[go: up one dir, main page]

login
Revision History for A289797 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing all changes.
p-INVERT of the triangular numbers (A000217), where p(S) = 1 - S - S^2.
(history; published version)
#10 by Harvey P. Dale at Fri Jul 10 13:32:14 EDT 2020
STATUS

editing

approved

#9 by Harvey P. Dale at Fri Jul 10 13:32:11 EDT 2020
MATHEMATICA

LinearRecurrence[{7, -17, 23, -16, 6, -1}, {1, 5, 21, 84, 330, 1291}, 30] (* Harvey P. Dale, Jul 10 2020 *)

STATUS

approved

editing

#8 by Alois P. Heinz at Sun Aug 13 23:02:04 EDT 2017
STATUS

proposed

approved

#7 by Clark Kimberling at Sun Aug 13 14:20:34 EDT 2017
STATUS

editing

proposed

#6 by Clark Kimberling at Sun Aug 13 14:14:52 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).

STATUS

proposed

editing

#5 by Jon E. Schoenfield at Sat Aug 12 10:41:29 EDT 2017
STATUS

editing

proposed

#4 by Jon E. Schoenfield at Sat Aug 12 10:41:25 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

#3 by Clark Kimberling at Sat Aug 12 09:19:17 EDT 2017
STATUS

editing

proposed

#2 by Clark Kimberling at Sat Aug 12 09:01:54 EDT 2017
NAME

allocated for Clark Kimberlingp-INVERT of the triangular numbers (A000217), where p(S) = 1 - S - S^2.

DATA

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

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="/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)

FORMULA

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).

MATHEMATICA

z = 60; s = x/(1 - x)^3; p = 1 - s - s^2;

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000217 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289797 *)

CROSSREFS
KEYWORD

allocated

nonn,easy

AUTHOR

Clark Kimberling, Aug 12 2017

STATUS

approved

editing

#1 by Clark Kimberling at Wed Jul 12 14:22:41 EDT 2017
NAME

allocated for Clark Kimberling

KEYWORD

allocated

STATUS

approved