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LinearRecurrence[{7, -17, 23, -16, 6, -1}, {1, 5, 21, 84, 330, 1291}, 30] (* Harvey P. Dale, Jul 10 2020 *)
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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^2 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).
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Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0) + c(1)*x + c(2)*x^2 + ... 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).
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allocated for Clark Kimberlingp-INVERT of the triangular numbers (A000217), where p(S) = 1 - S - S^2.
1, 5, 21, 84, 330, 1291, 5052, 19784, 77500, 303608, 1189372, 4659245, 18252027, 71500068, 280092848, 1097230105, 4298267549, 16837948391, 65960645632, 258392925744, 1012223324455, 3965263584006, 15533444957104, 60850409347588, 238374187312038
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Suppose s = (c(0), c(1), c(2),...) is a sequence and p(S) is a polynomial. Let S(x) = c(0) + c(1)*x + c(2)*x^2 + ... 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 A289780 for a guide to related sequences.
Clark Kimberling, <a href="/A289797/b289797.txt">Table of n, a(n) for n = 0..1000</a>
<a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (7, -17, 23, -16, 6, -1)
G.f.: (1 - 2 x + 3 x^2 - x^3)/(1 - 7 x + 17 x^2 - 23 x^3 + 16 x^4 - 6 x^5 + x^6).
a(n) = 7*a(n-1) - 17*a(n-2) + 23*a(n-3) - 16*a(n-4) + 6*a(n-5) - a(n-6).
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Clark Kimberling, Aug 12 2017
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