A290990 -id:A290990 - 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.

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

A290992 p-INVERT of (0,0,0,1,2,3,4,5,...), the nonnegative integers A000027 preceded by two zeros, where p(S) = 1 - S - S^2.

Original entry on oeis.org

0, 0, 0, 1, 2, 3, 4, 7, 14, 27, 48, 82, 140, 242, 420, 726, 1250, 2153, 3720, 6446, 11184, 19408, 33676, 58431, 101378, 175861, 304988, 528800, 916714, 1589091, 2754612, 4775074, 8277754, 14350253, 24878304, 43131381, 74777890, 129645147, 224770632

Offset: 0

Clark Kimberling, Aug 21 2017

See A290890 for a guide to related sequences.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,0,-2,1,0,1).

Cf. A000027, A033453, A289780, A290890, A290990, A290991.

R:=PowerSeriesRing(Integers(), 60); [0,0,0] cat Coefficients(R!( x^3*(1-2*x+x^2+x^4)/(1-4*x+6*x^2-4*x^3+2*x^5-x^6-x^8) )); // G. C. Greubel, Apr 12 2023

z = 60; s = x^4/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0,0,0,1,2,3,4,5,... *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290992 *)

concat(vector(3), Vec(x^3*(1 - 2*x + x^2 + x^4) / (1 - 4*x + 6*x^2 - 4*x^3 + 2*x^5 - x^6 - x^8) + O(x^50))) \\ Colin Barker, Aug 24 2017

def f(x): return x^3*(1-2*x+x^2+x^4)/(1-4*x+6*x^2-4*x^3+2*x^5-x^6-x^8)
def A290992_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( f(x) ).list()
A290992_list(60) # G. C. Greubel, Apr 12 2023

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - 2*a(n-5) + a(n-6) + a(n-8).

G.f.: x^3*(1 - 2*x + x^2 + x^4) / (1 - 4*x + 6*x^2 - 4*x^3 + 2*x^5 - x^6 - x^8). - Colin Barker, Aug 24 2017

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

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

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.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A290992 p-INVERT of (0,0,0,1,2,3,4,5,...), the nonnegative integers A000027 preceded by two zeros, where p(S) = 1 - S - S^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula