

A292326


pINVERT of (1,1,1,0,0,0,0,0,0,0,0,...), where p(S) = (1  S)^3.


1



3, 9, 25, 63, 153, 359, 819, 1830, 4018, 8694, 18582, 39298, 82350, 171186, 353338, 724719, 1478061, 2999175, 6057687, 12183945, 24411935, 48740193, 96998325, 192459996, 380812692, 751557756, 1479686972, 2906717460, 5698014924, 11147786740, 21769549380
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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 pINVERT 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 pINVERT is a generalization of the "INVERT" transform (e.g., A033453).


LINKS



FORMULA

G.f.: (((1 + x + x^2) (3  3 x  2 x^2  x^3 + 3 x^4 + 2 x^5 + x^6))/(1 + x + x^2 + x^3)^3).
a(n) = 3*a(n1)  2*a(n3)  6*a(n4) + 4*a(n6) + 6*a(n7) + 3*a(n8) + a(n9) for n >= 10.


MATHEMATICA

z = 60; s = x + x^2 + x^3; p = (1  s)^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292326 *)
LinearRecurrence[{3, 0, 2, 6, 0, 4, 6, 3, 1}, {3, 9, 25, 63, 153, 359, 819, 1830, 4018}, 40] (* Harvey P. Dale, Nov 01 2019 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



