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A291739
p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^3 - S^6.
2
0, 0, 1, 0, 3, 2, 3, 12, 4, 30, 27, 45, 108, 90, 260, 342, 498, 1115, 1218, 2709, 3913, 5949, 11469, 15262, 28461, 44556, 68028, 123243, 178650, 311337, 498114, 777996, 1340603, 2052765, 3435906, 5569902, 8800392, 14783823, 23242761, 38249550, 62156709
OFFSET
0,5
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 A291728 for a guide to related sequences.
LINKS
Index entries for linear recurrences with constant coefficients, signature (0, 0, 1, 0, 3, 1, 3, 6, 1, 15, 0, 20, 0, 15, 0, 6, 0, 1)
FORMULA
G.f.: -((x^2 (1 + x^2)^3 (1 + x + x^2) (1 + x + x^3) (1 - 2 x + 2 x^2 - x^3 + x^4))/(-1 + x^3 + 3 x^5 + x^6 + 3 x^7 + 6 x^8 + x^9 + 15 x^10 + 20 x^12 + 15 x^14 + 6 x^16 + x^18)).
a(n) = a(n-3) + 3*a(n-5) + a(n-6) + 3*a(n-7) + 6*a(n-8) + a(n-9) + 15*a(n-10) + 20 *a(n-12) + 15*a(n-14) + 6*a(n-16) + a(n-18) for n >= 19.
MATHEMATICA
z = 60; s = x + x^3; p = 1 - s^3 - s^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291739 *)
CROSSREFS
Sequence in context: A078017 A343170 A169816 * A057053 A081850 A247237
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Sep 11 2017
STATUS
approved