A290890 -id:A290890 - cp's OEIS frontend

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

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.

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

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

A113067 Expansion of -x/((x^2+x+1)(x^2+3x+1)); invert transform gives signed version of tetrahedral numbers A000292.

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

Creighton Dement, Oct 13 2005

Invert(a(n)) gives (0, -1, 4, -10, 20, -35, ...) = A000292 (with alternating signs).
Binomial(a(n)) gives (0, -1, 2, -2, 4, -7, 10, ...) = A094686 (with alternating signs, from 2nd term).
Floretion Algebra Multiplication Program, FAMP Code: 2basei[C*F]; C = - .5'j + .5'k - .5j' + .5k' - 'ii' - .5'ij' - .5'ik' - .5'ji' - .5'ki'; F = + .5'i + .5'ii' + .5'ij' + .5'ik'

Creighton Dement, Floretion Integer Sequences (work in progress).

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

Cf. A000292, A113067, A113068, A094686, A001906, A109961, A290890.

-x/((x^2+x+1)*(x^2+3*x+1)) + O[x]^30 // CoefficientList[#, x]& (* Jean-François Alcover, Jun 15 2017 *)

concat(0, Vec(-x / ((1 + x + x^2)*(1 + 3*x + x^2)) + O(x^30))) \\ Colin Barker, May 11 2019

[((lucas_number1(n,3,1)-lucas_number1(n,1,1)))/(-2) for n in range(1,32)] # Zerinvary Lajos, Jul 06 2008

a(n) + a(n+1) + a(n+2) = (-1)^n *A001906(n+2) = (-1)^n*F(2n+4).
a(n) + 3*a(n+1) + 3*a(n+2) + a(n+3) = ((-1)^(n+1))*A109961(n+2).
(|a(n)|) = A290890(n) for n >= 0, this being the p-INVERT of (1,2,3,4,...), where p(S) = 1 - S^2.  - Clark Kimberling, Aug 21 2017
a(n) = -4*a(n-1) - 5*a(n-2) - 4*a(n-3) - a(n-4) for n > 3. - Colin Barker, May 11 2019
2*a(n) = (-1)^n*A001906(n+1) - A049347(n). - R. J. Mathar, Sep 20 2020

A289976 p-INVERT of (0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by two zeros, where p(S) = 1 - S - S^2.

Original entry on oeis.org

0, 0, 1, 1, 2, 5, 9, 18, 36, 70, 137, 268, 522, 1017, 1980, 3852, 7492, 14568, 28321, 55051, 106999, 207952, 404134, 785366, 1526186, 2965752, 5763103, 11198858, 21761463, 42286357, 82169547, 159668921, 310262351, 602888757, 1171506956, 2276419286

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, 1, -1, -2, -1, 1)

Cf. A000045, A289975, A289780.

z = 60; s = x^3/(1 - x - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0,0,1,2,3,5,... *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289976 *)

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

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

A290902 p-INVERT of the positive integers, where p(S) = 1 - 3*S.

Original entry on oeis.org

3, 15, 72, 345, 1653, 7920, 37947, 181815, 871128, 4173825, 19997997, 95816160, 459082803, 2199597855, 10538906472, 50494934505, 241935766053, 1159183895760, 5553983712747, 26610734667975, 127499689627128, 610887713467665, 2926938877711197, 14023806675088320

Offset: 0

Clark Kimberling, Aug 17 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.

			s = (1,2,3,4,...), p(S) = 1-3*S;
S(x) = x + 2 x^2 + 3 x^3 + ... ;
p(S(x)) = 1 - 3(x + 2 x^2 + 3 x^3 + ...);
1/p(S(x)) = 1 + 3 x + 15 x^2 + 72 x^3 + ... ;
(-p(0) + 1/p(S(x)))/x = 3 + 15 x + 72 x^2 + ... ;
t(s) = (3, 15, 72, ...), with offset 0.

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

Cf. A000027, A004254, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 3 s;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290902 *)
u/3 (* A004254 shifted *)

G.f.: 3/(1 - 5 x + x^2).
a(n) = 5*a(n-1) - a(n-2).
a(n) = 3*A004254(n+1) for n >= 0.

A290903 p-INVERT of the positive integers, where p(S) = 1 - 5*S.

Original entry on oeis.org

5, 35, 240, 1645, 11275, 77280, 529685, 3630515, 24883920, 170556925, 1169014555, 8012544960, 54918800165, 376419056195, 2580014593200, 17683683096205, 121205767080235, 830756686465440, 5694091038177845, 39027880580779475, 267501073027278480

Offset: 0

Clark Kimberling, Aug 17 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.

			(See the example at A290902.)

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

Cf. A000027, A004187, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 5 s;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290903 *)
u/5 (* A004187 shifted *)

G.f.: 5/(1 - 7 x + x^2).
a(n) = 7*a(n-1) - a(n-2).
a(n) = 5*A004187(n+1) for n >= 0.

cp's OEIS Frontend

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

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

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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.

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A113067 Expansion of -x/((x^2+x+1)*(x^2+3*x+1)); invert transform gives signed version of tetrahedral numbers A000292.

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A289976 p-INVERT of (0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by two zeros, where p(S) = 1 - S - S^2.

A113067 Expansion of -x/((x^2+x+1)(x^2+3x+1)); invert transform gives signed version of tetrahedral numbers A000292.