A291008 -id:A291008 - cp's OEIS frontend

A291000 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3.

1, 3, 9, 26, 74, 210, 596, 1692, 4804, 13640, 38728, 109960, 312208, 886448, 2516880, 7146144, 20289952, 57608992, 163568448, 464417728, 1318615104, 3743926400, 10630080640, 30181847168, 85694918912, 243312448256, 690833811712, 1961475291648, 5569190816256

Offset: 0

Clark Kimberling, Aug 22 2017

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).
In the following guide to p-INVERT sequences using s = (1,1,1,1,1,...) = A000012, in some cases t(1,1,1,1,1,...) is a shifted version of the cited sequence:
p(S)                             t(1,1,1,1,1,...)
1 - S                                A000079
1 - S^2                              A000079
1 - S^3                              A024495
1 - S^4                              A000749
1 - S^5                              A139761
1 - S^6                              A290993
1 - S^7                              A290994
1 - S^8                              A290995
1 - S - S^2                          A001906
1 - S - S^3                          A116703
1 - S - S^4                          A290996
1 - S^3 - S^6                        A290997
1 - S^2 - S^3                        A095263
1 - S^3 - S^4                        A290998
1 - 2 S^2                            A052542
1 - 3 S^2                            A002605
1 - 4 S^2                            A015518
1 - 5 S^2                            A163305
1 - 6 S^2                            A290999
1 - 7 S^2                            A291008
1 - 8 S^2                            A291001
(1 - S)^2                            A045623
(1 - S)^3                            A058396
(1 - S)^4                            A062109
(1 - S)^5                            A169792
(1 - S)^6                            A169793
(1 - S^2)^2                          A024007
1 - 2 S - 2 S^2                      A052530
1 - 3 S - 2 S^2                      A060801
(1 - S)(1 - 2 S)                     A053581
(1 - 2 S)(1 - 3 S)                   A291002
(1 - S)(1 - 2 S)(1 - 3 S)(1 - 4 S)   A291003
(1 - 2 S)^2                          A120926
(1 - 3 S)^2                          A291004
1 + S - S^2                          A000045  (Fibonacci numbers starting with -1)
1 - S - S^2 - S^3                    A291000
1 - S - S^2 - S^3 - S^4              A291006
1 - S - S^2 - S^3 - S^4 - S^5        A291007
1 - S^2 - S^4                        A290990
(1 - S)(1 - 3 S)                     A291009
(1 - S)(1 - 2 S)(1 - 3 S)            A291010
(1 - S)^2 (1 - 2 S)                  A291011
(1 - S^2)(1 - 2 S)                   A291012
(1 - S^2)^3                          A291013
(1 - S^3)^2                          A291014
1 - S - S^2 + S^3                    A045891
1 - 2 S - S^2 + S^3                  A291015
1 - 3 S + S^2                        A136775
1 - 4 S + S^2                        A291016
1 - 5 S + S^2                        A291017
1 - 6 S + S^2                        A291018
1 - S - S^2 - S^3 + S^4              A291019
1 - S - S^2 - S^3 - S^4 + S^5        A291020
1 - S - S^2 - S^3 + S^4 + S^5        A291021
1 - S - 2 S^2 + 2 S^3                A175658
1 - 3 S^2 + 2 S^3                    A291023
(1 - 2 S^2)^2                        A291024
(1 - S^3)^3                          A291143
(1 - S - S^2)^2                      A209917

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, -4, 2)

Cf. A000012, A289780.

z = 60; s = x/(1 - x); p = 1 - s - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291000 *)

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

a(n) = 4*a(n-1) - 4*a(n-2) + 2*a(n-3) for n >= 4.

A083099 a(n) = 2a(n-1) + 6a(n-2), a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, 10, 32, 124, 440, 1624, 5888, 21520, 78368, 285856, 1041920, 3798976, 13849472, 50492800, 184082432, 671121664, 2446737920, 8920205824, 32520839168, 118562913280, 432250861568, 1575879202816, 5745263575040

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Apr 22 2003

a(n+1) = a(n) + A083098(n+1). A083098(n+1)/a(n) converges to sqrt(7).
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 7 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(7). - Cino Hilliard, Sep 25 2005
Pisano period lengths: 1, 1, 2, 1, 12, 2, 7, 1, 6, 12, 60, 2,168, 7, 12, 1,288, 6, 18, 12, ... - R. J. Mathar, Aug 10 2012
a(n) is divisible by 2^ceiling(n/2), see formula below. - Ralf Stephan, Dec 24 2013
Connect the center of a regular hexagon with side length 1 with its six vertices. a(n) is the number of paths of length n from the center to any of its vertices. Number of paths of length n from the center to itself is 6*a(n-1). - Jianing Song, Apr 20 2019

John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Project Euler, Problem 752, sequence beta(n).
Index entries for linear recurrences with constant coefficients, signature (2,6).

The following sequences (and others) belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A063727, A083098, A083099, A083100, A084057.

Cf. A154245, A291008.

[n le 2 select n-1 else 2*Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

A083099 := proc(n)
    option remember;
    if n <= 1 then
        n;
    else
        2*procname(n-1)+6*procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Sep 23 2016

CoefficientList[Series[x/(1-2x-6x^2), {x, 0, 25}], x] (* Adapted for offset 0 by Vincenzo Librandi, Feb 07 2014 *)
Expand[Table[((1 + Sqrt[7])^n - (1 - Sqrt[7])^n)7/(14Sqrt[7]), {n, 0, 25}]] (* Zerinvary Lajos, Mar 22 2007 *)
LinearRecurrence[{2,6}, {0,1}, 25] (* Sture Sjöstedt, Dec 06 2011 *)

a(n)=([0,1; 6,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, May 10 2016

my(x='x+O('x^30)); concat([0], Vec(x/(1-2*x-6*x^2))) \\ G. C. Greubel, Jan 24 2018

[lucas_number1(n,2,-6) for n in range(0, 25)] # Zerinvary Lajos, Apr 22 2009

A083099=BinaryRecurrenceSequence(2,6,0,1)
[A083099(n) for n in range(41)] # G. C. Greubel, Jun 01 2023

G.f.: x/(1 - 2*x - 6*x^2).
From Paul Barry, Sep 29 2004: (Start)
E.g.f.: (d/dx)(exp(x)*sinh(sqrt(7)*x)/sqrt(7));
a(n-1) = Sum_{k=0..n} binomial(n, 2k+1)*7^k. (End)
Simplified formula: a(n) = ((1 + sqrt(7))^n - (1 - sqrt(7))^n)/sqrt(28). - Al Hakanson (hawkuu(AT)gmail.com), Jan 05 2009
G.f.: G(0)*x/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(7*k-1)/(x*(7*k+6) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(2n) = 2^n * A154245(n), a(2n+1) = 2^n * (5*A154245(n) - 9*A154245(n-1)). - Ralf Stephan, Dec 24 2013
a(n) = Sum_{k=1,3,5,...<=n} binomial(n,k)*7^((k-1)/2). - Vladimir Shevelev, Feb 06 2014
a(n) = i^(n-1)*6^((n-1)/2)*ChebyshevU(n-1, -i/sqrt(6)). - G. C. Greubel, Jun 01 2023

cp's OEIS Frontend

A291000 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3.

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A083099 a(n) = 2a(n-1) + 6a(n-2), a(0) = 0, a(1) = 1.

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cp's OEIS Frontend

A291000 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A083099 a(n) = 2*a(n-1) + 6*a(n-2), a(0) = 0, a(1) = 1.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Sage

SageMath

Formula

A083099 a(n) = 2a(n-1) + 6a(n-2), a(0) = 0, a(1) = 1.