A291728 -id:A291728 - cp's OEIS frontend

A291038 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - 2 S)^2.

4, 12, 32, 84, 216, 544, 1348, 3300, 8000, 19236, 45936, 109056, 257604, 605820, 1419232, 3313396, 7711944, 17900320, 41445764, 95746260, 220735616, 507934276, 1166792160, 2676017408, 6128381316, 14015556588, 32012831648, 73033858964, 166434905016

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 (4, -4, 2, -4, 0, -1)

Cf. A079978, A290616, A291039.

z = 60; s = x/(x - x^3); p = (1 - 2 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291038 *)
u/4 (* A291039 *)

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

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

A291723 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^3.

Original entry on oeis.org

0, 0, 1, 0, 3, 1, 3, 6, 2, 15, 9, 21, 36, 27, 85, 72, 141, 222, 231, 513, 540, 945, 1422, 1741, 3222, 3876, 6337, 9339, 12447, 20809, 27135, 42546, 62195, 86709, 136866, 187278, 286113, 417303, 595852, 910431, 1281810, 1926984, 2810883, 4064571, 6097464

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).

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, 0, 3, 0, 1)

Cf. A154272, A291728.

z = 60; s = x + x^3; p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291723 *)
LinearRecurrence[{0,0,1,0,3,0,3,0,1},{0,0,1,0,3,1,3,6,2},60] (* Harvey P. Dale, Jun 07 2022 *)

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))).

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

A291724 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^5.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 5, 0, 10, 1, 10, 10, 5, 45, 2, 120, 15, 210, 105, 253, 455, 230, 1365, 310, 3004, 1185, 5030, 4855, 6735, 15506, 8735, 38790, 17655, 77955, 56134, 130030, 178500, 195365, 481750, 327263, 1088225, 761775, 2095350, 2162550, 3593394, 5940325

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).

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, 0, 0, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 1)

Cf. A154272, A291728.

z = 60; s = x + x^3; p = 1 - s^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291724 *)
LinearRecurrence[{0, 0, 0, 0, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 1}, {0, 0, 0, 0, 1, 0, 5, 0, 10, 1, 10, 10, 5, 45, 2}, 50] (* Vincenzo Librandi, Sep 10 2017 *)

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

a(n) = a(n-5) + 5*a(n-7) + 10*a(n-9) + 10*a(n-11) + 5*a(n-13) + a(n-15) for n >= 15.

A291725 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^2.

Original entry on oeis.org

2, 3, 6, 11, 18, 30, 50, 81, 130, 208, 330, 520, 816, 1275, 1984, 3077, 4758, 7337, 11286, 17322, 26532, 40563, 61908, 94336, 143540, 218112, 331008, 501749, 759726, 1149159, 1736534, 2621751, 3954826, 5960902, 8977686, 13511461, 20320854, 30542064, 45875998

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).

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. A154272, A291728.

z = 60; s = x + x^3; p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291725 *)
LinearRecurrence[{2, -1, 2, -2, 0, -1}, {2, 3, 6, 11, 18, 30}, 40] (* Vincenzo Librandi, Sep 10 2017 *)

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

A291726 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^3.

Original entry on oeis.org

3, 6, 13, 27, 51, 94, 171, 303, 527, 906, 1539, 2586, 4308, 7122, 11692, 19077, 30957, 49986, 80349, 128628, 205146, 326058, 516594, 816076, 1285674, 2020380, 3167464, 4954887, 7734993, 12051616, 18743037, 29099781, 45106223, 69810162, 107887629, 166505313

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).

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 (3, -3, 4, -6, 3, -3, 3, 0, 1)

Cf. A154272, A291728.

z = 60; s = x + x^3; p = (1 - s)^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291726 *)

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

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

A291727 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^4.

Original entry on oeis.org

4, 10, 24, 55, 116, 234, 460, 879, 1640, 3006, 5424, 9650, 16964, 29510, 50852, 86893, 147360, 248198, 415440, 691428, 1144772, 1886270, 3094292, 5055140, 8227084, 13341756, 21564360, 34746331, 55823080, 89439056, 142928424, 227851285, 362396564, 575135150

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).

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, -6, 8, -13, 12, -10, 12, -6, 4, -4, 0, -1)

Cf. A154272, A291728.

z = 60; s = x + x^3; p = (1 - s)^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291727 *)
LinearRecurrence[{4,-6,8,-13,12,-10,12,-6,4,-4,0,-1},{4,10,24,55,116,234,460,879,1640,3006,5424,9650},40] (* Harvey P. Dale, Mar 13 2025 *)

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

a(n) = 4*a(n-1) - 6*a(n-2) + 8*a(n-3) - 13*a(n-4) + 12*a(n-5) - 10*a(n-6) + 12*a(n-7) - 6*a(n-8) +4*a(n-9) - 4*a(n-10) - a(n-12) for n >= 13.

A291729 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2.

Original entry on oeis.org

2, 5, 14, 39, 106, 290, 794, 2173, 5946, 16272, 44530, 121860, 333480, 912597, 2497400, 6834349, 18702782, 51181767, 140063294, 383295214, 1048920220, 2870460125, 7855260268, 21496593296, 58827270844, 160985870984, 440551640160, 1205607339709, 3299247863502

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, 1, 2, 2, 0, 1)

Cf. A154272, A291728.

z = 60; s = x + x^3; p = 1 - 2 s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291729 *)

G.f.: (-2 - x - 2 x^2 - 2 x^3 - x^5)/(-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.

A291731 a(n) = (1/2)*A291730(n).

Original entry on oeis.org

1, 3, 9, 28, 84, 255, 772, 2340, 7088, 21476, 65064, 197126, 597228, 1809420, 5481980, 16608712, 50319264, 152451844, 461882016, 1399359936, 4239628608, 12844765920, 38915675520, 117902483528, 357208128400, 1082230358448, 3278823900048, 9933824239744

Offset: 0

Clark Kimberling, Sep 11 2017

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, A291730.

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

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

A291733 a(n) = (1/4)*A291732(n).

Original entry on oeis.org

1, 3, 9, 26, 72, 195, 520, 1368, 3560, 9184, 23520, 59860, 151536, 381840, 958256, 2396192, 5972736, 14845040, 36801792, 91021056, 224642304, 553347072, 1360598016, 3340024384, 8186748160, 20038426368, 48983457024, 119593531904, 291657627648, 710522702592

Offset: 0

Clark Kimberling, Sep 11 2017

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, A291732.

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.: -(((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.

A291734 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 2 S).

Original entry on oeis.org

3, 7, 18, 45, 108, 258, 615, 1459, 3453, 8164, 19287, 45540, 107496, 253695, 598659, 1412587, 3332970, 7863853, 18553752, 43774722, 103279023, 243668295, 574890057, 1356344056, 3200033343, 7549859464, 17812425600, 42024945087, 99149648967, 233924207559

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 (3, -2, 3, -4, 0, -2)

Cf. A154272, A291728.

z = 60; s = x + x^3; p = (1 - s) (1 - 2 s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291734 *)

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

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

cp's OEIS Frontend

A291038 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - 2 S)^2.

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A291723 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^3.

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A291724 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^5.

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A291725 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^2.

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A291726 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^3.

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A291727 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^4.

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A291729 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2.

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A291731 a(n) = (1/2)*A291730(n).

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