A291382 -id:A291382 - cp's OEIS frontend

A291399 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^2 - S^3 - S^4.

1, 3, 8, 22, 59, 156, 412, 1093, 2903, 7707, 20453, 54275, 144035, 382255, 1014469, 2692284, 7144989, 18961928, 50322686, 133550412, 354426839, 940606403, 2496256771, 6624766692, 17581338025, 46658767166, 123826784175, 328621466028, 872122042693

Offset: 0

Clark Kimberling, Sep 06 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 A291382 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, 3, 5, 7, 7, 4, 1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = 1 - s - s^2 - s^3 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291399 *)
LinearRecurrence[{1,2,3,5,7,7,4,1},{1,3,8,22,59,156,412,1093},40] (* Harvey P. Dale, Oct 06 2018 *)

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

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

A291400 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^2 - S^4.

Original entry on oeis.org

0, 1, 2, 3, 8, 15, 26, 52, 100, 193, 378, 726, 1396, 2699, 5210, 10065, 19444, 37528, 72448, 139890, 270104, 521547, 1007026, 1944336, 3754132, 7248558, 13995676, 27023186, 52176848, 100743849, 194517966, 375578833, 725174524, 1400180233, 2703493026

Offset: 0

Clark Kimberling, Sep 06 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 A291382 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, 2, 4, 6, 4, 1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = 1 - s^2 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291400 *)

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

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

A291401 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^4.

Original entry on oeis.org

1, 2, 3, 6, 14, 32, 67, 134, 266, 538, 1110, 2304, 4760, 9770, 19991, 40931, 83976, 172519, 354452, 727830, 1493768, 3065341, 6291208, 12914136, 26511196, 54423052, 111715200, 229312168, 470697488, 966192481, 1983312305, 4071174986, 8356928055, 17154242334

Offset: 0

Clark Kimberling, Sep 06 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 A291382 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, 1, 4, 6, 4, 1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = 1 - s - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291401 *)

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

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

A291402 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^3 - S^4.

Original entry on oeis.org

0, 0, 1, 4, 7, 8, 12, 31, 71, 125, 201, 367, 749, 1471, 2679, 4814, 9014, 17304, 32739, 60683, 112444, 210938, 397800, 746347, 1392898, 2601701, 4876692, 9149911, 17138518, 32060349, 60002060, 112404852, 210600344, 394370928, 738281497, 1382360598

Offset: 0

Clark Kimberling, Sep 06 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 A291382 for a guide to related sequences.

Index entries for linear recurrences with constant coefficients, signature (0, 0, 1, 4, 7, 7, 4, 1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = 1 - s^3 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291402 *)
LinearRecurrence[{0,0,1,4,7,7,4,1},{0,0,1,4,7,8,12,31},40] (* Harvey P. Dale, Feb 20 2020 *)

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

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

A291403 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S^2 - S^4.

Original entry on oeis.org

0, 2, 4, 7, 20, 42, 92, 214, 472, 1062, 2396, 5361, 12052, 27074, 60764, 136497, 306520, 688292, 1545768, 3471224, 7795184, 17505588, 39311608, 88280985, 198250312, 445204610, 999783508, 2245185343, 5041947516, 11322557726, 25426742788, 57100105470

Offset: 0

Clark Kimberling, Sep 07 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 A291382 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, 2, 4, 3, 4, 6, 4, 1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = 1 - 2 s^2 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291403 *)

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

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

A291404 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^2 - 2 S^4.

Original entry on oeis.org

0, 1, 2, 4, 12, 23, 42, 89, 188, 404, 856, 1763, 3652, 7641, 16030, 33612, 70252, 146623, 306334, 640637, 1340024, 2802056, 5857264, 12243403, 25596040, 53515853, 111889138, 233922392, 489039852, 1022399607, 2137493106, 4468804953, 9342779572, 19532522316

Offset: 0

Clark Kimberling, Sep 07 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 A291382 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, 3, 8, 12, 8, 2)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = 1 - s^2 - 2 s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291404 *)

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

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

A291406 a(n) = (1/2)A291405(n).

Original entry on oeis.org

0, 1, 2, 4, 12, 26, 60, 145, 336, 796, 1880, 4412, 10400, 24488, 57648, 135794, 319744, 752908, 1773016, 4175032, 9831472, 23151400, 54516976, 128377380, 302304640, 711870368, 1676321856, 3947423264, 9295440640, 21889019712, 51544533376, 121377706184

Offset: 0

Clark Kimberling, Sep 07 2017

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0, 2, 4, 4, 8, 12, 8, 2)

Cf. A019590, A291382, A291405.

z = 60; s = x + x^2; p = 1 - 2 s^2 - 2 s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291405 *)
u / 2 (* A291406 *)

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

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

A291407 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^3 - S^6.

Original entry on oeis.org

0, 0, 1, 3, 3, 3, 12, 30, 43, 57, 120, 259, 438, 708, 1360, 2703, 4827, 8276, 15114, 28488, 51769, 92031, 167334, 309341, 564237, 1016487, 1844115, 3374343, 6151563, 11151098, 20253876, 36931695, 67280308, 122243430, 222174201, 404488108, 736378968

Offset: 0

Clark Kimberling, Sep 07 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 A291382 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, 3, 3, 2, 6, 15, 20, 15, 6, 1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = 1 - s^3 - s^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291407 *)

G.f.: -((x^2 (1 + x)^3 (1 + x + x^2) (1 - x + 2 x^3 + x^4))/(-1 + x^3 + 3 x^4 + 3 x^5 + 2 x^6 + 6 x^7 + 15 x^8 + 20 x^9 + 15 x^10 + 6 x^11 + x^12)).

a(n) = a(n-3) + 3*a(n-4) + 3*a(n-5) + 2*a(n-6) + 6*a(n-7) + 15*a(n-8) + 20*a(n-9) + 15*a(n-10) + 6*a(n-11) + a(n-12) for n >= 13.

A291409 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S^2)(1 - S)^2.

Original entry on oeis.org

2, 6, 14, 31, 66, 136, 272, 534, 1030, 1958, 3678, 6837, 12594, 23016, 41768, 75325, 135084, 241032, 428112, 757236, 1334292, 2342892, 4100676, 7155937, 12453170, 21616242, 37432010, 64675099, 111512574, 191893120, 329605760, 565166682, 967491754, 1653659282

Offset: 0

Clark Kimberling, Sep 07 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 A291382 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, -5, -2, 4, 4, 1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = (1 - s^2)(1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291409 *)
LinearRecurrence[{2,2,-2,-5,-2,4,4,1},{2,6,14,31,66,136,272,534},40] (* Harvey P. Dale, May 12 2024 *)

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

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

A291410 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^2 - 2 S^3.

Original entry on oeis.org

1, 3, 9, 26, 69, 186, 511, 1401, 3824, 10438, 28521, 77938, 212928, 581700, 1589231, 4341911, 11862339, 32408429, 88541424, 241899801, 660882666, 1805564823, 4932894579, 13476918898, 36819627664, 100593093135, 274825440378, 750837066710, 2051325016200

Offset: 0

Clark Kimberling, Sep 07 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 A291382 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, 4, 7, 6, 2)

Cf. A019590, A291382.

a:=[1,3,9,26,69,186];;  for n in [7..10^2] do a[n]:=a[n-1]+2*a[n-2]+4*a[n-3]+7*a[n-4]+6*a[n-5]+2*a[n-6]; od; a; # Muniru A Asiru, Sep 12 2017

z = 60; s = x + x^2; p = 1 - s - s^2 - 2 s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291410 *)
CoefficientList[ Series[-(2x^5 +6x^4 +7x^3 +4x^2 +2x +1)/(2x^6 +6x^5 +7x^4 +4x^3 +2x^2 +x- 1), {x, 0, 28}], x] (* or *)
LinearRecurrence[{1, 2, 4, 7, 6, 2}, {1, 3, 9, 26, 69, 186}, 29] (* Robert G. Wilson v, Sep 25 2017 *)

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

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

cp's OEIS Frontend

A291399 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^2 - S^3 - S^4.

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A291400 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^2 - S^4.

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A291401 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^4.

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A291402 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^3 - S^4.

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A291403 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S^2 - S^4.

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A291404 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^2 - 2 S^4.

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A291406 a(n) = (1/2)A291405(n).

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A291407 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^3 - S^6.

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