A152745 -id:A152745 - cp's OEIS frontend

A152746 Six times hexagonal numbers: 6n(2*n-1).

0, 6, 36, 90, 168, 270, 396, 546, 720, 918, 1140, 1386, 1656, 1950, 2268, 2610, 2976, 3366, 3780, 4218, 4680, 5166, 5676, 6210, 6768, 7350, 7956, 8586, 9240, 9918, 10620, 11346, 12096, 12870, 13668, 14490, 15336, 16206, 17100

Offset: 0

Omar E. Pol, Dec 12 2008

Sequence found by reading the line from 0, in the direction 0, 6, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Sep 18 2011

a(n) is the number of walks on a cubic lattice of n dimensions that return to the origin, not necessarily for the first time, after 4 steps. - Shel Kaphan, Mar 20 2023

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000384, A001082, A002939, A094159, A085250, A152745.

Column n=2 of A287318.

[6*n*(2*n-1): n in [0..50]]; // G. C. Greubel, Sep 01 2018

6*PolygonalNumber[6,Range[0,40]] (* The program uses the PolygonalNumber function from Mathematica version 10 *) (* Harvey P. Dale, Mar 04 2016 *)
LinearRecurrence[{3,-3,1}, {0,6,36}, 50] (* or *) Table[6*n*(2*n-1), {n,0,50}] (* G. C. Greubel, Sep 01 2018 *)

a(n)=6*n*(2*n-1) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 12*n^2 - 6*n = A000384(n)*6 = A002939(n)*3 = A094159(n)*2.
a(n) = a(n-1) + 24*n - 18 (with a(0)=0). - Vincenzo Librandi, Nov 26 2010
From G. C. Greubel, Sep 01 2018: (Start)
G.f.: 6*x*(1+3*x)/(1-x)^3.
E.g.f.: 6*x*(1+2*x)*exp(x). (End)
From Amiram Eldar, Mar 30 2023: (Start)
Sum_{n>=1} 1/a(n) = log(2)/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/12 - log(2)/6. (End)

A153784 4 times heptagonal numbers: a(n) = 2n(5*n-3).

Original entry on oeis.org

0, 4, 28, 72, 136, 220, 324, 448, 592, 756, 940, 1144, 1368, 1612, 1876, 2160, 2464, 2788, 3132, 3496, 3880, 4284, 4708, 5152, 5616, 6100, 6604, 7128, 7672, 8236, 8820, 9424, 10048, 10692, 11356, 12040, 12744, 13468, 14212, 14976, 15760, 16564, 17388, 18232, 19096

Offset: 0

Omar E. Pol, Jan 02 2009

Sequence found by reading the line from 0, in the direction 0, 4, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - Omar E. Pol, Jul 18 2012

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000566, A085787, A087348, A135706, A152745, A152773.

s=0;lst={s};Do[s+=n;AppendTo[lst,s],{n,4,6!,20}];lst (* Vladimir Joseph Stephan Orlovsky, Apr 02 2009 *)
Table[2n(5n-3),{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,4,28},50] (* Harvey P. Dale, Mar 19 2015 *)

a(n)=2*n*(5*n-3) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 10*n^2 - 6*n = 4*A000566(n) = 2*A135706(n).
a(n) = 20*n + a(n-1) - 16 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
a(n) = A087348(n) - 1, n >= 1. - Omar E. Pol, Jul 18 2012
a(0)=0, a(1)=4, a(2)=28, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Mar 19 2015
From Elmo R. Oliveira, Dec 15 2024: (Start)
G.f.: 4*x*(1 + 4*x)/(1 - x)^3.
E.g.f.: 2*exp(x)*x*(2 + 5*x).
a(n) = A152745(n) - n. (End)

A154617 Eleven times hexagonal numbers: a(n) = 11n(2*n-1).

Original entry on oeis.org

0, 11, 66, 165, 308, 495, 726, 1001, 1320, 1683, 2090, 2541, 3036, 3575, 4158, 4785, 5456, 6171, 6930, 7733, 8580, 9471, 10406, 11385, 12408, 13475, 14586, 15741, 16940, 18183, 19470, 20801, 22176, 23595, 25058, 26565, 28116, 29711, 31350, 33033, 34760, 36531, 38346

Offset: 0

Omar E. Pol, Jan 13 2009

Sequence found by reading the line from 0, in the direction 0, 11, ..., in the square spiral whose vertices are the generalized tridecagonal numbers A195313. - Omar E. Pol, Sep 18 2011

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000384, A002939, A094159, A152745, A153781, A195313.

a(n)=11*n*(2*n-1) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 22*n^2 - 11*n = 11*A000384(n).
a(n) = a(n-1) + 44*n - 33 (with a(0)=0). - Vincenzo Librandi, Dec 15 2010
From Elmo R. Oliveira, Dec 15 2024: (Start)
G.f.: 11*x*(1 + 3*x)/(1 - x)^3.
E.g.f.: 11*x*(1 + 2*x)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

A383466 a(0) = 1; thereafter a(n) = 10n^2 - 5n + 2.

Original entry on oeis.org

1, 7, 32, 77, 142, 227, 332, 457, 602, 767, 952, 1157, 1382, 1627, 1892, 2177, 2482, 2807, 3152, 3517, 3902, 4307, 4732, 5177, 5642, 6127, 6632, 7157, 7702, 8267, 8852, 9457, 10082, 10727, 11392, 12077, 12782, 13507, 14252, 15017, 15802, 16607, 17432, 18277, 19142, 20027, 20932, 21857, 22802, 23767, 24752, 25757, 26782, 27827

Offset: 0

Scott R. Shannon and N. J. A. Sloane, Jul 22 2025

Definition: A regular pentagram of radius R is formed by placing five equally-spaced points P_0 .. P_4 around the boundary of a circle of radius R, and drawing line segments P_0 - P_2 - P_4 - P_1 - P_3 - P_0.
Theorem 1: a(n) is the maximum number of regions that can be formed in the plane by drawing n regular pentagrams with the same radius and the same center.
Conjecture 2: a(n) is the maximum number of regions that can be formed in the plane by drawing n regular pentagrams with any radii and any centers.
The following construction works for any n >= 1. Take 5*n equally-spaced points P_i around a circle, and draw a pentagram through P_i, P_{i+n}, P_{i+2*n}, P_{i+3*n}, P_{i+4*n} for i = 0, ..., n-1.
The resulting planar graph decomposes into 5*n triangular regions each with 2*n-1 cells (see the red triangle in "Illustration for a(n)..."), plus the interior and exterior regions, for a total of 10*n^2 - 5*n + 2 regions. There are 10*n^2 vertices (10 for n=1, 40 for n=2, and so on).

Paolo Xausa, Table of n, a(n) for n = 0..10000
Scott R. Shannon, Illustration for a(1) = 7. [Note that the cell counts shown on these four figures do not include the black exterior region, so the totals are off by 1]
Scott R. Shannon, Illustration for a(2) = 32.
Scott R. Shannon, Illustration for a(3) = 77.
Scott R. Shannon, Illustration for a(8) = 602.
N. J. A. Sloane, Illustration for a(1) = 7.
N. J. A. Sloane, Illustration for a(2) = 32.
N. J. A. Sloane, Illustration for a(n), n >= 1, showing a(3) = 77.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

See A077588, A069894, and A386477 for analogous sequences based on triangles, squares, and hexagrams.

Without the "+2" in the definition, the sequence is A152745.

A383466[n_] := If[n == 0, 1, 5*n*(2*n - 1) + 2]; Array[A383466, 50, 0] (* or *)
Join[{1}, 5*PolygonalNumber[6, Range[49]] + 2] (* or *)
LinearRecurrence[{3, -3, 1}, {1, 7, 32, 77}, 50] (* Paolo Xausa, Jul 22 2025 *)

From Elmo R. Oliveira, Sep 03 2025: (Start)
G.f.: (1 + 4*x + 14*x^2 + x^3)/(1 - x)^3.
E.g.f.: exp(x)*(2 + 5*x + 10*x^2) - 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A144560 Ten times hexagonal numbers: 10n(2*n-1).

Original entry on oeis.org

0, 10, 60, 150, 280, 450, 660, 910, 1200, 1530, 1900, 2310, 2760, 3250, 3780, 4350, 4960, 5610, 6300, 7030, 7800, 8610, 9460, 10350, 11280, 12250, 13260, 14310, 15400, 16530, 17700, 18910, 20160, 21450, 22780, 24150, 25560, 27010

Offset: 0

Omar E. Pol, Jan 01 2009

Sequence found by reading the line from 0, in the direction 0, 10,..., in the square spiral whose vertices are the generalized dodecagonal numbers A195162. - Omar E. Pol, Sep 18 2011

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000384, A002939, A152745, A152994.

[10*n*(2*n-1): n in [0..40]]; // G. C. Greubel, May 30 2024

10*Binomial[2*Range[0,40], 2] (* G. C. Greubel, May 30 2024 *)

a(n)=10*n*(2*n-1) \\ Charles R Greathouse IV, Jun 17 2017

[10*n*(2*n-1) for n in range(41)] # G. C. Greubel, May 30 2024

a(n) = 20*n^2 - 10*n = 10*A000384(n) = 5*A002939(n) = 2*A152745(n).
a(n) = a(n-1) +40*n -30  (with a(0)=0). - Vincenzo Librandi, Dec 14 2010
From G. C. Greubel, May 30 2024: (Start)
G.f.: 10*x*(1 + 3*x)/(1-x)^3.
E.g.f.: 10*x*(1 + 2*x)*exp(x). (End)

A194713 13 times hexagonal numbers: a(n) = 13n(2*n-1).

Original entry on oeis.org

0, 13, 78, 195, 364, 585, 858, 1183, 1560, 1989, 2470, 3003, 3588, 4225, 4914, 5655, 6448, 7293, 8190, 9139, 10140, 11193, 12298, 13455, 14664, 15925, 17238, 18603, 20020, 21489, 23010, 24583, 26208, 27885, 29614, 31395, 33228, 35113, 37050, 39039, 41080, 43173

Offset: 0

Omar E. Pol, Oct 02 2011

Sequence found by reading the line from 0, in the direction 0, 13, ..., in the square spiral whose vertices are the generalized 15-gonal numbers.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000384, A002939, A094159, A143698, A152745, A154617, A195320.

[13*n*(2*n-1): n in [0..50]]; // Vincenzo Librandi, Oct 03 2011

a(n)=13*n*(2*n-1) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 26*n^2 - 13*n = 13*A000384(n).
a(n) = a(n-1) + 52*n - 39, a(0)=0. - Vincenzo Librandi, Oct 03 2011
From Elmo R. Oliveira, Dec 15 2024: (Start)
G.f.: 13*x*(1 + 3*x)/(1 - x)^3.
E.g.f.: 13*x*(1 + 2*x)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

cp's OEIS Frontend

A152746 Six times hexagonal numbers: 6*n*(2*n-1).

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A153784 4 times heptagonal numbers: a(n) = 2*n*(5*n-3).

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A154617 Eleven times hexagonal numbers: a(n) = 11*n*(2*n-1).

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A383466 a(0) = 1; thereafter a(n) = 10*n^2 - 5*n + 2.

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A144560 Ten times hexagonal numbers: 10*n*(2*n-1).

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A194713 13 times hexagonal numbers: a(n) = 13*n*(2*n-1).

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Magma

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A152746 Six times hexagonal numbers: 6n(2*n-1).

A153784 4 times heptagonal numbers: a(n) = 2n(5*n-3).

A154617 Eleven times hexagonal numbers: a(n) = 11n(2*n-1).

A383466 a(0) = 1; thereafter a(n) = 10n^2 - 5n + 2.

A144560 Ten times hexagonal numbers: 10n(2*n-1).

A194713 13 times hexagonal numbers: a(n) = 13n(2*n-1).