A154272 -id:A154272 - cp's OEIS frontend

A291730 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - 2 S - 2 S^2.

2, 6, 18, 56, 168, 510, 1544, 4680, 14176, 42952, 130128, 394252, 1194456, 3618840, 10963960, 33217424, 100638528, 304903688, 923764032, 2798719872, 8479257216, 25689531840, 77831351040, 235804967056, 714416256800, 2164460716896, 6557647800096

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, 2, 2, 4, 0, 2)

Cf. A154272, A291728, A291731.

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

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

A291732 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - 2 S)^2.

Original entry on oeis.org

4, 12, 36, 104, 288, 780, 2080, 5472, 14240, 36736, 94080, 239440, 606144, 1527360, 3833024, 9584768, 23890944, 59380160, 147207168, 364084224, 898569216, 2213388288, 5442392064, 13360097536, 32746992640, 80153705472, 195933828096, 478374127616

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 (4, -4, 4, -8, 0, -4)

Cf. A154272, A291728, A291733.

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.: -((4 (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.

A157235 Number of primitive inequivalent oblique sublattices of hexagonal (triangular) lattice of index n (equivalence and symmetry of sublattices are determined using only parent lattice symmetries).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 1, 2, 1, 2, 1, 3, 2, 2, 2, 5, 2, 4, 3, 5, 3, 4, 4, 6, 5, 6, 4, 10, 4, 6, 6, 8, 6, 10, 5, 9, 7, 8, 6, 14, 6, 10, 10, 11, 7, 12, 8, 14, 10, 12, 8, 17, 10, 12, 11, 14, 9, 20, 9, 15, 14, 14, 12, 22, 10, 16, 14, 22, 11, 20, 11, 18, 18, 18

Offset: 1

N. J. A. Sloane, Feb 25 2009

nonn

John S. Rutherford, Sublattice enumeration. IV. Equivalence classes of plane sublattices by parent Patterson symmetry and colour lattice group type, Acta Cryst. (2009). A65, 156-163. [See Table 5.]

Cf. A003051 (all sublattices), A003050 (all primitive sublattices), A154272 (primitive sublattices fully inheriting the parent lattice symmetry, inlcuding the orientation of the mirrors), A000086 (primitive rotation-symmetric sublattices, counting mirror images as distinct), A060594 (primitive mirror-symmetric sublattices), A145377 (all sublattices inheriting the parent lattice symmetry), A304182.

a(n) = A003050(n) - (A000086(n)-A154272(n))/2 - A060594(n). - Andrey Zabolotskiy, Mar 19 2021

New name and a(1)=0 prepended by Andrey Zabolotskiy, May 09 2018

Terms a(31) and beyond from Andrey Zabolotskiy, Mar 19 2021

A278105 a(n) = floor(3/n).

Original entry on oeis.org

3, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Jason Kimberley, Nov 23 2016

Index entries for linear recurrences with constant coefficients, signature (1).

Cf. A033322, A033324, ..., A033420, A033421, A033422, A033427, A033426, A033425, A033424, A033423.

This sequence is (ignoring the trailing zeros) the third row of A010766.

Magma
```
[3 div n: n in[1..100]];
```

Table[Floor[3/n], {n, 105}] (* Michael De Vlieger, Nov 24 2016 *)

a(n) = A033322(n)+A154272(n). - R. J. Mathar, Jun 21 2025

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.

cp's OEIS Frontend

A291730 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - 2 S - 2 S^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A291732 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - 2 S)^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A157235 Number of primitive inequivalent oblique sublattices of hexagonal (triangular) lattice of index n (equivalence and symmetry of sublattices are determined using only parent lattice symmetries).

Views

Author

Keywords

Links

Crossrefs

Formula

Extensions

A278105 a(n) = floor(3/n).

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

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