearRecurrence[{0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1}, {0, 1, 0, 1, 0, 1, 2, 1, 4, 1, 6, 2}, 60] (* Vincenzo Librandi, Oct 01 2017 *)
PROG
(Magma) I:=[0, 1, 0, 1, 0, 1, 2, 1, 4, 1, 6, 2]; [n le 12 select I[n] else Self(n-2)+2*Self(n-7)+Self(n-12): n in [1..60]]; // Vincenzo Librandi, Oct 01 2017
CROSSREFS
Cf. A292402.
Sequence in context: A375277 A305812 A341857 * A271773 A339602 A277127
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Sep 30 2017
STATUS
approved