A291382 -id:A291382 - cp's OEIS frontend

A291411 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2 + S^3.

2, 7, 21, 63, 189, 567, 1699, 5092, 15260, 45731, 137046, 410697, 1230768, 3688339, 11053134, 33123790, 99264648, 297474121, 891463923, 2671519536, 8005951162, 23992058879, 71898875923, 215464974683, 645700711159, 1935021731510, 5798830691535, 17377808652745

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, 3, 1, -2, -3, -1)

Cf. A019590, A291382.

a:=[2,7,21,63,189,567];;  for n in [7..10^2] do a[n]:=2*a[n-1]+3*a[n-2]+a[n-3]-2*a[n-4]-3*a[n-5]-a[n-6]; od; a; # Muniru A Asiru, Sep 12 2017

z = 60; s = x + x^2; p = 1 - 2 s - s^2 + s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291411 *)

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

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

A291412 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - 2 S^2 + S^3.

Original entry on oeis.org

1, 4, 10, 24, 62, 156, 391, 987, 2484, 6252, 15744, 39636, 99788, 251237, 632525, 1592480, 4009326, 10094104, 25413498, 63982496, 161086011, 405559431, 1021059816, 2570679048, 6472089792, 16294506424, 41023988824, 103284359545, 260034658537, 654678248796

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, 3, 3, -1, -3, -1)

Cf. A019590, A291382.

a:=[1,4,10,24,62,156];;  for n in [7..10^2] do a[n]:=a[n-1]+3*a[n-2]+3*a[n-3]-a[n-4]-3*a[n-5]-a[n-6]; od; a; # Muniru A Asiru, Sep 12 2017

z = 60; s = x + x^2; p = 1 - s - 2 s^2 + s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291412 *)

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

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

A291413 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 3 S + S^2 + S^3.

Original entry on oeis.org

3, 11, 36, 117, 375, 1197, 3810, 12112, 38478, 122198, 388008, 1231911, 3911097, 12416751, 39419610, 125145175, 397296363, 1261288403, 4004182620, 12711979296, 40356397332, 128118414852, 406734209280, 1291248512101, 4099293000471, 13013918567075

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 (3, 2, -3, -4, -3, -1)

Cf. A019590, A291382.

a:=[3,11,36,117,375,1197];;   for n in [7..10^3] do a[n]:=3*a[n-1]+
2*a[n-2]-3*a[n-3]-4*a[n-4]-3*a[n-5]-a[n-6]; od; a; # Muniru A Asiru, Sep 12 2017

z = 60; s = x + x^2; p = 1 - 3 s + s^2 + s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291413 *)

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

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

A291414 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S + S^3.

Original entry on oeis.org

2, 6, 15, 37, 89, 212, 500, 1173, 2741, 6388, 14860, 34524, 80138, 185904, 431075, 999280, 2315960, 5366755, 12435075, 28810731, 66748062, 154635086, 358234115, 829886167, 1922494024, 4453566092, 10316878400, 23899399028, 55363614076, 128251081293

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, -1, -3, -3, -1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = 1 - 2 s + s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291414 *)

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

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

A291415 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 3 S + S^2.

Original entry on oeis.org

3, 11, 37, 126, 427, 1448, 4909, 16643, 56424, 191292, 648529, 2198680, 7454090, 25271280, 85676131, 290464093, 984747891, 3338548317, 11318536416, 38372746007, 130093466328, 441050269849, 1495273713773, 5069362002354, 17186439428582, 58266444593059

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 (3, 2, -2, -1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = 1 - 3 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291415 *)

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

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

A291416 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 4 S + S^2.

Original entry on oeis.org

4, 19, 86, 392, 1784, 8121, 36966, 168267, 765940, 3486508, 15870352, 72240785, 328835240, 1496836103, 6813498210, 31014589884, 141176346720, 642625324009, 2925187658218, 13315259321575, 60610173266216, 275893470389144, 1255848695053856, 5716539585528849

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 (4, 3, -2, -1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = 1 - 4 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291416 *)

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

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

A291462 a(n) = (1/2)*A291417(n).

Original entry on oeis.org

2, 9, 38, 161, 680, 2872, 12128, 51214, 216264, 913228, 3856328, 16284284, 68764352, 290374208, 1226175744, 5177825416, 21864627616, 92328709904, 389880441696, 1646365025296, 6952176889728, 29357258422912, 123968166486528, 523485745182432, 2210545926231680

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 (4, 2, -4, -2)

Cf. A019590, A291382, A291417.

z = 60; s = x + x^2; p = 1 - 4 s + 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291417 *)
u / 2  (*A291462)

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

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

A291380 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^5.

Original entry on oeis.org

0, 0, 0, 0, 1, 5, 10, 10, 5, 2, 10, 45, 120, 210, 253, 225, 225, 500, 1375, 3005, 5025, 6625, 7575, 9850, 18508, 40150, 78275, 128375, 180625, 237888, 345090, 607105, 1163155, 2109140, 3426771, 5056055, 7237835, 11059960, 18816930, 33638409, 58293475

Offset: 0

Clark Kimberling, Sep 04 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, 0, 0, 1, 5, 10, 10, 5, 1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = (1 - s)^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291380 *)

G.f.: -((x^4 (1 + x)^5)/((-1 + x + x^2) (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-5) + 5*a(n-6) + 10*a(n-7) + 10*a(n-8) + 5*a(n-9) + a(n-10) for n >= 11.

A291381 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^6.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 6, 15, 20, 15, 6, 2, 12, 66, 220, 495, 792, 925, 810, 648, 1036, 3126, 8580, 18566, 31848, 44034, 50644, 54384, 74328, 153161, 354702, 738966, 1312380, 1988814, 2638668, 3297933, 4531980, 7814811, 15621794, 30839391, 55350396, 88575614

Offset: 0

Clark Kimberling, Sep 04 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, 0, 0, 0, 1, 6, 15, 20, 15, 6, 1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = (1 - s)^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291381 *)

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

a(n) = 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.

A291386 a(n) = (1/3)*A099432(n+1).

Original entry on oeis.org

2, 11, 54, 252, 1134, 4977, 21438, 91017, 381996, 1588248, 6552252, 26853687, 109438938, 443837799, 1792373346, 7211142612, 28915704810, 115603540605, 460942202070, 1833459620517, 7276826042712, 28823185892016, 113957884236024, 449793742386627

Offset: 0

Clark Kimberling, Sep 04 2017

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6, -3, -18, -9)

Cf. A019590, A291382, A099432.

z = 60; s = x + x^2; p = (1 - 3 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A099432 *)
u / 3  (* A291386 *)
LinearRecurrence[{6,-3,-18,-9},{2,11,54,252},30] (* Harvey P. Dale, Oct 06 2017 *)

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

a(n) = 6*a(n-1) - 3*a(n-2) - 18*a(n-3) - 9*a(n-4) for n >= 5.

cp's OEIS Frontend

A291411 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2 + S^3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Mathematica

Formula

A291412 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - 2 S^2 + S^3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Mathematica

Formula

A291413 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 3 S + S^2 + S^3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Mathematica

Formula

A291414 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S + S^3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A291415 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 3 S + S^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A291416 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 4 S + S^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A291462 a(n) = (1/2)*A291417(n).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A291380 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^5.

Views

Author

Keywords

Comments

Links

Crossrefs