A033453 -id:A033453 - cp's OEIS frontend

A289780 p-INVERT of the positive integers (A000027), where p(S) = 1 - S - S^2.

1, 4, 14, 47, 156, 517, 1714, 5684, 18851, 62520, 207349, 687676, 2280686, 7563923, 25085844, 83197513, 275925586, 915110636, 3034975799, 10065534960, 33382471801, 110713382644, 367182309614, 1217764693607, 4038731742156, 13394504020957, 44423039068114

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).
Guide to p-INVERT sequences using p(S) = 1 - S - S^2:
t(A000012) = t(1,1,1,1,1,1,1,...)    = A001906
t(A000290) = t(1,4,9,16,25,36,...)   = A289779
t(A000027) = t(1,2,3,4,5,6,7,8,...)  = A289780
t(A000045) = t(1,2,3,5,8,13,21,...)  = A289781
t(A000032) = t(2,1,3,4,7,11,14,...)  = A289782
t(A000244) = t(1,3,9,27,81,243,...)  = A289783
t(A000302) = t(1,4,16,64,256,...)    = A289784
t(A000351) = t(1,5,25,125,625,...)   = A289785
t(A005408) = t(1,3,5,7,9,11,13,...)  = A289786
t(A005843) = t(2,4,6,8,10,12,14,...) = A289787
t(A016777) = t(1,4,7,10,13,16,...)   = A289789
t(A016789) = t(2,5,8,11,14,17,...)   = A289790
t(A008585) = t(3,6,9,12,15,18,...)   = A289795
t(A000217) = t(1,3,6,10,15,21,...)   = A289797
t(A000225) = t(1,3,7,15,31,63,...)   = A289798
t(A000578) = t(1,8,27,64,625,...)    = A289799
t(A000984) = t(1,2,6,20,70,252,...)  = A289800
t(A000292) = t(1,4,10,20,35,56,...)  = A289801
t(A002620) = t(1,2,4,6,9,12,16,...)  = A289802
t(A001906) = t(1,3,8,21,55,144,...)  = A289803
t(A001519) = t(1,1,2,5,13,34,...)    = A289804
t(A103889) = t(2,1,4,3,6,5,8,7,,...) = A289805
t(A008619) = t(1,1,2,2,3,3,4,4,...)  = A289806
t(A080513) = t(1,2,2,3,3,4,4,5,...)  = A289807
t(A133622) = t(1,2,1,3,1,4,1,5,...)  = A289809
t(A000108) = t(1,1,2,5,14,42,...)    = A081696
t(A081696) = t(1,1,3,9,29,97,...)    = A289810
t(A027656) = t(1,0,2,0,3,0,4,0,5...) = A289843
t(A175676) = t(1,0,0,2,0,0,3,0,...)  = A289844
t(A079977) = t(1,0,1,0,2,0,3,...)    = A289845
t(A059841) = t(1,0,1,0,1,0,1,...)    = A289846
t(A000040) = t(2,3,5,7,11,13,...)    = A289847
t(A008578) = t(1,2,3,5,7,11,13,...)  = A289828
t(A000142) = t(1!, 2!, 3!, 4!, ...)  = A289924
t(A000201) = t(1,3,4,6,8,9,11,...)   = A289925
t(A001950) = t(2,5,7,10,13,15,...)   = A289926
t(A014217) = t(1,2,4,6,11,17,29,...) = A289927
t(A000045*) = t(0,1,1,2,3,5,...)     = A289975 (* indicates prepended 0's)
t(A000045*) = t(0,0,1,1,2,3,5,...)   = A289976
t(A000045*) = t(0,0,0,1,1,2,3,5,...) = A289977
t(A290990*) = t(0,1,2,3,4,5,...)     = A290990
t(A290990*) = t(0,0,1,2,3,4,5,...)   = A290991
t(A290990*) = t(0,0,01,2,3,4,5,...)  = A290992

			Example 1:  s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S.
S(x) = x + 2x^2 + 3x^3 + 4x^4 + ...
p(S(x)) = 1 - (x + 2x^2 + 3x^3 + 4x^4 + ... )
- p(0) + 1/p(S(x)) = -1 + 1 + x + 3x^2 + 8x^3 + 21x^4 + ...
T(x) = 1 + 3x + 8x^2 + 21x^3 + ...
t(s) = (1,3,8,21,...) = A001906.
***
Example 2:  s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S - S^2.
S(x) =  x + 2x^2 + 3x^3 + 4x^4 + ...
p(S(x)) = 1 - ( x + 2x^2 + 3x^3 + 4x^4 + ...) - ( x + 2x^2 + 3x^3 + 4x^4 + ...)^2
- p(0) + 1/p(S(x)) = -1 + 1 + x + 4x^2 + 14x^3 + 47x^4 + ...
T(x) = 1 + 4x + 14x^2 + 47x^3 + ...
t(s) = (1,4,14,47,...) = A289780.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5, -7, 5, -1)

Cf. A000027.

P:=[1,4,14,47];; for n in [5..10^2] do P[n]:=5*P[n-1]-7*P[n-2]+5*P[n-3]-P[n-4]; od; P; # Muniru A Asiru, Sep 03 2017

z = 60; s = x/(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] (* A289780 *)

x='x+O('x^99); Vec((1-x+x^2)/(1-5*x+7*x^2-5*x^3+x^4)) \\ Altug Alkan, Aug 13 2017

G.f.: (1 - x + x^2)/(1 - 5 x + 7 x^2 - 5 x^3 + x^4).

a(n) = 5*a(n-1) - 7*a(n-2) + 5*a(n-3) - a(n-4).

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

Original entry on oeis.org

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.

A291219 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^3.

Original entry on oeis.org

1, 1, 3, 5, 11, 21, 42, 83, 163, 323, 635, 1255, 2473, 4880, 9625, 18985, 37451, 73869, 145715, 287421, 566954, 1118331, 2205947, 4351307, 8583091, 16930447, 33395857, 65874464, 129939569, 256310161, 505580371, 997274197, 1967156763, 3880282533, 7653987242

Offset: 0

Clark Kimberling, Aug 24 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 = (1,0,1,0,1,...) = A000035, in some cases t(1,0,1,0,1,...) is a shifted version of the indicated sequence.
p(S)                             t(1,0,1,0,1,...)
1 - S                                A000045 (Fibonacci numbers)
1 - S^2                              A147600
1 - S^3                              A291217
1 - S^5                              A291218
1 - S - S^2                          A289846
1 - S - S^3                          A291219
1 - S - S^4                          A291220
1 - S^3- S^6                         A291221
1 - S^2- S^3                         A291222
1 - S^3- S^4                         A291223
1 - 2S                               A052542
1 - 3S                               A006190
(1 - S)^2                            A239342
(1 - S)^3                            A276129
(1 - S)^4                            A291224
(1 - S)^5                            A291225
(1 - S)^6                            A291226
1 - S - 2 S^2                        A291227
1 - 2 S - 2 S^2                      A291228
1 - 3 S - 2 S^2                      A060801
(1 - S)(1 - 2 S)                     A291229
(1 - S)(1 - 2 S)(1 - 3 S)            A291230
(1 - S)(1 - 2 S)(1 - 3 S)( 1 - 4 S)  A291231
(1 - 2 S)^2                          A291264
(1 - 3 S)^2                          A291232
1 - S - S^2 - S^3                    A291233
1 - S - S^2 - S^3 - S^4              A291234
1 - S - S^2 - S^3 - S^4 - S^5        A291235
(1 - S)(1 - 3 S)                     A291236
(1 - S)(1 - 2S)( 1 - 4S)             A291237
(1 - S)^2 (1 - 2S)                   A291238
(1 - S^2) (1 - 2S)                   A291239
(1 - S^3)^2                          A291240
1 - S - S^2 + S^3                    A291241
1 - 2 S - S^2 + S^3                  A291242
1 - 3 S + S^2                        A291243
1 - 4 S + S^2                        A291244
1 - 5 S + S^2                        A291245
1 - 6 S + S^2                        A291246
1 - S - S^2 - S^3 + S^4              A291247
1 - S - S^2 - S^3 - S^4 + S^5        A291248
1 - S - S^2 - S^3 + S^4 + S^5        A291249
1 - S - 2 S^2 + 2 S^3                A291250
1 - 3 S^2 + 2 S^3                    A291251 (includes negative terms)
(1 - S^3)^3                          A291252
(1 - S - S^2)^2                      A291253
(1 - 2 S - S^2)^2                    A291254
(1 - S - 2 S^2)^2                    A291255

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

Cf. A000035, A290890, A291000.

I:=[1,1,3,5,11,21]; [n le 6 select I[n] else Self(n-1)+3*Self(n-2)-Self(n-3)-3*Self(n-4)+Self(n-5)+Self(n-6): n in [1..45]]; // Vincenzo Librandi, Aug 25 2017

z = 60; s = x/(1 - x^2); p = 1 - s - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291219 *)
LinearRecurrence[{1, 3, -1, -3, 1, 1}, {1, 1, 3, 5, 11, 21}, 50] (* Vincenzo Librandi, Aug 25 2017 *)

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

a(n) = a(n-1) + 3*a(n-2) - a(n-3) - 3*a(n-4) + a(n-5) + a(n-6) for n >= 7.

A291382 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2.

Original entry on oeis.org

2, 7, 22, 70, 222, 705, 2238, 7105, 22556, 71608, 227332, 721705, 2291178, 7273743, 23091762, 73308814, 232731578, 738846865, 2345597854, 7446508273, 23640235416, 75050038224, 238259397096, 756395887969, 2401310279090, 7623377054503, 24201736119310

Offset: 0

Clark Kimberling, Sep 04 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 = (1,1,0,0,0,...) = A019590, in some cases t(1,1,0,0,0,...) is a shifted version of the cited sequence:
p(S)                     t(1,1,0,0,0,...)
1 - S                       A000045 (Fibonacci numbers)
1 - S^2                     A094686
1 - S^3                     A115055
1 - S^4                     A291379
1 - S^5                     A281380
1 - S^6                     A281381
1 - 2 S                     A002605
1 - 3 S                     A125145
(1 - S)^2                   A001629
(1 - S)^3                   A001628
(1 - S)^4                   A001629
(1 - S)^5                   A001873
(1 - S)^6                   A001874
1 - S - S^2                 A123392
1 - 2 S - S^2               A291382
1 - S - 2 S^2               A124861
1 - 2 S - S^2               A291383
(1 - 2 S)^2                 A073388
(1 - 3 S)^2                 A291387
(1 - 5 S)^2                 A291389
(1 - 6 S)^2                 A291391
(1 - S)(1 - 2 S)            A291393
(1 - S)(1 - 3 S)            A291394
(1 - 2 S)(1 - 3 S)          A291395
(1 - S)(1 - 2 S)            A291393
(1 - S)(1 - 2 S)(1 - 3 S)   A291396
1 - S - S^3                 A291397
1 - S^2 - S^3               A291398
1 - S - S^2 - S^3           A186812
1 - S - S^2 - S^3 - S^4     A291399
1 - S^2 - S^4               A291400
1 - S - S^4                 A291401
1 - S^3 - S^4               A291402
1 - 2 S^2 - S^4             A291403
1 - S^2 - 2 S^4             A291404
1 - 2 S^2 - 2 S^4           A291405
1 - S^3 - S^6               A291407
(1 - S)(1 - S^2)            A291408
(1 - S^2)(1 - S)^2          A291409
1 - S - S^2 - 2 S^3         A291410
1 - 2 S - S^2 + S^3         A291411
1 - S - 2 S^2 + S^3         A291412
1 - 3 S + S^2 + S^3         A291413
1 - 2 S + S^3               A291414
1 - 3 S + S^2               A291415
1 - 4 S + S^2               A291416
1 - 4 S + 2 S^2             A291417

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

Cf. A019590, A290890, A291000, A291219, A291728, A292479, A292480.

z = 60; s = x + x^2; p = 1 - 2 s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291382 *)

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

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

A291728 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 2, 4, 9, 17, 35, 70, 142, 285, 576, 1160, 2340, 4716, 9510, 19171, 38653, 77926, 157110, 316747, 638599, 1287479, 2595698, 5233196, 10550681, 21271280, 42885152, 86460984, 174314476, 351436368, 708532813, 1428476905, 2879960190, 5806303628, 11706120825

Offset: 0

Clark Kimberling, Sep 08 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 = (1,0,1,0,0,0,0,...) = A154272, in some cases t(1,0,1,0,0,0,0,...) is a shifted (or differently indexed) version of the indicated sequence:
***
p(S)             t(1,0,1,0,0,0,0,...)
1 - S                A000930 (Narayana's cows sequence)
1 - S^2              A002478 (except for 0's)
1 - S^3              A291723
1 - S^5              A291724
(1 - S)^2            A291725
(1 - S)^3            A291726
(1 - S)^4            A291727
1 - S - S^2          A291728
1 - 2S - S^2         A291729
1 - 2S - 2S^2        A291730
(1 - 2S)^2           A291732
(1 - S)(1 - 2S)      A291734
1 - S - S^3          A291735
1 - S^2 - S^3        A291736
1 - S - S^2 - S^3    A291737
1 - S - S^4          A291738
1 - S^3 - S^6        A291739
(1 - S)(1 - S^2)     A291740
(1 - S)(1 + S^2)     A291741

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

Cf. A154272, A290890, A291000, A291382, A291219, A291382.

z = 60; s = x + x^3; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291728 *)

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

a(n) = a(n-1) + a(n-2) + a(n-3) + 2*a(n-4) + a(n-6) for n >= 7.

A292480 p-INVERT of the odd positive integers, where p(S) = 1 - S^2.

Original entry on oeis.org

0, 1, 6, 20, 56, 160, 480, 1456, 4384, 13136, 39360, 118064, 354272, 1062928, 3188736, 9565936, 28697632, 86093264, 258280512, 774841520, 2324523104, 6973567888, 20920705152, 62762119792, 188286360736, 564859074896, 1694577214656, 5083731648560

Offset: 0

Clark Kimberling, Oct 02 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 = (1,3,5,7,9,...) = A005408, in some cases t(1,3,5,7,9,...) is a shifted (or differently indexed) version of the cited sequence:
p(S) *********** t(1,3,5,7,9,...)
- S               A003946
- S^2             A292480
- S^3             (not yet in OEIS)
(1 - S)^2           (not yet in OEIS)
(1 - S)^3           (not yet in OEIS)
- S - S^2         A289786
+ S - S^2         A289484
- S - 2 S^2       A289785
- S - 3 S^2       A289786
- S - 4 S^2       A289787
- S - 5 S^2       A289788
- S - 6 S^2       A289789
- S - 7 S^2       A289790
+ S - 2 S^2       A289791
- S + S^2 - S^3   A289792
+ S - 3 S^2       A289793
- S - S^2 - S^3   A289794

			s = (1,3,5,7,9,...), S(x) = x + 3 x^2 + 5 x^3 + 7 x^4 + ...,
p(S(x)) = 1 - ( x + 3 x^2 + 5 x^3 + 7 x^4 + ...)^2,
1/p(S(x)) = 1 + x^2 + 6 x^3 + 20 x^4 + 56 x^5 + ...
T(x) = (-1 + 1/p(S(x)))/x = x + 6 x^2 + 20 x^3 + 56 x^4 + ...
t(s) = (0,1,2,20,56,...).

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

Cf. A005408, A292479.

I:=[0,1,6,20]; [n le 4 select I[n] else 4*Self(n-1)- 5*Self(n-2)+6*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Oct 03 2017

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292480 *)
Join[{0}, LinearRecurrence[{4, -5, 6}, {1, 6, 20}, 30]] (* Vincenzo Librandi, Oct 03 2017 *)

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

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

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

A292324 p-INVERT of (1,0,0,1,0,0,0,0,0,...), where p(S) = (1 - S)^2.

Original entry on oeis.org

2, 3, 4, 7, 12, 19, 28, 42, 64, 97, 144, 212, 312, 459, 672, 979, 1422, 2062, 2984, 4308, 6206, 8925, 12816, 18376, 26310, 37620, 53728, 76648, 109230, 155507, 221184, 314325, 446320, 633249, 897804, 1271993, 1800942, 2548242, 3603468, 5092747, 7193604

Offset: 0

Clark Kimberling, Sep 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).

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

Cf. A079978, A292322.

z = 60; s = x + x^4; p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292324 *)

G.f.: -(((-1 + x) (1 + x) (1 - x + x^2) (2 + x + x^2 + x^3))/(-1 + x + x^4)^2).
a(n) = 2*a(n-1) - a(n-2) + 2*a(n-4) - 2*a(n-5) - a(n-8) for n >= 9.
a(n) = a(n-1)+a(n-4)+A003269(n+2). - R. J. Mathar, Mar 19 2024

A292479 p-INVERT of the positive squares, where p(S) = 1 - S^2.

Original entry on oeis.org

0, 1, 8, 35, 120, 392, 1336, 4725, 16792, 59191, 207536, 727440, 2553264, 8968569, 31502248, 110627195, 388451624, 1364010648, 4789766120, 16819647565, 59063332152, 207403715119, 728306773600, 2557481457440, 8980717116000, 31536219644721, 110740934436168

Offset: 0

Clark Kimberling, Oct 02 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 = (1,4,9,16,...) = A000290, in some cases t(1,4,9,16,...) is a shifted (or differently indexed) version of the cited sequence:
** p(S) ********** t(1, 4, 9, 16,...)
1 - S                A033453
1 - S^2              A292479
1 - S^3              (not yet in OEIS)
(1 - S)^2            (not yet in OEIS)
1 - S - S^2          A289779
1 + S - S^2          (not yet in OEIS)
1 + S - 2 S^2        (not yet in OEIS)
1 + S - 3 S^2        (not yet in OEIS)

			s = (1,4,9,16,25,...), S(x) = x + 4 x^2 + 9 x^3 + 16 x^4 + ...,
p(S(x)) = 1 - (x + 4 x^2 + 9 x^3 + 16 x^4 + ...)^2,
1/p(S(x)) = 1 + x^2 + 8*x^3 + 35*x^4 + 120*x^5 + ...
T(x) = (-1 + 1/p(S(x)))/x = x + 8 x^2 + 35 x^3 + 120 x^4 + ...
t(s) = (0, 1, 8, 35, 120, ...).

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-14,22,-14,6,-1)

Cf. A000290, A292480.

I:=[0,1,8,35,120,392]; [n le 6 select I[n] else 6*Self(n-1)-14*Self(n-2)+22*Self(n-3)-14*Self(n-4)+6*Self(n-5)- Self(n-6): n in [1..30]]; // Vincenzo Librandi, Oct 03 2017

z = 60; s = x (x + 1)/(1 - x)^3; p = 1 - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000290 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292479 *)
LinearRecurrence[{6, -14, 22, -14, 6, -1}, {0, 1, 8, 35, 120, 392}, 30] (* Vincenzo Librandi, Oct 03 2017 *)

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

a(n) = 6*a(n-1) - 14*a(n-2) + 22*a(n-3) - 14*a(n-4) + 6*a(n-5) - a(n-6) for n >= 7.

cp's OEIS Frontend

A289780 p-INVERT of the positive integers (A000027), where p(S) = 1 - S - S^2.

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

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

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A291219 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^3.

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A291382 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2.

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A291728 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^2.

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A292480 p-INVERT of the odd positive integers, where p(S) = 1 - S^2.

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

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