A045949 -id:A045949 - cp's OEIS frontend

A229620 Incorrect version of A045949.

0, 6, 38, 116, 256, 478, 798, 1236, 1808, 2534, 3430, 4516, 5808, 7326, 9086, 11108, 13408, 16006, 18918, 22164, 25760, 29726, 34078, 38836, 44016, 49638, 55718, 62276, 69328, 76894, 84990, 93636, 102848, 112646, 123046, 134068, 145728, 158046, 171038, 184724, 199120, 214246, 230118, 246756, 264176, 282398, 301438, 321316, 342048, 363654

Offset: 0

Max Alekseyev, Sep 26 2013

Arises from the formula in Problem 11 of Zhuravlev and Samovol (2012) paper, which incorrectly claims it to produce sequence A045949. Terms a(n) for n<=3 match those of A045949 but afterwards the two sequences diverge. Nevertheless these sequences satisfy the same linear recurrent relation.

V. Zhuravlev and P. Samovol, Mathematical enigmas of king Solomon's stamp, Kvant 1 (2012), 40-43. (in Russian)
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Cf. A045949.

{ a(n) = if(n%2, (n+1)*(6*n^2+3*n+1)/2- 4*n, n*(6*n^2+9*n-4)/2 ) }

For even n, a(n) = n*(6*n^2+9*n-4)/2; for odd n, a(n) = (n+1)*(6*n^2+3*n+1)/2 - 4*n.
For n>=4, a(n) = 3*a(n-1) - 2*a(n-2) - 2*a(n-2) + 3*a(n-3) - a(n-4).
a(n) = (1-(-1)^n-8*n+18*n^2+12*n^3)/4. G.f.: -2*x*(2*x+1)*(x^2-4*x-3) / ((x-1)^4*(x+1)). - Colin Barker, Sep 29 2013
E.g.f.: (x*(11 + 27*x + 6*x^2)*cosh(x) + (1 + 11*x + 27*x^2 + 6*x^3)*sinh(x))/2. - Stefano Spezia, Mar 20 2022

A033581 a(n) = 6*n^2.

Original entry on oeis.org

0, 6, 24, 54, 96, 150, 216, 294, 384, 486, 600, 726, 864, 1014, 1176, 1350, 1536, 1734, 1944, 2166, 2400, 2646, 2904, 3174, 3456, 3750, 4056, 4374, 4704, 5046, 5400, 5766, 6144, 6534, 6936, 7350, 7776, 8214, 8664, 9126, 9600, 10086, 10584, 11094, 11616

Offset: 0

N. J. A. Sloane

Number of edges of a complete 4-partite graph of order 4n, K_n,n,n,n. - Roberto E. Martinez II, Oct 18 2001
Number of edges of the complete bipartite graph of order 7n, K_n, 6n. - Roberto E. Martinez II, Jan 07 2002
Number of edges in the line graph of the product of two cycle graphs, each of order n, L(C_n x C_n). - Roberto E. Martinez II, Jan 07 2002
Total surface area of a cube of edge length n. See A000578 for cube volume. See A070169 and A071399 for surface area and volume of a regular tetrahedron and links for the other Platonic solids. - Rick L. Shepherd, Apr 24 2002
a(n) can represented as n concentric hexagons (see example). - Omar E. Pol, Aug 21 2011
Sequence found by reading the line from 0, in the direction 0, 6, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Opposite numbers to the members of A003154 in the same spiral. - Omar E. Pol, Sep 08 2011
Together with 1, numbers m such that floor(2*m/3) and floor(3*m/2) are both squares. Example: floor(2*150/3) = 100 and floor(3*150/2) = 225 are both squares, so 150 is in the sequence. - Bruno Berselli, Sep 15 2014
a(n+1) gives the number of vertices in a hexagon-like honeycomb built from A003215(n) congruent regular hexagons (see link). Example: a hexagon-like honeycomb consisting of 7 congruent regular hexagons has 1 core hexagon inside a perimeter of six hexagons. The perimeter has 18 vertices. The core hexagon has 6 vertices. a(2) = 18 + 6 = 24 is the total number of vertices. - Ivan N. Ianakiev, Mar 11 2015
a(n) is the area of the Pythagorean triangle whose sides are (3n, 4n, 5n). - Sergey Pavlov, Mar 31 2017
More generally, if k >= 5 we have that the sequence whose formula is a(n) = (2*k - 4)*n^2 is also the sequence found by reading the line from 0, in the direction 0, (2*k - 4), ..., in the square spiral whose vertices are the generalized k-gonal numbers. In this case k = 5. - Omar E. Pol, May 13 2018
The sequence also gives the number of size=1 triangles within a match-made hexagon of size n. - John King, Mar 31 2019
For hexagons, the number of matches required is A045945; thus number of size=1 triangles is A033581; number of larger triangles is A307253 and total number of triangles is A045949. See A045943 for analogs for Triangles; see A045946 for analogs for Stars. - John King, Apr 04 2019

			From _Omar E. Pol_, Aug 21 2011: (Start)
Illustration of initial terms as concentric hexagons:
.
.                                 o o o o o o
.                                o           o
.              o o o o          o   o o o o   o
.             o       o        o   o       o   o
.   o o      o   o o   o      o   o   o o   o   o
.  o   o    o   o   o   o    o   o   o   o   o   o
.   o o      o   o o   o      o   o   o o   o   o
.             o       o        o   o       o   o
.              o o o o          o   o o o o   o
.                                o           o
.                                 o o o o o o
.
.    6            24                   54
.
(End)

Nathaniel Johnston, Table of n, a(n) for n = 0..10000
Bruno Berselli, An interpretation of initial terms.
Ivan N. Ianakiev, Hexagon-like honeycomb built form regular congruent hexagons.
Leo Tavares, Illustration: Diamond Star Rays
Eric Weisstein's World of Mathematics, Platonic Solid.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A032528. Central column of triangle A001283.
Cf. A000217, A000290, A001105, A009111, A033428, A033583, A085250, A086729, A152734, A152751.
Cf. A000578, A001318, A003215, A005897, A045943, A045945, A045949, A070169, A071399.
Cf. A017593 (first differences).

a033581 = (* 6) . (^ 2)  -- Reinhard Zumkeller, Apr 27 2014

seq(6*n^2,n=0..44); # Nathaniel Johnston, Jun 26 2011

6 Range[44]^2 (* Michael De Vlieger, Apr 02 2017 *)
LinearRecurrence[{3,-3,1},{0,6,24},50] (* Harvey P. Dale, Jul 03 2017 *)

vector(100,n,6*(n-1)^2) \\ Derek Orr, Mar 11 2015

a(n) = A000290(n)*6. - Omar E. Pol, Dec 11 2008
a(n) = A001105(n)*3 = A033428(n)*2. - Omar E. Pol, Dec 13 2008
a(n) = 12*n + a(n-1) - 6, with a(0)=0. - Vincenzo Librandi, Aug 05 2010
G.f.: 6*x*(1+x)/(1-x)^3. - Colin Barker, Feb 14 2012
For n > 0: a(n) = A005897(n) - 2. - Reinhard Zumkeller, Apr 27 2014
a(n) = 3*floor(1/(1-cos(1/n))) = floor(1/(1-n*sin(1/n))) for n > 0. - Clark Kimberling, Oct 08 2014
a(n) = t(4*n) - 4*t(n), where t(i) = i*(i+k)/2 for any k. Special case (k=1): a(n) = A000217(4*n) - 4*A000217(n). - Bruno Berselli, Aug 31 2017
From Amiram Eldar, Feb 03 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/36.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/72 (A086729).
Product_{n>=1} (1 + 1/a(n)) = sqrt(6)*sinh(Pi/sqrt(6))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(6)*sin(Pi/sqrt(6))/Pi. (End)
E.g.f.: 6*exp(x)*x*(1 + x). - Stefano Spezia, Aug 19 2022

More terms from Larry Reeves (larryr(AT)acm.org), Nov 08 2001

A045945 Hexagonal matchstick numbers: a(n) = 3n(3*n+1).

Original entry on oeis.org

0, 12, 42, 90, 156, 240, 342, 462, 600, 756, 930, 1122, 1332, 1560, 1806, 2070, 2352, 2652, 2970, 3306, 3660, 4032, 4422, 4830, 5256, 5700, 6162, 6642, 7140, 7656, 8190, 8742, 9312, 9900, 10506, 11130, 11772, 12432, 13110, 13806, 14520, 15252, 16002, 16770, 17556

Offset: 0

R. K. Guy

This may also be construed as the number of line segments illustrating the isometric projection of a cube of side length n. Moreover, a(n) equals the number of rods making a cube of side length n+1 minus the number of rods making a cube of side length n. See the illustration in the links and formula below.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Peter M. Chema, Illustration of initial terms as the first difference of number of rods required to make a 3-D cube.
Craig Knecht, Number of positions a frame shifted H1 hexagon can occupy in a hexagon of order n.
Amelia Carolina Sparavigna, The groupoids of Mersenne, Fermat, Cullen, Woodall and other Numbers and their representations by means of integer sequences, Politecnico di Torino, Italy (2019), [math.NT].
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000290, A005449, A033580, A059986.

The hexagon matchstick sequences are: Number of matchsticks: this sequence; size=1 triangles: A033581; larger triangles: A307253; total triangles: A045949. Analog for triangles: A045943; analog for stars: A045946. - John King, Apr 05 2019

a:= n-> 3*n*(3*n+1): seq(a(n), n=0..42); # Zerinvary Lajos, May 03 2007

f[n_]:=3*n*(3*n+1);f[Range[0,60]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011 *)

a(n) = 3*n*(3*n+1) \\ Charles R Greathouse IV, Feb 27 2017

def a(n): return 3*n*(3*n+1) # Indranil Ghosh, Mar 26 2017

a(n) = a(n-1) + 6*(3*n-1) (with a(0)=0). - Vincenzo Librandi, Nov 18 2010
G.f.: 6*x*(2+x)/(1-x)^3. - Colin Barker, Feb 12 2012
a(n) = 6*A005449(n). - R. J. Mathar, Feb 13 2016
a(n) = A059986(n) - A059986(n-1). - Peter M. Chema, Feb 26 2017
a(n) = 6*(A000217(n) + A000290(n)). - Peter M. Chema, Mar 26 2017
From Amiram Eldar, Jan 14 2021: (Start)
Sum_{n>=1} 1/a(n) = 1 - Pi/(6*sqrt(3)) - log(3)/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = -1 + Pi/(3*sqrt(3)) + 2*log(2)/3. (End)
From Elmo R. Oliveira, Dec 12 2024: (Start)
E.g.f.: 3*exp(x)*x*(4 + 3*x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

A008893 Number of equilateral triangles formed by triples of points taken from a hexagonal chunk of side n in the hexagonal lattice.

Original entry on oeis.org

0, 8, 66, 258, 710, 1590, 3108, 5516, 9108, 14220, 21230, 30558, 42666, 58058, 77280, 100920, 129608, 164016, 204858, 252890, 308910, 373758, 448316, 533508, 630300, 739700, 862758, 1000566, 1154258, 1325010, 1514040, 1722608, 1952016, 2203608, 2478770

Offset: 0

N. J. A. Sloane, R. K. Guy

The hexagonal lattice is the familiar 2-dimensional lattice in which each point has 6 neighbors. This is sometimes called the triangular lattice. Here we consider a hexagonal chunk of the lattice in which each bounding edge contains n+1 points.

Nathaniel Johnston, Table of n, a(n) for n = 0..10000
Gabriele Nebe and N. J. A. Sloane, Home page for hexagonal (or triangular) lattice A2.
N. J. A. Sloane, Illustration for a(1)=8. [The drawing was made for a different offset, so it says a(2)=8.]
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A045949, A152041.

A008893[n_] := n*(n + 1)*(7*n*(n + 1) + 2)/4; Array[A008893, 50, 0] (* or *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 8, 66, 258, 710}, 50] (* Paolo Xausa, Aug 16 2025 *)

A008893(n):=n*(n+1)*(7*n^2+7*n+2)/4$
makelist(A008893(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

a(n) = n*(n+1)*(7*n^2+7*n+2)/4.
G.f.: -2*x*(4*x^2+13*x+4)/(x-1)^5 [From Maksym Voznyy (voznyy(AT)mail.ru), Aug 10 2009]
From Elmo R. Oliveira, Aug 15 2025: (Start)
E.g.f.: exp(x)*x*(2 + x)*(16 + 42*x + 7*x^2)/4.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = 2*A152041(n). (End)

Edited May 29 2012 by N. J. A. Sloane, May 29 2012

A307253 Number of triangles larger than size=1 in a matchstick-made hexagon with side length n.

Original entry on oeis.org

0, 0, 14, 62, 166, 346, 624, 1020, 1556, 2252, 3130, 4210, 5514, 7062, 8876, 10976, 13384, 16120, 19206, 22662, 26510, 30770, 35464, 40612, 46236, 52356, 58994, 66170, 73906, 82222, 91140, 100680, 110864, 121712, 133246, 145486, 158454, 172170, 186656, 201932

Offset: 0

John King, Mar 31 2019

nonn

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Cf. A033581 (number of size=1 triangles), A045949 (total number of triangles).

The hexagon matchstick sequences are as follows: number of matchsticks: A045945; for T1 triangles: A033581; for larger triangles: this sequence and for total triangles: A045949. There are analogs for triangles (see A045943) and stars (see A045946).

LinearRecurrence[{3,-2,-2,3,-1},{0, 0, 14, 62, 166},166] (* Metin Sariyar, Oct 27 2019 *)

concat([0,0], Vec(2*x^2*(7 + 10*x + 4*x^2) / ((1 - x)^4*(1 + x)) + O(x^40))) \\ Colin Barker, Apr 02 2019

a(n) = floor(n*(14*n^2+9*n+2)/4)-6*n^2.
G.f.: 2*x^2*(4*x^2+10*x+7)/((x+1)*(x-1)^4).
a(n) = A045949(n) - A033581(n).
a(n) = 3*a(n-1) - 2*a(n-2) - 2*a(n-3) + 3*a(n-4) - a(n-5) for n>4. - Colin Barker, Apr 02 2019

A299965 Number of triangles in a Star of David of size n.

Original entry on oeis.org

0, 20, 118, 354, 788, 1480, 2490, 3878, 5704, 8028, 10910, 14410, 18588, 23504, 29218, 35790, 43280, 51748, 61254, 71858, 83620, 96600, 110858, 126454, 143448, 161900, 181870, 203418, 226604, 251488, 278130, 306590, 336928, 369204, 403478, 439810, 478260, 518888

Offset: 0

John King, Feb 22 2018

In a Star of David of size n, there are A135453(n) "size=1" triangles.

The number of matchstick units is A045946.

			For n=1, there are 12 (size=1) + 6 (size=4) + 2 (size=9) = 20 triangles.

Paolo Xausa, Table of n, a(n) for n = 0..10000 (corrected and extended original b-file from Colin Barker after data change).
John King, Star a=6, 84 matches, 118 triangles
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A045950, A045946, A135453,

For the total number of triangles in a different arrangement, see A002717 (for triangular matchstick), A045949 (for hexagonal matchstick).

A299965[n_] := n*(n*(10*n + 9) + 1); Array[A299965, 50, 0] (* or *)
LinearRecurrence[{4, -6, 4, -1}, {0, 20, 118, 354}, 50] (* Paolo Xausa, Sep 18 2024 *)

concat(0, Vec(2*x*(10 + 19*x + x^2) / (1 - x)^4 + O(x^40))) \\ Colin Barker, Apr 04 2019

a(n) = n*(10*n^2+9*n+1) = 2*A045950(n).
From Colin Barker, Apr 04 2019: (Start)
G.f.: 2*x*(10 + 19*x + x^2) / (1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. (End)
E.g.f.: exp(x)*x*(20 + 39*x + 10*x^2). -  Stefano Spezia, Sep 20 2024

Corrected by John King, Stefano Spezia, and Paolo Xausa, Sep 20 2024

cp's OEIS Frontend

A229620 Incorrect version of A045949.

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A033581 a(n) = 6*n^2.

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A045945 Hexagonal matchstick numbers: a(n) = 3*n*(3*n+1).

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A008893 Number of equilateral triangles formed by triples of points taken from a hexagonal chunk of side n in the hexagonal lattice.

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A307253 Number of triangles larger than size=1 in a matchstick-made hexagon with side length n.

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A299965 Number of triangles in a Star of David of size n.

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A045945 Hexagonal matchstick numbers: a(n) = 3n(3*n+1).