A292480 -id:A292480 - cp's OEIS frontend

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

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.

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.

A292481 p-INVERT of the odd positive integers, where p(S) = 1 - S^3.

Original entry on oeis.org

0, 0, 1, 9, 42, 139, 381, 984, 2685, 8061, 25434, 79695, 242577, 721584, 2131785, 6333633, 18984618, 57194883, 172319157, 517851144, 1552599333, 4651054101, 13939132698, 41810229351, 125475990057, 376585031520, 1129975049169, 3389800055481, 10168040440746

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).

See A292480 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, -15, 21, -12, 9)

Cf. A005408, A292480.

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292481 *)
LinearRecurrence[{6,-15,21,-12,9},{0,0,1,9,42,139},30] (* Harvey P. Dale, Jun 06 2024 *)

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

a(n) = 6*a(n-1) - 25*a(n-2) + 21*a(n-3) - 12*a(n-4) + 9*a(n-5) for n >= 6.

A292482 p-INVERT of the odd positive integers, where p(S) = (1 - S)^2.

Original entry on oeis.org

2, 9, 32, 112, 384, 1296, 4320, 14256, 46656, 151632, 489888, 1574640, 5038848, 16061328, 51018336, 161558064, 510183360, 1607077584, 5050815264, 15841193328, 49589822592, 154968195600, 483500770272, 1506290861232, 4686238234944, 14560811658576

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).

See A292480 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, -9)

Cf. A005408, A292480.

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292482 *)

G.f.: ((1 + x) (2 - 5 x + x^2))/(-1 + 3 x)^2.
a(n) = 6*a(n-1) - 9*a(n-2) for n >= 3.
a(n) = 16*3^(n-3)*(4 + n) for n>1. - Colin Barker, Oct 03 2017

A292483 p-INVERT of the odd positive integers, where p(S) = (1 - S)^3.

Original entry on oeis.org

3, 15, 61, 240, 912, 3376, 12240, 43632, 153360, 532656, 1831248, 6240240, 21100176, 70858800, 236510928, 785115504, 2593432080, 8528565168, 27932538960, 91144257264, 296391022992, 960802812720, 3105562639824, 10010945435760, 32189993590032, 103264606820016

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).

See A292480 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 (9, -27, 27)

Cf. A005408, A292480.

z = 60; s = x (x + 1)/(1 - x)^2; p = (1 - s)^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292483 *)

G.f.: -(((1 + x) (3 - 15 x + 22 x^2 - 7 x^3 + x^4))/(-1 + 3 x)^3).
a(n) = 9*a(n-1) - 27*a(n-2) + 27*a(n-3)  for n >= 6.
a(n) = 16*3^(n-5)*(51 + 22*n + 2*n^2) for n>2. - Colin Barker, Oct 03 2017

A292484 p-INVERT of the odd positive integers, where p(S) = 1 + S - S^2.

Original entry on oeis.org

-1, -1, 4, 9, 5, 8, 63, 183, 348, 745, 2061, 5456, 12991, 30831, 76660, 192137, 472597, 1155032, 2843007, 7024935, 17315404, 42592489, 104847389, 258355104, 636507775, 1567442143, 3859933668, 9507231753, 23417547813, 57675809960, 142047927231, 349856144791

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).

See A292480 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 (3, -4, 7, -1)

Cf. A005408, A292480.

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 + s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292484 *)
LinearRecurrence[{3,-4,7,-1},{-1,-1,4,9},40] (* Harvey P. Dale, Sep 22 2024 *)

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

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

A292485 p-INVERT of the odd positive integers, where p(S) = 1 - S - 2 S^2.

Original entry on oeis.org

1, 6, 28, 120, 504, 2128, 9016, 38208, 161864, 685648, 2904408, 12303264, 52117544, 220773552, 935211704, 3961620096, 16781691912, 71088388112, 301135245080, 1275629368416, 5403652717288, 22890240236144, 96964613663352, 410748694893888, 1739959393240264

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).

See A292480 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 (5, -5, 7, 2)

Cf. A005408, A292480.

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292485 *)
LinearRecurrence[{5,-5,7,2},{1,6,28,120},30] (* Harvey P. Dale, Oct 14 2023 *)

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

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

A292486 p-INVERT of the odd positive integers, where p(S) = 1 - S - 3 S^2.

Original entry on oeis.org

1, 7, 36, 165, 747, 3420, 15705, 72063, 330516, 1515933, 6953283, 31893516, 146289393, 671000247, 3077745156, 14117009877, 64751939163, 297004363452, 1362300384969, 6248602953135, 28661108314356, 131462846314317, 602994126047283, 2765815028667756

Offset: 0

Clark Kimberling, Oct 03 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 A292480 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 (5, -4, 9, 3)

Cf. A005408, A292480.

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 3 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292486 *)
LinearRecurrence[{5,-4,9,3},{1,7,36,165},30] (* Harvey P. Dale, Sep 29 2024 *)

x='x+O('x^99); Vec(((1+x)*(1+x+4*x^2))/(1-5*x+4*x^2-9*x^3-3*x^4)) \\ Altug Alkan, Oct 03 2017

G.f.: -(((1 + x) (1 + x + 4 x^2))/(-1 + 5 x - 4 x^2 + 9 x^3 + 3 x^4)).

a(n) = 5*a(n-1) - 4*a(n-2) + 9*a(n-3) + 3*a(n-4) for n >= 5.

A292487 p-INVERT of the odd positive integers, where p(S) = 1 - S - 4 S^2.

Original entry on oeis.org

1, 8, 44, 212, 1020, 4980, 24348, 118868, 580156, 2831924, 13824092, 67481876, 329408892, 1607991540, 7849328028, 38316090836, 187038012604, 913016364980, 4456842098396, 21755843899028, 106200025265148, 518409923170932, 2530591191342108, 12352949840710484

Offset: 0

Clark Kimberling, Oct 03 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 A292480 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 (5, -3, 11, 4)

Cf. A005408, A292480.

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 4 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292487 *)

x='x+O('x^99); Vec(((1+x)*(1+2*x+5*x^2))/(1-5*x+3*x^2-11*x^3-4*x^4)) \\ Altug Alkan, Oct 03 2017

G.f.: -(((1 + x) (1 + 2 x + 5 x^2))/(-1 + 5 x - 3 x^2 + 11 x^3 + 4 x^4)).

a(n) = 5*a(n-1) - 3*a(n-2) + 11*a(n-3) + 4*a(n-4) for n >= 5.

A292488 p-INVERT of the odd positive integers, where p(S) = 1 - S - 5 S^2.

Original entry on oeis.org

1, 9, 52, 261, 1323, 6814, 35077, 180261, 926348, 4761289, 24472527, 125783886, 646502873, 3322895889, 17079026852, 87782799261, 451186103523, 2319006747614, 11919233055677, 61262485125261, 314876977751548, 1618404981969089, 8318279426249127

Offset: 0

Clark Kimberling, Oct 03 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 A292480 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 (5, -2, 13, 5)

Cf. A005408, A292480.

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 5 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292488 *)
LinearRecurrence[{5,-2,13,5},{1,9,52,261},30] (* Harvey P. Dale, Jul 31 2025 *)

x='x+O('x^99); Vec(((1+x)*(1+3*x+6*x^2))/(1-5*x+2*x^2-13*x^3-5*x^4)) \\ Altug Alkan, Oct 03 2017

G.f.: -(((1 + x) (1 + 3 x + 6 x^2))/(-1 + 5 x - 2 x^2 + 13 x^3 + 5 x^4)).

a(n) = 5*a(n-1) - 2*a(n-2) + 13*a(n-3) + 5*a(n-4) for n >= 5.

cp's OEIS Frontend

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

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

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A292481 p-INVERT of the odd positive integers, where p(S) = 1 - S^3.

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A292482 p-INVERT of the odd positive integers, where p(S) = (1 - S)^2.

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A292483 p-INVERT of the odd positive integers, where p(S) = (1 - S)^3.

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A292484 p-INVERT of the odd positive integers, where p(S) = 1 + S - S^2.

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A292485 p-INVERT of the odd positive integers, where p(S) = 1 - S - 2 S^2.

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A292486 p-INVERT of the odd positive integers, where p(S) = 1 - S - 3 S^2.

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