cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Aug 22 2017

Keywords

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^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

Links

Crossrefs

Cf. A000012, A289780.

Programs

  • Mathematica
    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 *)

Formula

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.

A290998 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^3 - S^4.

Original entry on oeis.org

0, 0, 1, 4, 10, 21, 43, 92, 205, 462, 1035, 2301, 5099, 11303, 25088, 55728, 123800, 274969, 610628, 1355970, 3011157, 6686979, 14850196, 32978725, 73237462, 162641499, 361184653, 802098203, 1781254927, 3955712256, 8784625824, 19508406192, 43323176177
Offset: 0

Views

Author

Clark Kimberling, Aug 22 2017

Keywords

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^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).
See A291000 for a guide to related sequences.
For n >= 1, a(n-1) is the number of ways to split [n] into an unspecified number of intervals and then choose 3 blocks (i.e., subintervals) from each interval. For example, for n=9, a(8)=205 since the number of ways to split [9] into intervals and then select 3 blocks from each interval is C(9,3) + C(6,3)*C(3,3) + C(5,3)*C(4,3) + C(4,3)*C(5,3) + C(3,3)*C(6,3) + C(3,3)*C(3,3)*C(3,3) for a total of 205 ways. - Enrique Navarrete, Dec 23 2023
a(n-1) is also the number of compositions of n using parts of size at least 3 where there are binomial(i,3) types of i, n>=1, i>=3 (see example). - Enrique Navarrete, Dec 25 2023

Examples

			From _Enrique Navarrete_, Dec 25 2023: (Start)
Since there are binomial(3,3) = 1 type of 3, binomial(4,3) = 4 types of 4, binomial(5,3) = 10 types of 5, binomial(6,3) = 20 types of 6, and binomial(9,3) = 84 types of 9, we can write 9 in the following ways:
 9 in 84 ways;
 6+3 in 20 ways;
 5+4 in 40 ways;
 4+5 in 40 ways;
 3+6 in 20 ways;
 3+3+3 in 1 way, for a total of 205 ways. (End)

Links

Crossrefs

Cf. A000012, A033453, A289780, A290997, A291000.

Programs

  • Magma
    I:=[0,0,1,4]; [n le 4 select I[n] else 4*Self(n-1) -6*Self(n-2) +5*Self(n-3) -Self(n-4): n in [1..41]]; // G. C. Greubel, Apr 25 2023
  • Mathematica
    z = 60; s = x/(1 - x); p = 1 - s^3 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* this sequence *)
LinearRecurrence[{4,-6,5,-1}, {0,0,1,4}, 41] (* G. C. Greubel, Apr 25 2023 *)
  • PARI
    concat(vector(2), Vec(x^2 / (1 - 4*x + 6*x^2 - 5*x^3 + x^4) + O(x^50))) \\ Colin Barker, Aug 22 2017
  • SageMath
    def A290998_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^2/(1-4*x+6*x^2-5*x^3+x^4) ).list()
A290998_list(40) # G. C. Greubel, Apr 25 2023

Formula

a(n) = 4*a(n-1) - 6*a(n-2) + 5*a(n-3) - a(n-4) for n >= 5.
G.f.: x^2 / (1 - 4*x + 6*x^2 - 5*x^3 + x^4). - Colin Barker, Aug 22 2017
G.f.: 1/(x*(1-Sum_{k>=3} binomial(k,3)*x^k)) - 1/x. - Enrique Navarrete, Dec 26 2023

A290999 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 6*S^2.

Original entry on oeis.org

0, 6, 12, 54, 168, 606, 2052, 7134, 24528, 84726, 292092, 1007814, 3476088, 11991246, 41362932, 142682094, 492178848, 1697768166, 5856430572, 20201701974, 69685556808, 240379623486, 829187031012, 2860272179454, 9866479513968, 34034319925206, 117401037420252
Offset: 0

Views

Author

Clark Kimberling, Aug 22 2017

Keywords

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^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).
See A291000 for a guide to related sequences.

Links

Crossrefs

Cf. A000012, A002532, A033453, A289780, A290997, A290998, A291000.

Programs

  • Magma
    [n le 2 select 6*(n-1) else 2*Self(n-1) +5*Self(n-2): n in [1..41]]; // G. C. Greubel, Apr 25 2023
  • Mathematica
    z = 60; s = x/(1 - x); p = 1 - s^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290999 *)
LinearRecurrence[{2,5},{0,6},30] (* Harvey P. Dale, Mar 25 2018 *)
  • SageMath
    A290999=BinaryRecurrenceSequence(2,5,0,6)
[A290999(n) for n in range(41)] # G. C. Greubel, Apr 25 2023

Formula

G.f.: 6*x/(1 - 2*x - 5*x^2).
a(n) = 2*a(n-1) + 5*a(n-2) for n >= 3.
a(n) = 6*A002532(n).
a(n) = sqrt(3/2)*((1+sqrt(6))^n - (1-sqrt(6))^n). - Colin Barker, Aug 23 2017
Showing 1-3 of 3 results.

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