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Revision History for A289843

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p-INVERT of (1,0,2,0,3,0,4,0,5,...) (A027656), where p(S) = 1 - S - S^2.
(history; published version)
#9 by Alois P. Heinz at Sun Aug 13 23:00:56 EDT 2017
STATUS

proposed

approved

#8 by Clark Kimberling at Sun Aug 13 14:24:20 EDT 2017
STATUS

editing

proposed

#7 by Clark Kimberling at Sun Aug 13 14:20:21 EDT 2017
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^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).

LINKS

<a href="/index/Rec#order_0408">Index entries for linear recurrences with constant coefficients</a>, signature (1, 5, -2, -6, 1, 4, 0, -1)

STATUS

proposed

editing

#6 by Jon E. Schoenfield at Sat Aug 12 17:50:22 EDT 2017
STATUS

editing

proposed

#5 by Jon E. Schoenfield at Sat Aug 12 17:50:18 EDT 2017
COMMENTS

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).

STATUS

proposed

editing

#4 by Clark Kimberling at Sat Aug 12 16:22:02 EDT 2017
STATUS

editing

proposed

#3 by Clark Kimberling at Sat Aug 12 16:18:35 EDT 2017
NAME

allocated for Clark Kimberling

p-INVERT of (1,0,2,0,3,0,4,0,5,...) (A027656), where p(S) = 1 - S - S^2.

DATA

1, 2, 5, 13, 29, 73, 168, 410, 962, 2317, 5483, 13131, 31193, 74509, 177311, 423025, 1007505, 2402354, 5723761, 13644587, 32514730, 77501115, 184698088, 440216833, 1049148789, 2500520812, 5959478837, 14203542282, 33851496564, 80679640434, 192285583548

OFFSET

0,2

COMMENTS

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.

LINKS

Clark Kimberling, <a href="/A289843/b289843.txt">Table of n, a(n) for n = 0..1000</a>

<a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1, 5, -2, -6, 1, 4, 0, -1)

FORMULA

G.f.: (1 + x - 2 x^2 + x^4)/(1 - x - 5 x^2 + 2 x^3 + 6 x^4 - x^5 - 4 x^6 + x^8).

a(n) = a(n-1) + 5*a(n-2) - 2*a(n-3) - 6*a(n-4) + a(n-5) + 4*a(n-6) - a(n-8).

MATHEMATICA

z = 60; s = x/(1 - x^2)^2; p = 1 - s - s^2;

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A027656 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289843 *)

CROSSREFS
KEYWORD

allocated

nonn,easy

AUTHOR

Clark Kimberling, Aug 12 2017

STATUS

approved

editing

#2 by Clark Kimberling at Thu Jul 13 19:47:58 EDT 2017
KEYWORD

allocating

allocated

#1 by Clark Kimberling at Thu Jul 13 19:47:58 EDT 2017
NAME

allocated for Clark Kimberling

KEYWORD

allocating

STATUS

approved