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%I #6 Sep 04 2017 19:30:37
%S 0,0,0,0,0,1,6,15,20,15,6,2,12,66,220,495,792,925,810,648,1036,3126,
%T 8580,18566,31848,44034,50644,54384,74328,153161,354702,738966,
%U 1312380,1988814,2638668,3297933,4531980,7814811,15621794,30839391,55350396,88575614
%N p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^6.
%C 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).
%C See A291382 for a guide to related sequences.
%H Clark Kimberling, <a href="/A291381/b291381.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 0, 0, 1, 6, 15, 20, 15, 6, 1)
%F G.f.: -((x^5 (1 + x)^6)/((-1 + x + x^2) (1 + x + x^2) (1 - x + 2 x^3 + x^4) (1 + x + 2 x^2 + 2 x^3 + x^4)))
%F a(n) = a(n-6) + 6*a(n-7) + 15*a(n-8) + 20*a(n-9) + 15*a(n-10) + 6*a(n-11) + a(n-12) for n >= 13.
%t z = 60; s = x + x^2; p = (1 - s)^6;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A019590 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291381 *)
%Y Cf. A019590, A291382.
%K nonn,easy
%O 0,7
%A _Clark Kimberling_, Sep 04 2017
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