A291728 -id:A291728 - cp's OEIS frontend

A291735 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^3.

1, 1, 3, 5, 10, 19, 35, 67, 124, 234, 441, 827, 1558, 2927, 5503, 10349, 19453, 36580, 68774, 129304, 243119, 457093, 859415, 1615837, 3038024, 5711986, 10739431, 20191855, 37963921, 71378219, 134202491, 252322113, 474405911, 891958973, 1677025407

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 (1, 0, 2, 0, 3, 0, 3, 0, 1)

Cf. A154272, A291728.

z = 60; s = x + x^3; p = 1 - s - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291735 *)
LinearRecurrence[{1,0,2,0,3,0,3,0,1},{1,1,3,5,10,19,35,67,124},40] (* Harvey P. Dale, Aug 25 2024 *)

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

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

A291736 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^2 - S^3.

Original entry on oeis.org

0, 1, 1, 3, 5, 7, 16, 22, 47, 73, 137, 235, 410, 734, 1258, 2255, 3895, 6904, 12056, 21184, 37210, 65172, 114612, 200765, 352779, 618598, 1085950, 1905601, 3343713, 5868895, 10297254, 18073207, 31712887, 55655620, 97666401, 171392667, 300776956, 527817651

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 (0, 1, 1, 2, 3, 1, 3, 0, 1)

Cf. A154272, A291728.

z = 60; s = x + x^3; p = 1 - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291736 *)

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

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

A291737 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^2 - S^3.

Original entry on oeis.org

1, 2, 5, 11, 25, 54, 121, 267, 591, 1310, 2899, 6422, 14218, 31486, 69722, 154389, 341881, 757050, 1676405, 3712200, 8220236, 18202762, 40307892, 89257156, 197649588, 437672056, 969173912, 2146123007, 4752340053, 10523504828, 23303078705, 51601960101

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 (1, 1, 2, 2, 3, 1, 3, 0, 1)

Cf. A154272, A291728.

z = 60; s = x + x^3; p = 1 - s - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291737 *)

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

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

A291738 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^4.

Original entry on oeis.org

1, 1, 2, 4, 6, 13, 23, 43, 76, 138, 244, 444, 795, 1444, 2600, 4705, 8474, 15307, 27583, 49797, 89800, 162088, 292388, 527663, 951922, 1717692, 3098937, 5591589, 10088361, 18202665, 32841990, 59256835, 106914493, 192904396, 348050363, 627980316, 1133045001

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 (1, 0, 1, 1, 0, 4, 0, 6, 0, 4, 0, 1)

Cf. A154272, A291728.

z = 60; s = x + x^3; p = 1 - s - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291738 *)

G.f.: -(((1 + x^2) (1 + x + x^2) (1 + x + x^3) (1 - 2 x + 2 x^2 - x^3 + x^4))/(-1 + x + x^3 + x^4 + 4 x^6 + 6 x^8 + 4 x^10 + x^12)).

a(n) = a(n-1) + a(n-3) + a(n-4) + 4* a(n-6) + 6*a(n-8) + 4*a(n-10) + a(n-12) for n >= 13.

A291739 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^3 - S^6.

Original entry on oeis.org

0, 0, 1, 0, 3, 2, 3, 12, 4, 30, 27, 45, 108, 90, 260, 342, 498, 1115, 1218, 2709, 3913, 5949, 11469, 15262, 28461, 44556, 68028, 123243, 178650, 311337, 498114, 777996, 1340603, 2052765, 3435906, 5569902, 8800392, 14783823, 23242761, 38249550, 62156709

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 (0, 0, 1, 0, 3, 1, 3, 6, 1, 15, 0, 20, 0, 15, 0, 6, 0, 1)

Cf. A154272, A291728.

z = 60; s = x + x^3; p = 1 - s^3 - s^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291739 *)

G.f.: -((x^2 (1 + x^2)^3 (1 + x + x^2) (1 + x + x^3) (1 - 2 x + 2 x^2 - x^3 + x^4))/(-1 + x^3 + 3 x^5 + x^6 + 3 x^7 + 6 x^8 + x^9 + 15 x^10 + 20 x^12 + 15 x^14 + 6 x^16 + x^18)).

a(n) = a(n-3) + 3*a(n-5) + a(n-6) + 3*a(n-7) + 6*a(n-8) + a(n-9) + 15*a(n-10) + 20 *a(n-12) + 15*a(n-14) + 6*a(n-16) + a(n-18) for n >= 19.

A291740 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)(1 - S^2).

Original entry on oeis.org

1, 2, 3, 7, 9, 18, 25, 47, 65, 118, 165, 290, 408, 702, 992, 1677, 2379, 3966, 5643, 9300, 13266, 21654, 30954, 50116, 71770, 115388, 165504, 264475, 379863, 603792, 868267, 1373621, 1977413, 3115222, 4488843, 7045205, 10160427, 15892794, 22937999, 35769390

Offset: 0

Clark Kimberling, Sep 12 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, 0, 2, -3, 1, -3, 0, -1)

Cf. A154272, A291728.

z = 60; s = x + x^3; p = (1 - s) (1 - s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291740 *)

x='x+O('x^99); Vec(((1+x^2)*(1+x-x^2+x^3-2*x^4-x^6))/((-1+x+x^3)^2*(1+x+x^3))) \\ Altug Alkan, Oct 04 2017

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

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

A291741 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)(1 + S^2).

Original entry on oeis.org

1, 0, 1, 1, 1, 4, 5, 7, 11, 12, 19, 30, 42, 68, 98, 137, 205, 292, 429, 644, 936, 1380, 2024, 2936, 4316, 6324, 9260, 13625, 19949, 29216, 42841, 62701, 91917, 134784, 197485, 289547, 424331, 621708, 911255, 1335378, 1957086, 2868620, 4203998, 6161329

Offset: 0

Clark Kimberling, Sep 12 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, 2, -2, 3, -1, 3, 0, 1)

Cf. A154272, A291728.

z = 60; s = x + x^3; p = (1 - s) (1 + s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291741 *)

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

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

A289920 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 2, 3, 6, 12, 22, 42, 80, 151, 287, 544, 1031, 1956, 3708, 7031, 13333, 25280, 47936, 90895, 172350, 326806, 619677, 1175008, 2228011, 4224672, 8010672, 15189552, 28801880, 54613096, 103555397, 196358029, 372327066, 705993241, 1338679088, 2538355336

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, 1, 2, -1, 0, -1)

Cf. A079978, A289918.

z = 60; s = x/(x - x^3); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289920 *)

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

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

A292321 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S^2 - S^3.

Original entry on oeis.org

0, 1, 1, 1, 4, 5, 7, 17, 23, 38, 75, 109, 190, 339, 524, 917, 1563, 2519, 4360, 7305, 12056, 20621, 34407, 57452, 97423, 162672, 272961, 460454, 770281, 1294575, 2177777, 3649129, 6134192, 10306017, 17287962, 29054244, 48790024, 81894794, 137592982

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 (0, 1, 4, 0, -1, -3, 0, 0, 1)

Cf. A079978, A292320.

z = 60; s = x/(x - x^3); p = 1 - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292321 *)

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

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

A292323 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - S)(1 + S^2).

Original entry on oeis.org

1, 0, 0, 2, 1, 0, 5, 6, 1, 11, 23, 10, 22, 71, 57, 50, 191, 243, 164, 474, 860, 676, 1175, 2674, 2758, 3225, 7626, 10256, 10313, 20882, 34642, 36384, 57921, 108270, 130025, 170606, 321415, 448093, 540825, 934958, 1468860, 1798559, 2750605, 4605556, 6042649

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, A292322.

z = 60; s = x/(x - x^3); p = (1 - s)(1 + s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292323 *)

x='x+O('x^99); Vec((1-x+x^2-2*x^3+x^4+x^6)/((1-x-x^3)*(1+x^2-2*x^3+x^6))) \\ Altug Alkan, Oct 05 2017

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

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.

cp's OEIS Frontend

A291735 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^3.

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A291736 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^2 - S^3.

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A291737 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^2 - S^3.

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A291738 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^4.

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A291739 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^3 - S^6.

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A291740 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)(1 - S^2).

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A291741 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)(1 + S^2).

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A289920 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - S^2.

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