A291219 -id:A291219 - cp's OEIS frontend

A291218 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^5.

0, 0, 0, 0, 1, 0, 5, 0, 15, 1, 35, 10, 70, 55, 127, 220, 225, 715, 450, 2003, 1175, 5025, 3775, 11650, 12630, 25850, 40150, 57475, 118425, 134883, 325075, 345090, 840725, 952195, 2083888, 2722455, 5056055, 7765010, 12293890, 21615771, 30591685, 58293475

Offset: 0

Clark Kimberling, Aug 24 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.

Andrew Howroyd, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,5, 0,-10,1,10,0,-5,0,1)

Cf. A000035, A291000, A291219.

I:=[0,0,0,0,1,0,5,0,15,1]; [n le 10 select I[n] else 5*Self(n-2)-10*Self(n-4)+Self(n-5)+10*Self(n-6)-5*Self(n-8)+Self(n-10): n in [1..45]]; // Vincenzo Librandi, Aug 25 2017

z = 60; s = x/(1 - x^2); p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291218 *)
LinearRecurrence[{0, 5, 0, -10, 1, 10, 0, -5, 0, 1}, {0, 0, 0, 0, 1, 0, 5, 0, 15, 1}, 50] (* Vincenzo Librandi, Aug 25 2017 *)

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

a(n) = 5*a(n-2) - 10*a(n-4) + a(n-5) + 10* a(n-6) - 5*a(n-8) + a(n-10) for n >= 11.

A291220 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^4.

Original entry on oeis.org

1, 1, 2, 4, 7, 15, 27, 55, 101, 199, 370, 718, 1347, 2595, 4898, 9397, 17803, 34066, 64682, 123561, 234917, 448289, 852979, 1626689, 3096695, 5903316, 11241426, 21424775, 40805833, 77759648, 148118585, 282229961, 537636210, 1024373916, 1951472023, 3718072991

Offset: 0

Clark Kimberling, Aug 24 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, -5, 3, 4, -1, -1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - s - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291220 *)

Vec((1 + x - x^2)*(1 - x - x^2 + x^3 + x^4) / ((1 - x - 2*x^2 + x^4)*(1 - 2*x^2 + x^3 + x^4)) + O(x^30)) \\ Colin Barker, Aug 25 2017

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

G.f.: (1 + x - x^2)*(1 - x - x^2 + x^3 + x^4) / ((1 - x - 2*x^2 + x^4)*(1 - 2*x^2 + x^3 + x^4)). - Colin Barker, Aug 25 2017

A291221 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^3 - S^6.

Original entry on oeis.org

0, 0, 1, 0, 3, 2, 6, 12, 13, 42, 42, 117, 156, 312, 531, 894, 1641, 2757, 4866, 8643, 14525, 26637, 44292, 80738, 136563, 243747, 420347, 739188, 1286250, 2252976, 3921546, 6879438, 11951510, 20993796, 36461328, 64002901, 111314775, 195060591, 339831254

Offset: 0

Clark Kimberling, Aug 24 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, 1, -15, -3, 21, 3, -15, -1, 6, 0, -1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - s^3 - s^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291221 *)

concat(vector(2), Vec(x^2*(1 + x - x^2)*(1 - x - x^2 + x^3 + x^4) / (1 - 6*x^2 - x^3 + 15*x^4 + 3*x^5 - 21*x^6 - 3*x^7 + 15*x^8 + x^9 - 6*x^10 + x^12) + O(x^60))) \\ Colin Barker, Aug 25 2017

a(n) = a(n-2) + 6*a(n-2) - 15*a(n-3) - 3*a(n-4) + 21*a(n-5) + 3*a(n-6) - 15*a(n-7) - a(n-8) - a(n-10) for n >= 11.

G.f.: x^2*(1 + x - x^2)*(1 - x - x^2 + x^3 + x^4) / (1 - 6*x^2 - x^3 + 15*x^4 + 3*x^5 - 21*x^6 - 3*x^7 + 15*x^8 + x^9 - 6*x^10 + x^12). - Colin Barker, Aug 25 2017

A291222 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^2 - S^3.

Original entry on oeis.org

0, 1, 1, 3, 5, 9, 19, 30, 66, 106, 223, 379, 753, 1345, 2565, 4723, 8816, 16456, 30480, 57093, 105677, 197751, 366697, 684765, 1272311, 2371846, 4412898, 8218386, 15300891, 28483823, 53042669, 98734485, 183863833, 342263703, 637320032, 1186464528, 2209131168

Offset: 0

Clark Kimberling, Aug 24 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, 4, 1, -4, 0, 1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291222 *)

concat(0, Vec(x*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 4*x^4 - x^6) + O(x^40))) \\ Colin Barker, Aug 25 2017

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

G.f.: x*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 4*x^4 - x^6). - Colin Barker, Aug 25 2017

A291223 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^3 - S^4.

Original entry on oeis.org

0, 0, 1, 1, 3, 5, 8, 17, 25, 52, 83, 159, 271, 497, 868, 1572, 2762, 4984, 8784, 15799, 27939, 50089, 88831, 158880, 282293, 504179, 896780, 1600335, 2848339, 5080363, 9045953, 16129172, 28726972, 51209648, 91223508, 162594868, 289675121, 516264093

Offset: 0

Clark Kimberling, Aug 24 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, 4, 1, -5, -1, 4, 0, -1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - s^3 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291223 *)

concat(vector(2), Vec(x^2*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 5*x^4 + x^5 - 4*x^6 + x^8) + O(x^50))) \\ Colin Barker, Aug 25 2017

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

G.f.: x^2*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 5*x^4 + x^5 - 4*x^6 + x^8). - Colin Barker, Aug 25 2017

A291224 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^4.

Original entry on oeis.org

4, 10, 24, 55, 120, 254, 524, 1059, 2104, 4120, 7968, 15244, 28888, 54284, 101240, 187537, 345268, 632122, 1151408, 2087485, 3768280, 6775322, 12136940, 21666712, 38555100, 68401582, 121011800, 213521067, 375813760, 659910710, 1156204452, 2021495767

Offset: 0

Clark Kimberling, Aug 24 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, -8, 5, 8, -2, -4, -1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = (1 - s)^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291224 *)
LinearRecurrence[{4,-2,-8,5,8,-2,-4,-1},{4,10,24,55,120,254,524,1059},40] (* Harvey P. Dale, Mar 22 2025 *)

Vec((2 - x - 2*x^2)*(2 - 2*x - 3*x^2 + 2*x^3 + 2*x^4) / (1 - x - x^2)^4 + O(x^40)) \\ Colin Barker, Aug 25 2017

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

G.f.: (2 - x - 2*x^2)*(2 - 2*x - 3*x^2 + 2*x^3 + 2*x^4) / (1 - x - x^2)^4. - Colin Barker, Aug 25 2017

A291225 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^5.

Original entry on oeis.org

5, 15, 40, 100, 236, 535, 1175, 2515, 5270, 10846, 21980, 43950, 86850, 169840, 329042, 632135, 1205205, 2281925, 4293270, 8030558, 14940700, 27659095, 50968455, 93518940, 170905555, 311159365, 564521620, 1020800470, 1840124050, 3307314163, 5927828905

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 (5, -5, -10, 15, 11, -15, -10, 5, 5, 1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = (1 - s)^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291225 *)

Vec((5 - 10*x - 10*x^2 + 25*x^3 + 11*x^4 - 25*x^5 - 10*x^6 + 10*x^7 + 5*x^8) / (1 - x - x^2)^5 + O(x^40)) \\ Colin Barker, Aug 28 2017

a(n) = 5*a(n-1) - 5*a(n-2) - 10*a(n-3) + 15*a(n-4) + 11*a(n-5) - 15*a(n-6) - 10*a(n-7) + 5*a(n-8) + 5*a(n-9) + a(n-10) for n >= 11.

G.f.: (5 - 10*x - 10*x^2 + 25*x^3 + 11*x^4 - 25*x^5 - 10*x^6 + 10*x^7 + 5*x^8) / (1 - x - x^2)^5. - Colin Barker, Aug 28 2017

A291226 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^6.

Original entry on oeis.org

6, 21, 62, 168, 426, 1029, 2394, 5403, 11892, 25626, 54228, 112958, 232056, 470904, 945152, 1878351, 3699666, 7227807, 14015538, 26991978, 51654946, 98275461, 185958162, 350093468, 655988730, 1223722623, 2273327418, 4206691146, 7755620994, 14248825833

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 (6, -9, -10, 30, 6, -41, -6, 30, 10, -9, -6, -1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = (1 - s)^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291226 *)

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

a(n) = 6*a(n-1) - 9*a(n-2) - 10*a(n-3) + 30*a(n-4) + 6*a(n-5) - 41*a(n-6) - 6*a(n-7) + 30*a(n-8) + 10*a(n-9) - 9*a(n-10) - 6*a(n-11) - a(n-12) for n >= 13.

A291229 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S).

Original entry on oeis.org

3, 7, 18, 45, 111, 272, 663, 1611, 3906, 9457, 22875, 55296, 133611, 322751, 779490, 1882341, 4545159, 10974256, 26496255, 63970947, 154444914, 372871721, 900206067, 2173312512, 5246877459, 12667142455, 30581283762, 73829906397, 178241414367, 430313249360

Offset: 0

Clark Kimberling, Aug 25 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 (3, 0, -3, -1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 2 s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291229 *)

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

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

A291230 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S)(1 - 3 S).

Original entry on oeis.org

6, 25, 96, 351, 1242, 4304, 14706, 49761, 167232, 559303, 1864110, 6197472, 20567262, 68166713, 225713280, 746866143, 2470077378, 8166190192, 26990599050, 89190984033, 294691499808, 973574384231, 3216160413654, 10623856065984, 35092075282998, 115910575744921

Offset: 0

Clark Kimberling, Aug 25 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 (6, -8, -6, 8, 6, 1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 2 s)(1 - 3 s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291230 *)

G.f.: (-6 + 11 x + 6 x^2 - 11 x^3 - 6 x^4)/(-1 + 6 x - 8 x^2 - 6 x^3 + 8 x^4 + 6 x^5 + x^6).

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

cp's OEIS Frontend

A291218 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^5.

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A291220 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^4.

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A291221 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^3 - S^6.

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A291222 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^2 - S^3.

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A291223 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^3 - S^4.

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A291224 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^4.

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A291225 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^5.

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A291226 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^6.

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