A291728 -id:A291728 - 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.

A290616 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - 2 S - S^2.

Original entry on oeis.org

2, 5, 12, 31, 80, 205, 526, 1350, 3464, 8889, 22810, 58532, 150198, 385420, 989018, 2537899, 6512450, 16711463, 42882940, 110041025, 282373998, 724594076, 1859365870, 4771280299, 12243483684, 31417750230, 80620439004, 206878440932, 530866488090

Offset: 0

Clark Kimberling, Sep 14 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 A291728 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, 2, -2, 0, -1)

Cf. A079978, A289918, A291035.

z = 60; s = x/(x - x^3); p = 1 - 2 s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290616 *)

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

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

A289918 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - S^2).

Original entry on oeis.org

0, 1, 0, 1, 2, 1, 4, 4, 6, 11, 12, 22, 30, 42, 68, 91, 140, 205, 292, 443, 634, 936, 1380, 1999, 2960, 4316, 6324, 9300, 13576, 19949, 29216, 42785, 62790, 91917, 134784, 197548, 289402, 424331, 621708, 911218, 1335586, 1957086, 2868620, 4203927, 6161084

Offset: 0

Clark Kimberling, Sep 14 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 A291728 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 (0, 1, 2, 0, 0, -1)

Cf. A079978, A289919.

z = 60; s = x/(x - x^3); p = (1 - s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289918 *)

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

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

Recurrence corrected by Colin Barker, Sep 14 2017

A292322 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - S^2 - S^3.

Original entry on oeis.org

1, 2, 4, 8, 17, 36, 73, 152, 317, 653, 1355, 2812, 5818, 12061, 25001, 51786, 107323, 222409, 460824, 954942, 1978840, 4100398, 8496827, 17606974, 36484494, 75602461, 156661630, 324629762, 672690133, 1393931744, 2888469094, 5985414154, 12402824741

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).

See A291728 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 (1, 1, 4, -2, -1, -3, 1, 0, 1)

Cf. A079978, A292320.

z = 60; s = x/(x - x^3); p = 1 - s - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292322 *)

G.f.: -((1 + x + x^2 - 2 x^3 - x^4 + x^6)/(-1 + x + x^2 + 4 x^3 - 2 x^4 - x^5 - 3 x^6 + x^7 + x^9)).

a(n) = a(n-1) + a(n-2) + 4*a(n-3) - 2*a(n-4) - a(n-5) - 3*a(n-6) + a(n-7) + a(n-9) for n >= 10.

A291730 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - 2 S - 2 S^2.

Original entry on oeis.org

2, 6, 18, 56, 168, 510, 1544, 4680, 14176, 42952, 130128, 394252, 1194456, 3618840, 10963960, 33217424, 100638528, 304903688, 923764032, 2798719872, 8479257216, 25689531840, 77831351040, 235804967056, 714416256800, 2164460716896, 6557647800096

Offset: 0

Clark Kimberling, Sep 11 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 A291728 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, 2, 4, 0, 2)

Cf. A154272, A291728, A291731.

z = 60; s = x + x^3; p = 1 - 2 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291730 *)
u / 2  (* A291731 *)

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

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

A291732 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - 2 S)^2.

Original entry on oeis.org

4, 12, 36, 104, 288, 780, 2080, 5472, 14240, 36736, 94080, 239440, 606144, 1527360, 3833024, 9584768, 23890944, 59380160, 147207168, 364084224, 898569216, 2213388288, 5442392064, 13360097536, 32746992640, 80153705472, 195933828096, 478374127616

Offset: 0

Clark Kimberling, Sep 11 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 A291728 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, -4, 4, -8, 0, -4)

Cf. A154272, A291728, A291733.

z = 60; s = x + x^3; p = (1 - 2 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291732 *)
u / 4  (*A291733)

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

a(n) = 4*a(n-1) - 4*a(n-2) + 4*a(n-3) - 8*a(n-4) - 4*a(n-6) for n >= 7.

A292320 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - S^3.

Original entry on oeis.org

1, 1, 2, 4, 6, 12, 22, 36, 67, 122, 209, 377, 681, 1193, 2130, 3823, 6764, 12043, 21531, 38252, 68076, 121456, 216126, 384691, 685636, 1220767, 2173346, 3871747, 6894873, 12276852, 21866387, 38941846, 69344928, 123500513, 219943018, 391676701, 697538335

Offset: 0

Clark Kimberling, Sep 14 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 A291728 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 (1, 0, 4, -2, 0, -3, 1, 0, 1)

Cf. A079978, A290616.

z = 60; s = x/(x - x^3); p = 1 - s - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292320 *)

G.f.: -((1 + x^2 - 2 x^3 + x^6)/(-1 + x + 4 x^3 - 2 x^4 - 3 x^6 + x^7 + x^9)).

a(n) = a(n-1) + 4*a(n-3) - 2*a(n-4) - 3*a(n-6) + a(n-7) + a(n-9) for n >= 10.

A289919 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - S)^2.

Original entry on oeis.org

2, 3, 4, 7, 12, 19, 30, 48, 76, 119, 186, 290, 450, 696, 1074, 1653, 2538, 3889, 5948, 9081, 13842, 21068, 32022, 48609, 73700, 111618, 168868, 255232, 385410, 581479, 876576, 1320411, 1987516, 2989583, 4493910, 6750968, 10135584, 15208443, 22807902

Offset: 0

Clark Kimberling, Sep 14 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 A291728 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, 2, -2, 0, -1)

Cf. A079978, A289918.

z = 60; s = x/(x - x^3); p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289919 *)

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

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

A291035 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - 2 S^2.

Original entry on oeis.org

1, 3, 5, 12, 27, 58, 130, 285, 629, 1389, 3060, 6753, 14892, 32844, 72445, 159775, 352401, 777244, 1714259, 3780930, 8339090, 18392449, 40565829, 89470733, 197333940, 435233685, 959938112, 2117210180, 4669654005, 10299246171, 22715702489, 50101058976

Offset: 0

Clark Kimberling, Sep 14 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 A291728 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 (1,2,2,-1,0,-1).

Cf. A079978, A289918, A290616.

z = 60; s = x/(x - x^3); p = 1 - s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291035 *)

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

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

A291036 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - 2 S - 2 S^2.

Original entry on oeis.org

2, 6, 16, 46, 132, 376, 1074, 3066, 8752, 24986, 71328, 203624, 581298, 1659462, 4737360, 13524006, 38607732, 110215648, 314638754, 898216794, 2564189568, 7320134930, 20897197344, 59656394448, 170304435554, 486177568038, 1387918211824, 3962167507006

Offset: 0

Clark Kimberling, Sep 14 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 A291728 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, 2, -2, 0, -1)

Cf. A079978, A290616, A291037.

z = 60; s = x/(x - x^3); p = 1 - 2 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291036 *)
u/2 (* A291037 *)

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

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

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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A290616 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - 2 S - S^2.

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A289918 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - S^2).

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A292322 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - S^2 - S^3.

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A291730 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - 2 S - 2 S^2.

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A291732 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - 2 S)^2.

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A292320 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - S^3.

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A289919 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - S)^2.

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