A289780 -id:A289780 - cp's OEIS frontend

A290890 p-INVERT of the positive integers, where p(S) = 1 - S^2.

0, 1, 4, 11, 28, 72, 188, 493, 1292, 3383, 8856, 23184, 60696, 158905, 416020, 1089155, 2851444, 7465176, 19544084, 51167077, 133957148, 350704367, 918155952, 2403763488, 6293134512, 16475640049, 43133785636, 112925716859, 295643364940, 774004377960

Offset: 0

Clark Kimberling, Aug 15 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).
Note that in A290890, s = (1,2,3,4,...); i.e., A000027(n+1) for n>=0, whereas in A290990, s = (0,1,2,3,4,...); i.e., A000027(n) for n>=0.
Guide to p-INVERT sequences using s = (1,2,3,4,5,...) = A000027:
p(S)                 t(1,2,3,4,5,...)
- S                    A001906
- S^2                  A290890; see A113067 for signed version
- S^3                  A290891
- S^4                  A290892
- S^5                  A290893
- S^6                  A290894
- S^7                  A290895
- S^8                  A290896
- S - S^2              A289780
- S - S^3              A290897
- S - S^4              A290898
- S^2 - S^4            A290899
- S^2 - S^3            A290900
- S^3 - S^4            A290901
- 2S                   A052530;  (1/2)*A052530 = A001353
- 3S                   A290902;  (1/3)*A290902 = A004254
- 4S                   A003319;  (1/4)*A003319 = A001109
- 5S                   A290903;  (1/5)*A290903 = A004187
- 2*S^2                A290904;  (1/2)*A290904 = A290905
- 3*S^2                A290906;  (1/3)*A290906 = A290907
- 4*S^2                A290908;  (1/4)*A290908 = A099486
- 5*S^2                A290909;  (1/5)*A290909 = A290910
- 6*S^2                A290911;  (1/6)*A290911 = A290912
- 7*S^2                A290913;  (1/7)*A290913 = A290914
- 8*S^2                A290915;  (1/8)*A290915 = A290916
(1 - S)^2                A290917
(1 - S)^3                A290918
(1 - S)^4                A290919
(1 - S)^5                A290920
(1 - S)^6                A290921
- S - 2*S^2            A290922
- 2*S - 2*S^2          A290923;  (1/2)*A290923 = A290924
- 3*S - 2*S^2          A290925
(1 - S^2)^2              A290926
(1 - S^2)^3              A290927
(1 - S^3)^2              A290928
(1 - S)(1 - S^2)         A290929
(1 - S^2)(1 - S^4)       A290930
- 3 S + S^2            A291025
- 4 S + S^2            A291026
- 5 S + S^2            A291027
- 6 S + S^2            A291028
- S - S^2 - S^3        A291029
- S - S^2 - S^3 - S^4  A201030
- 3 S + 2 S^3          A291031
- S - S^2 - S^3 + S^4  A291032
- 6 S                  A291033
- 7 S                  A291034
- 8 S                  A291181
- 3 S + 2 S^3          A291031
- 3 S + 2 S^2          A291182
- 4 S + 2 S^3          A291183
- 4 S + 3 S^3          A291184

			(See the examples at A289780.)

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, -5, 4, -1)

Cf. A000027, A113067, A289780, A113067 (signed version of same sequence).

z = 60; s = x/(1 - x)^2; p = 1 - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290890 *)

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

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

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

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.

A081696 Expansion of 1/(x + sqrt(1-4x)).

Original entry on oeis.org

1, 1, 3, 9, 29, 97, 333, 1165, 4135, 14845, 53791, 196417, 721887, 2667941, 9907851, 36950465, 138320021, 519515209, 1957091277, 7392602917, 27992976565, 106236268337, 404005515873, 1539293204549, 5875059106769, 22459721336977

Offset: 0

Emanuele Munarini, Apr 02 2003

Number of irreducible ordered pairs of compositions of n. A pair of compositions of n into the same number of (positive) parts, say n = a1 + ... + ak and n = b1 + ... + bk, is irreducible if for all j < k, a1 + ... + aj is not equal to b1 + ... + bj. E.g., a(3)=3 because the irreducible pairs are (1+2,2+1), (2+1,1+2), (3,3). - Herbert S. Wilf, May 22 2004
Hankel transform is 2^n. - Paul Barry, Nov 22 2007
Equals left border of triangle A152229. - Gary W. Adamson, Nov 29 2008
Equals INVERTi transform of A000984: (1, 2, 6, 20, 70, 252, ...). - Gary W. Adamson, May 15 2009
(1 + 3x + 9x^2 + 29x^3 + ...) * 1/(1 + x + 3x^2 + 9x^3 + 29x^4 + ...) = (1 + 2x + 4x^2 + 10x^3 + 28x^4 + ...); where A068875 = (1, 2, 4, 10, 28, ...). - Gary W. Adamson, Nov 21 2011
Apparently the number of grand Motzkin paths of length n with two types of flat step (F,f) and avoiding F at level 0. - David Scambler, Jul 04 2013
Starting at n=1 p-INVERT of Catalan numbers (A000108, starting at n=0), where p(S) = 1 - S - S^2; see A289780. - Clark Kimberling, Aug 12 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Ron M. Adin, Arkady Berenstein, Jacob Greenstein, Jian-Rong Li, Avichai Marmor, and Yuval Roichman, Transitive and Gallai colorings, arXiv:2309.11203 [math.CO], 2023. See p. 25.
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.
Paul Barry, Chebyshev moments and Riordan involutions, arXiv:1912.11845 [math.CO], 2019.
Paul Barry and Arnauld Mesinga Mwafise, Classical and Semi-Classical Orthogonal Polynomials Defined by Riordan Arrays, and Their Moment Sequences, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.5.
Edward A. Bender, Gregory F. Lawler, Robin Pemantle and Herbert S. Wilf, Irreducible compositions and the first return to the origin of a random walk, arXiv:math/0404253 [math.CO], 2004.
Edward A. Bender, Gregory F. Lawler, Robin Pemantle and Herbert S. Wilf, Irreducible compositions and the first return to the origin of a random walk, Sem. Lothar. 50 (2004) B50h.
David Callan, An identity for the central binomial coefficient, arXiv preprint arXiv:1206.3174 [math.CO], 2012. - From _N. J. A. Sloane_, Nov 25 2012
G. Chatel and V. Pilaud, Cambrian Hopf Algebras, arXiv:1411.3704 [math.CO], 2014-2015.
Ivan Dimitrov, Cole Gigliotti, Etan Ossip, Charles Paquette, and David Wehlau, Inversion Sets and Quotient Root Systems, arXiv:2310.16767 [math.CO], 2023.
A. Umar, Some combinatorial problems in the theory of symmetric ..., Algebra Disc. Math. 9 (2010) 115-126.

Cf. A000045, A000984, A081698, A152229, A189675.

Cf. A068875.

y[n_] := y[n] = (2*(4*n - 3)*y[n - 1] - (15*n - 24)*y[n - 2] - (4*n - 6)*y[n - 3])/n; y[0] = 1; y[1] = 1; y[2] = 3; (* corrected by Wouter Meeussen, Apr 30 2011 *)
CoefficientList[Series[1/(x+Sqrt[1-4x] ),{x,0,30}],x] (* Harvey P. Dale, May 05 2021 *)

makelist(sum(binomial(2*n-k,n+k)*(3*k+1)/(n+k+1),k,0,n),n,0,12); /* Emanuele Munarini, Apr 02 2011 */

x='x+O('x^66); Vec(1/(x+sqrt(1-4*x))) \\ Joerg Arndt, Jul 06 2013

G.f.: 1/(x + sqrt(1-4*x)).
D-finite with recurrence: n*a(n) + 2*(-4*n+3)*a(n-1) + 3*(5*n-8)*a(n-2) + 2*(2*n-3)*a(n-3) = 0.
a(n) = Sum_{k=0..n} binomial(2n-k,n+k)*(3k+1)/(n+k+1). - Emanuele Munarini, Apr 02 2011
From Paul Barry, Dec 18 2004: (Start)
A Catalan transform of the Fibonacci numbers F(n+1) under the mapping G(x) -> G(xc(x)), c(x) the g.f. of A000108. The inverse mapping is H(x) -> H(x(1-x)).
a(n) = Sum{k=0..n} (k/(2n-k))binomial(2n-k, n-k)F(k+1). (End)
From Bill Gosper, May 14 2011: (Start)
We have (per Wouter Meeussen): a(n) = (Sum_{k=1..n} k*Fibonacci(k+1)*(-1)^(n+k)*binomial(-n,n-k))/n = (Sum_{k=1..n} k*Fibonacci(k+1)*binomial(2*n-k-1,n-1))/n.
If we introduce an alternating sign, defining b(n) = (Sum_{k=1..n} k*Fibonacci(k+1)*binomial(-n,n-k))/n = (Sum_{k=1..n} k*Fibonacci(k+1)*(-1)^(n+k)*binomial(2*n-k-1,n-1))/n, then b(n) = 1 for all n. (Not obvious--I proved it satisfies b(n+2) = ((17*n^2 + 37*n + 18)*b(n+1) - 4*(2*n+1)*(2*n+3)*b(n))/((n+2)*(n+3)).) (End)
G.f.: 1/(1-x-2x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction). - Paul Barry, Aug 03 2009
From Gary W. Adamson, Jul 11 2011: (Start)
a(n) = the upper left term in M^n, M = an infinite square production matrix in which a column of (1,2,2,2,...) is prepended to an infinite lower triangular matrix of all 1's and the rest zeros:
  1, 1, 0, 0, 0, 0, ...
  2, 1, 1, 0, 0, 0, ...
  2, 1, 1, 1, 0, 0, ...
  2, 1, 1, 1, 1, 0, ...
  2, 1, 1, 1, 1, 1, ...
  ... (End)

More terms from Paul Barry, Dec 18 2004

Wouter credited with first sums in Gosper's FORMULA Comment, which were mistyped by NJAS (caught by Julian Ziegler Hunts), May 14 2011

A039717 Row sums of convolution triangle A030523.

Original entry on oeis.org

1, 4, 15, 55, 200, 725, 2625, 9500, 34375, 124375, 450000, 1628125, 5890625, 21312500, 77109375, 278984375, 1009375000, 3651953125, 13212890625, 47804687500, 172958984375, 625771484375, 2264062500000, 8191455078125

Offset: 1

Wolfdieter Lang

Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 3, s(2n) = 5.
With offset 0 = INVERT transform of A001792: (1, 3, 8, 20, 48, 112, ...). - Gary W. Adamson, Oct 26 2010
From Tom Copeland, Nov 09 2014: (Start)
The array belongs to a family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the o.g.f. (1-sqrt(1-4x/(1+(1-t)x)))/2 and inverse x*(1-x)/(1 + (t-1)*x*(1-x)). See A091867 for more info on this family. Here t = -4 (mod signs in the results).
Let C(x) = (1 - sqrt(1-4x))/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1+t*x) with inverse P(x,-t).
O.g.f.: G(x) = x*(1-x)/(1 - 5x*(1-x)) = P(Cinv(x),-5).
Inverse O.g.f.: Ginv(x) = (1 - sqrt(1 - 4*x/(1+5x)))/2 = C(P(x,5)) (signed A026378). Cf. A030528. (End)
p-INVERT of (2^n), where p(s) = 1 - s - s^2; see A289780. - Clark Kimberling, Aug 10 2017

Michel Marcus, Table of n, a(n) for n = 1..1000
Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
Index entries for linear recurrences with constant coefficients, signature (5,-5).

Appears in A109106.

Cf. A000045, A000108, A001792, A005043, A091867, A026378, A030528.

CoefficientList[Series[(1 - x) / (1 - 5 x + 5 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 10 2014 *)

Vec(x*(1-x)/(1-5*x+5*x^2) + O(x^40)) \\ Altug Alkan, Nov 20 2015

G.f.: x*(1-x)/(1-5*x+5*x^2) = g1(3, x)/(1-g1(3, x)), g1(3, x) := x*(1-x)/(1-2*x)^2 (g.f. first column of A030523).
From Paul Barry, Apr 16 2004: (Start)
Binomial transform of Fibonacci(2n+2).
a(n) = (sqrt(5)/2 + 5/2)^n*(3*sqrt(5)/10 + 1/2) - (5/2 - sqrt(5)/2)^n*(3*sqrt(5)/10 - 1/2). (End)
a(n) = (1/5)*Sum_{r=1..9} sin(3*r*Pi/10)*sin(r*Pi/2)*(2*cos(r*Pi/10))^(2*n).
a(n) = 5*a(n-1) - 5*a(n-2).
a(n) = Sum_{k=0..n} Sum_{i=0..n} binomial(n, i)*binomial(k+i+1, 2k+1). - Paul Barry, Jun 22 2004
From Johannes W. Meijer, Jul 01 2010: (Start)
Limit_{k->oo} a(n+k)/a(k) = (A020876(n) + A093131(n)*sqrt(5))/2.
Limit_{n->oo} A020876(n)/A093131(n) = sqrt(5).
(End)
From Benito van der Zander, Nov 19 2015: (Start)
Limit_{k->oo} a(k+1)/a(k) = 1 + phi^2 = (5 + sqrt(5)) / 2.
a(n) = a(n-1) * 3 + A081567(n-2) (not proved).
(End)
E.g.f.: exp(x*5/2) * (cosh(x*sqrt(5)/2) + (3/sqrt(5))*sinh(x*sqrt(5)/2)). - Fabian Pereyra, Oct 29 2024

A290995 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^8.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 8, 36, 120, 330, 792, 1716, 3432, 6436, 11456, 19584, 32640, 54264, 93024, 170544, 341088, 735472, 1653632, 3749760, 8386560, 18289440, 38724480, 79594560, 159189120, 311058496, 597137408, 1133991936, 2147450880, 4089171840

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).

See A291000 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8).

Cf. A000012, A033453, A289780, A291000.

Sequences of the form 1/((1-x)^m - x^m): A000079 (m=1,2), A024495 (m=3), A000749 (m=4), A049016 (m=5), A192080 (m=6), A049017 (m=7), this sequence (m=8), A306939 (m=9).

R:=PowerSeriesRing(Integers(), 60); [0,0,0,0,0,0,0] cat Coefficients(R!( x^7/((1-x)^8 - x^8) )); // G. C. Greubel, Apr 11 2023

z = 60; s = x/(1 - x); p = 1 - s^8;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290995 *)

concat(vector(7), Vec(x^7 / ((1 - 2*x)*(1 - 2*x + 2*x^2)*(1 - 4*x + 6*x^2 - 4*x^3 + 2*x^4)) + O(x^50))) \\ Colin Barker, Aug 22 2017

def A290995_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^7/((1-x)^8 - x^8) ).list()
A290995_list(60) # G. C. Greubel, Apr 11 2023

a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) for n >= 9.
G.f.: x^7 / ((1 - 2*x)*(1 - 2*x + 2*x^2)*(1 - 4*x + 6*x^2 - 4*x^3 + 2*x^4)). - Colin Barker, Aug 22 2017
G.f.: x^7/((1-x)^8 - x^8). - G. C. Greubel, Apr 11 2023

A290990 p-INVERT of the nonnegative integers (A000027), where p(S) = 1 - S - S^2.

Original entry on oeis.org

0, 1, 2, 5, 12, 28, 64, 145, 328, 743, 1686, 3830, 8704, 19781, 44950, 102133, 232048, 527208, 1197808, 2721421, 6183108, 14048151, 31917714, 72517738, 164761792, 374342057, 850512458, 1932380869, 4390407092, 9975090996, 22663602720, 51492150953

Offset: 0

Clark Kimberling, Aug 21 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).

See A290890 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,2,1).

Cf. A000027, A033453, A289780, A290890, A290991.

I:=[0,1,2,5]; [n le 4 select I[n] else 4*Self(n-1) -5*Self(n-2) +2*Self(n-3) +Self(n-4): n in [1..50]]; // G. C. Greubel, Apr 12 2023

z = 60; s = x^2/(1-x)^2; p = 1 -s -s^2;
Drop[CoefficientList[Series[s, {x,0,z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x,0,z}], x], 1]  (* A290990 *)
LinearRecurrence[{4,-5,2,1}, {0,1,2,5}, 50] (* G. C. Greubel, Apr 12 2023 *)

concat(0, Vec(x*(1-2*x+2*x^2)/(1-4*x+5*x^2-2*x^3-x^4) + O(x^50))) \\ Colin Barker, Aug 24 2017

@CachedFunction
def a(n): # a = A290990
    if (n<4): return (0,1,2,5)[n]
    else: return 4*a(n-1) -5*a(n-2) +2*a(n-3) +a(n-4)
[a(n) for n in range(51)] # G. C. Greubel, Apr 12 2023

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

G.f.: x*(1 - 2*x + 2*x^2) / (1 - 4*x + 5*x^2 - 2*x^3 - x^4). - Colin Barker, Aug 24 2017

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

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).
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

			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)

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,5,-1).

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

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

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 *)

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

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

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

A289975 p-INVERT of the Fibonacci numbers (A000045, including 0), where p(S) = 1 - S - S^2.

Original entry on oeis.org

0, 1, 1, 4, 7, 18, 37, 85, 183, 407, 888, 1956, 4284, 9409, 20630, 45270, 99289, 217819, 477776, 1048053, 2298912, 5042783, 11061455, 24263687, 53223023, 116746272, 256086074, 561731936, 1232174181, 2702807740, 5928681960, 13004724921, 28526216361

Offset: 0

Clark Kimberling, Aug 21 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).

See A290890 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2, 2, -3, -1)

Cf. A000045, A289781, A289976, A289780.

z = 60; s = x^2/(1 - x - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289975 *)

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

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

A290991 p-INVERT of (0,0,1,2,3,4,5,...), the nonnegative integers A000027 preceded by one zero, where p(S) = 1 - S - S^2.

Original entry on oeis.org

0, 0, 1, 2, 3, 6, 13, 26, 50, 96, 184, 351, 669, 1278, 2447, 4692, 9004, 17285, 33182, 63687, 122208, 234461, 449774, 862776, 1655010, 3174766, 6090231, 11683285, 22413104, 42997349, 82486280, 158241688, 303570021, 582365698, 1117202719, 2143225358

Offset: 0

Clark Kimberling, Aug 21 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).

See A290890 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, -6, 5, -3, 1, 1)

Cf. A000027, A289780, A290990.

z = 60; s = x^3/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0,0,1,2,3,4,5,... *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290991 *)

concat(vector(2), Vec(x^2*(1 - 2*x + x^2 + x^3) / (1 - 4*x + 6*x^2 - 5*x^3 + 3*x^4 - x^5 - x^6) + O(x^40))) \\ Colin Barker, Aug 24 2017

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

G.f.: x^2*(1 - 2*x + x^2 + x^3) / (1 - 4*x + 6*x^2 - 5*x^3 + 3*x^4 - x^5 - x^6). - Colin Barker, Aug 24 2017

A289787 p-INVERT of the even positive integers (A005843), where p(S) = 1 - S - S^2.

Original entry on oeis.org

2, 12, 62, 312, 1570, 7908, 39838, 200688, 1010978, 5092860, 25655582, 129241512, 651061762, 3279762132, 16521995710, 83230530528, 419278719938, 2112141348588, 10640036959358, 53599815453720, 270012240337762, 1360202629711812, 6852101192007262

Offset: 0

Clark Kimberling, Aug 10 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).

See A289780 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6, -6, 6, -1)

Cf. A005843, A289780, A289788.

z = 60; s = 2*x/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005843 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289787 *)
u/2 (* A289788 *)

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

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

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A290995 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^8.

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A290990 p-INVERT of the nonnegative integers (A000027), where p(S) = 1 - S - S^2.

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

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