A291219 -id:A291219 - cp's OEIS frontend

A291231 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S)(1 - 3 S)(1 - 4 S).

10, 65, 360, 1831, 8830, 41104, 186710, 833401, 3672840, 16034303, 69506930, 299700192, 1287010850, 5509712833, 23531008200, 100312445063, 427025152550, 1815832379312, 7714875191470, 32756357939033, 139008007848360, 589672772732671, 2500620567692890

Offset: 0

Clark Kimberling, Aug 26 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 A291219 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 (10, -31, 20, 40, -20, -31, -10, -1)

Cf. A000035, A054447, A291219.

z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 2 s)(1 - 3 s)(1 - 4 s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291231 *)

G.f.: (10 - 35 x + 20 x^2 + 46 x^3 - 20 x^4 - 35 x^5 - 10 x^6)/(1 - 10 x + 31 x^2 - 20 x^3 - 40 x^4 + 20 x^5 + 31 x^6 + 10 x^7 + x^8).

a(n) = 10*a(n-1) - 31*a(n-2) + 20*a(n-3) + 40*a(n-4) - 20*a(n-5) -31*a(n-6) - 10*a(n-7) - a(n-8) for n >= 7.

A291233 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3.

Original entry on oeis.org

1, 2, 5, 11, 26, 58, 134, 303, 693, 1576, 3593, 8184, 18645, 42480, 96773, 220481, 502290, 1144350, 2607062, 5939501, 13531493, 30827806, 70232669, 160005808, 364529269, 830479602, 1892019493, 4310445875, 9820165646, 22372546322, 50969693930, 116120429167

Offset: 0

Clark Kimberling, Aug 26 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 A291219 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, 4, -1, -4, 1, 1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - s - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291233 *)

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

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

A291234 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4.

Original entry on oeis.org

1, 2, 5, 12, 28, 67, 156, 370, 866, 2044, 4799, 11304, 26574, 62547, 147108, 346149, 814270, 1915795, 4506952, 10603417, 24945414, 58687660, 138068915, 324824928, 764187814, 1797846170, 4229645000, 9950753025, 23410332344, 55075627972, 129572006209

Offset: 0

Clark Kimberling, Aug 26 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 A291219 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, 5, -2, -7, 2, 5, -1, -1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - s - s^2 - s^3 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291234 *)

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

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

A291235 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 - S^5.

Original entry on oeis.org

1, 2, 5, 12, 29, 69, 166, 394, 944, 2245, 5365, 12781, 30506, 72734, 173520, 413838, 987130, 2354465, 5615889, 13395047, 31949764, 76206828, 181768094, 433554067, 1034112065, 2466566144, 5883251633, 14032736684, 33470882601, 79834762768, 190421890053

Offset: 0

Clark Kimberling, Aug 28 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 A291219 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, 6, -3, -12, 5, 12, -3, -6, 1, 1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - s - s^2 - s^3 - s^4 - s^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291235 *)

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

a(n) = a(n-1) + 6*a(n-2) - 3*a(n-3) - 12*a(n-4) + 5*a(n-5) + 12*a(n-6) - 3*a(n-7) - 6*a(n-8) + a(n-9) + a(n-10) for n >= 11.

A291236 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 3 S).

Original entry on oeis.org

4, 13, 44, 147, 488, 1616, 5344, 17661, 58348, 192739, 636620, 2102688, 6944828, 22937405, 75757420, 250210275, 826389232, 2729379568, 9014530520, 29772975309, 98333463212, 324773375891, 1072653608596, 3542734230336, 11700856345972, 38645303343277

Offset: 0

Clark Kimberling, Aug 28 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 A291219 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, -1, -4, -1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 3s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291236 *)

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

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

A291237 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S)(1 - 4 S).

Original entry on oeis.org

7, 35, 162, 721, 3139, 13504, 57707, 245671, 1043634, 4428053, 18774815, 79573152, 337178159, 1428553243, 6052037010, 25638260873, 108608846171, 460082737472, 1948961747155, 8255982722783, 34973020586946, 148148373971341, 627567262233463, 2658419223345984

Offset: 0

Clark Kimberling, Aug 28 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 A291219 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 (7, -11, -6, 11, 7, 1).

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 2s)(1 - 4 s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291237 *)

G.f.: (-7 + 14 x + 6 x^2 - 14 x^3 - 7 x^4)/(-1 + 7 x - 11 x^2 - 6 x^3 + 11 x^4 + 7 x^5 + x^6).

a(n) = 7*a(n-1) - 11*a(n-2) - 6*a(n-3) + 11*a(n-4) + 7*a(n-5) + a(n-6) for n >= 7.

A291238 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^2 (1 - 2 S).

Original entry on oeis.org

4, 11, 30, 79, 202, 508, 1262, 3109, 7614, 18569, 45152, 109560, 265448, 642463, 1553790, 3755843, 9075302, 21923060, 52949458, 127869209, 308767326, 745537309, 1800065788, 4346043888, 10492781068, 25332654899, 61159842270, 147655261111, 356475233986

Offset: 0

Clark Kimberling, Aug 28 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 A291219 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, -2, -6, 2, 4, 1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = (1 - s)^2(1 - 2s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291238 *)

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

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

A291239 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S^2) (1 - 2 S).

Original entry on oeis.org

2, 5, 12, 31, 74, 184, 442, 1081, 2604, 6323, 15250, 36912, 89074, 215293, 519660, 1255223, 3030106, 7317032, 17664170, 42649553, 102963276, 248587051, 600137378, 1448890464, 3497918306, 8444802101, 20387522508, 49220043535, 118827609578, 286875776920

Offset: 0

Clark Kimberling, Aug 28 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 A291219 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, 4, -6, -4, 2, 1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = (1 - s^2)(1 - 2s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291239 *)

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

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

A291240 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S^3)^2.

Original entry on oeis.org

0, 0, 2, 0, 6, 3, 12, 18, 24, 63, 66, 173, 222, 438, 722, 1146, 2142, 3213, 5958, 9327, 16210, 26898, 44400, 75875, 123252, 210240, 344160, 578052, 958200, 1588404, 2650008, 4370292, 7285684, 12022704, 19960488, 33008505, 54594504, 90368550, 149168350

Offset: 0

Clark Kimberling, Aug 28 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 A291219 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 (0,6,2,-15,-6,19,6,-15,-2,6,0,-1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = (1 - s^3)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291240 *)
LinearRecurrence[{0, 6, 2, -15, -6, 19, 6, -15, -2, 6, 0, -1}, {0, 0, 2, 0, 6, 3, 12, 18, 24, 63, 66, 173}, 40] (* Vincenzo Librandi, Aug 29 2017 *)

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

a(n) = 6*a(n-2) + 2*a(n-3) - 15*a(n-4) - 6*a(n-5) + 19*a(n-6) + 6*a(n-7) - 15*a(n-8) - 2*a(n-9) + 6*a(n-10) - a(n-12) for n >= 13.

A291241 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 + S^3.

Original entry on oeis.org

1, 2, 3, 7, 10, 22, 32, 67, 99, 200, 299, 588, 887, 1708, 2595, 4913, 7508, 14018, 21526, 39725, 61251, 111922, 173173, 313752, 486925, 875702, 1362627, 2434747, 3797374, 6746350, 10543724, 18636343, 29180067, 51340988, 80521055, 141089508, 221610563

Offset: 0

Clark Kimberling, Aug 28 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 A291219 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 (1,4,-3,-4,1,1)

Cf. A000035, A291219.

I:=[1,2,3,7,10,22]; [n le 6 select I[n] else Self(n-1)+4*Self(n-2)-3*Self(n-3)-4*Self(n-4)+Self(n-5)+Self(n-6): n in [1..40]]; // Vincenzo Librandi, Aug 29 2017

z = 60; s = x/(1 - x^2); p = 1 - s - s^2 + s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291241 *)
LinearRecurrence[{1, 4, -3, -4, 1, 1}, {1, 2, 3, 7, 10, 22}, 40] (* Vincenzo Librandi, Aug 29 2017 *)

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

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

cp's OEIS Frontend

A291231 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S)(1 - 3 S)(1 - 4 S).

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A291233 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3.

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A291234 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4.

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A291235 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 - S^5.

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A291236 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 3 S).

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A291237 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S)(1 - 4 S).

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A291238 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^2 (1 - 2 S).

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A291239 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S^2) (1 - 2 S).

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