A045945 -id:A045945 - cp's OEIS frontend

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

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

A033580 Four times second pentagonal numbers: a(n) = 2n(3*n+1).

Original entry on oeis.org

0, 8, 28, 60, 104, 160, 228, 308, 400, 504, 620, 748, 888, 1040, 1204, 1380, 1568, 1768, 1980, 2204, 2440, 2688, 2948, 3220, 3504, 3800, 4108, 4428, 4760, 5104, 5460, 5828, 6208, 6600, 7004, 7420, 7848, 8288, 8740, 9204, 9680, 10168, 10668, 11180, 11704, 12240

Offset: 0

N. J. A. Sloane

Subsequence of A062717: A010052(6*a(n)+1) = 1. - Reinhard Zumkeller, Feb 21 2011
Sequence found by reading the line from 0, in the direction 0, 8,..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Opposite numbers to the members of A139267 in the same spiral - Omar E. Pol, Sep 09 2011
a(n) is the number of edges of the octagonal network O(n,n); O(m,n) is defined by Fig. 1 of the Siddiqui et al. reference. - Emeric Deutsch May 13 2018
The partial sums of this sequence give A035006. - Leo Tavares, Oct 03 2021

Ivan Panchenko, Table of n, a(n) for n = 0..1000
M. K. Siddiqui, M. Naeem, N. A. Rahman, and M. Imran, Computing topological indices of certain networks, J. of Optoelectronics and Advanced Materials, 18, No. 9-10 (2016), pp. 884-892.
Leo Tavares, Illustration: Crossed Stars
Leo Tavares, Illustration: Four Quarter Star Crosses
Leo Tavares, Illustration: Triangulated Star Crosses
Leo Tavares, Illustration: Oblong Star Crosses
Leo Tavares, Illustration: Crossed Diamond Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A033579, A049451, A045945, A186423.
Cf. sequences listed in A254963.
Cf. A003154, A016813.
Cf. A035006.
Cf. A005449, A046092, A033996, A002378, A016742, A000290, A016754, A001105.

List([0..50], n-> 2*n*(3*n+1)); # G. C. Greubel, Oct 09 2019

[2*n*(3*n+1): n in [0..50]]; // G. C. Greubel, Oct 09 2019

seq(2*n*(3*n+1), n=0..50); # G. C. Greubel, Oct 09 2019

4*Binomial[3*Range[50]-2, 2]/3 (* G. C. Greubel, Oct 09 2019 *)

a(n)=2*n*(3*n+1) \\ Charles R Greathouse IV, Sep 28 2015

[2*n*(3*n+1) for n in (0..50)] # G. C. Greubel, Oct 09 2019

a(n) = a(n-1) +12*n -4 (with a(0)=0). - Vincenzo Librandi, Aug 05 2010
G.f.: 4*x*(2+x)/(1-x)^3. - Colin Barker, Feb 13 2012
a(-n) = A033579(n). - Michael Somos, Jun 09 2014
E.g.f.: 2*x*(4 + 3*x)*exp(x). - G. C. Greubel, Oct 09 2019
From Amiram Eldar, Jan 14 2021: (Start)
Sum_{n>=1} 1/a(n) = 3/2 - Pi/(4*sqrt(3)) - 3*log(3)/4.
Sum_{n>=1} (-1)^(n+1)/a(n) =  -3/2 + Pi/(2*sqrt(3)) + log(2). (End)
From Leo Tavares, Oct 12 2021: (Start)
a(n) = A003154(n+1) - A016813(n). See Crossed Stars illustration.
a(n) = 4*A005449(n). See Four Quarter Star Crosses illustration.
a(n) = 2*A049451(n).
a(n) = A046092(n-1) + A033996(n). See Triangulated Star Crosses illustration.
a(n) = 4*A000217(n-1) + 8*A000217(n).
a(n) = 4*A000217(n-1) + 4*A002378. See Oblong Star Crosses illustration.
a(n) = A016754(n) + 4*A000217(n). See Crossed Diamond Stars illustration.
a(n) = 2*A001105(n) + 4*A000217(n).
a(n) = A016742(n) + A046092(n).
a(n) = 4*A000290(n) + 4*A000217(n). (End)

A082040 a(n) = 9n^2 + 3n + 1.

Original entry on oeis.org

1, 13, 43, 91, 157, 241, 343, 463, 601, 757, 931, 1123, 1333, 1561, 1807, 2071, 2353, 2653, 2971, 3307, 3661, 4033, 4423, 4831, 5257, 5701, 6163, 6643, 7141, 7657, 8191, 8743, 9313, 9901, 10507, 11131, 11773, 12433, 13111, 13807, 14521, 15253, 16003, 16771, 17557

Offset: 0

Paul Barry, Apr 02 2003

4th row of A082039, case k = 3 of family T(n,k) = k^2*n^2 + k*n + 1.
a(n)^2 = 81*n^4 + 54*n^3 + 27*n^2 + 6*n + 1 = (24*((3*((3*n^2 + n)/2)^2 + ((3*n^2 + n)/2))/2) + 1). Therefore, (a(n)^2 - 1)/24 is a second pentagonal number (A005449) of index number equal to the n-th second pentagonal number. For example, a(30) = 8191 and (8191^2 - 1)/24 = (67092481 - 1)/24 = 2795520, the 1365th second pentagonal number. 1365 is the 30th second pentagonal number. - Raphie Frank, Sep 19 2012
For n >= 1, a(n) is the number of vertices in the hex derived network HDN1(n+1) from the Manuel et al. reference (see HFN1(4) in Fig. 8). - Emeric Deutsch, May 21 2018
4*a(n) - 3 is a square. - Muniru A Asiru, May 24 2018

Muniru A Asiru, Table of n, a(n) for n = 0..5000
Nicolay Avilov, Graphic illustration of sequence members
P. Manuel, R. Bharati, I. Rajasingh, and Chris Monica M, On minimum metric dimension of honeycomb networks, Journal of Discrete Algorithms, Vol. 6 (2008), pp. 20-27.
Leo Tavares, Snowflake illustration.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002061, A005449, A045945, A054569, A082039.
Partial sums of A019557.
Cf. A000290, A003215.

List([0..50], n->9*n^2+3*n+1); # Muniru A Asiru, May 21 2018

seq(9*n^2+3*n+1,n=0..50); # Muniru A Asiru, May 21 2018

a(n)=9*n^2+3*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 18*n + a(n-1) - 6 with n > 0, a(0)=1. - Vincenzo Librandi, Aug 08 2010
a(n) = A045945(n) + 1: subsequence of A002061. - Muniru A Asiru, May 26 2018
a(n) = A003215(n) + 6*A000290(n). - Leo Tavares, Jul 14 2023
From Elmo R. Oliveira, Oct 23 2024: (Start)
G.f.: (1 + 10*x + 7*x^2)/(1 - x)^3.
E.g.f.: (1 + 12*x + 9*x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A045946 Star of David matchstick numbers: a(n) = 6n(3*n+1).

Original entry on oeis.org

0, 24, 84, 180, 312, 480, 684, 924, 1200, 1512, 1860, 2244, 2664, 3120, 3612, 4140, 4704, 5304, 5940, 6612, 7320, 8064, 8844, 9660, 10512, 11400, 12324, 13284, 14280, 15312, 16380, 17484, 18624, 19800, 21012, 22260, 23544, 24864, 26220, 27612, 29040, 30504, 32004

Offset: 0

R. K. Guy

Vertical spoke of triangular spiral in A051682. - Paul Barry, Mar 15 2003

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

Cf. A005449, A033580, A049451, A045945, A051682, A081266.

Table[6n(3n+1),{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,24,84},40] (* Harvey P. Dale, Nov 23 2012 *)

a(n)=18*n^2+6*n \\ Charles R Greathouse IV, Feb 19 2017

a(n) = 24*C(n,1) + 36*C(n,2); binomial transform of (0, 24, 36, 0, 0, 0, ...). - Paul Barry, Mar 15 2003
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=24, a(2)=84. - Harvey P. Dale, Nov 23 2012
G.f.: 12*x*(2+x)/(1-x)^3. - Ivan Panchenko, Nov 13 2013
a(n) = 2*A045945(n). - Michel Marcus, Nov 13 2013
a(n) = 12*A005449(n). - R. J. Mathar, Feb 08 2016
From Amiram Eldar, Jan 14 2021: (Start)
Sum_{n>=1} 1/a(n) = 1/2 - Pi/(12*sqrt(3)) - log(3)/4.
Sum_{n>=1} (-1)^(n+1)/a(n) = -1/2 + Pi/(6*sqrt(3)) + log(2)/3. (End)
From Elmo R. Oliveira, Dec 12 2024: (Start)
E.g.f.: 6*exp(x)*x*(4 + 3*x).
a(n) = 6*A049451(n) = 4*A081266(n) = 3*A033580(n). (End)

A045949 Number of equilateral triangles formed out of matches that can be found in a hexagonal chunk of side length n in hexagonal array of matchsticks.

Original entry on oeis.org

0, 6, 38, 116, 262, 496, 840, 1314, 1940, 2738, 3730, 4936, 6378, 8076, 10052, 12326, 14920, 17854, 21150, 24828, 28910, 33416, 38368, 43786, 49692, 56106, 63050, 70544, 78610, 87268, 96540, 106446, 117008, 128246, 140182, 152836, 166230, 180384, 195320, 211058

Offset: 0

R. K. Guy

nonn

N. J. A. Sloane, Illustration of a(1)=6
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1). [From _R. J. Mathar_, Sep 03 2010]

See A008893 for a related sequence.

For hexagons, the number of matches required is A045945, the number of size=1 triangles is A033581, the larger triangles is A307253 and the total number is A045949. For the analogs for triangles see A045943 and for stars see A045946. - John King, Apr 05 2019

List([0..40], n-> (28*n^3 +18*n^2 +4*n -1 +(-1)^n)/8); # G. C. Greubel, Apr 05 2019

[(28*n^3 +18*n^2 +4*n -1 +(-1)^n)/8: n in [0..40]]; // G. C. Greubel, Apr 05 2019

LinearRecurrence[{3,-2,-2,3,-1},{0,6,38,116,262},40] (* or *) CoefficientList[Series[(2*x*(x*(x+2)*(x+5)+3))/((x-1)^4*(x+1)),{x,0,40}],x] (* Harvey P. Dale, Jun 11 2011 *)

A045949(n):=floor(n*(14*n^2+9*n+2)/4)$
makelist(A045949(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

{a(n) = (28*n^3 +18*n^2 +4*n -1 +(-1)^n)/8}; \\ G. C. Greubel, Apr 05 2019

floor(1:25*(14*(1:25)^2+9*(1:25)+2)/4) # Christian N. K. Anderson, Apr 27 2013

[(28*n^3 +18*n^2 +4*n -1 +(-1)^n)/8 for n in (0..40)] # G. C. Greubel, Apr 05 2019

a(n) = floor(n*(14*n^2 + 9*n + 2)/4).
From R. J. Mathar, Sep 03 2010: (Start)
a(n) = +3*a(n-1) -2*a(n-2) -2*a(n-3) +3*a(n-4) -a(n-5).
G.f.: 2*x*(3+10*x+7*x^2+x^3) / ( (1+x)*(1-x)^4 ).
a(n) = (28*n^3 + 18*n^2 + 4*n - 1 + (-1)^n)/8. (End)
a(n) = A033581(n) + A307253(n). - John King, Apr 04 2019
E.g.f.: (x*(25 + 51*x + 14*x^2)*exp(x) - sinh(x))/4. - G. C. Greubel, Apr 05 2019

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

A059986 Number of rods required to make a 3-D cube of side length n.

Original entry on oeis.org

0, 12, 54, 144, 300, 540, 882, 1344, 1944, 2700, 3630, 4752, 6084, 7644, 9450, 11520, 13872, 16524, 19494, 22800, 26460, 30492, 34914, 39744, 45000, 50700, 56862, 63504, 70644, 78300, 86490, 95232, 104544, 114444, 124950, 136080, 147852, 160284, 173394

Offset: 0

Laura Twomey (sxe15(AT)hotmail.com), Mar 07 2001

Equals number of rods making a cube of side length n+1 minus the number of line segments illustrating the isometric projection of a cube of side length n+1 (i.e., the hexagonal matchstick numbers). See the illustration in the links and formula below. - Peter M. Chema, Mar 14 2017

a(n) is also the edge count and intersection number of the (n+1) X (n+1) X (n+1) grid graph. - Eric W. Weisstein, Mar 09 2024

			A 1 X 1 X 1 cube requires 12 rods.

Peter M. Chema, First difference are the hexagonal matchstick numbers or isometric projection of a cube.
Eric Weisstein's World of Mathematics, Edge Count.
Eric Weisstein's World of Mathematics, Grid Graph.
Eric Weisstein's World of Mathematics, Intersection Number.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A045945, A117227.

[3*n*(n+1)^2: n in [0..50]]; // Wesley Ivan Hurt, May 13 2014

A059986:=n->3*n*(n+1)^2; seq(A059986(n), n=0..50); # Wesley Ivan Hurt, May 13 2014

Table[EdgeCount[GridGraph[{n, n, n}]], {n, 39}] (* Geoffrey Critzer, May 17 2009 *)
Table[3 n (n + 1)^2, {n, 0, 50}] (* Wesley Ivan Hurt, May 13 2014 *)
LinearRecurrence[{4, -6, 4, -1}, {0, 12, 54, 144}, 20] (* Eric W. Weisstein, Mar 09 2024 *)
CoefficientList[Series[6 x (2 + x)/(-1 + x)^4, {x, 0, 20}], x] (* Eric W. Weisstein, Mar 09 2024 *)

a(n) = 3*n*(n+1)^2 \\ Charles R Greathouse IV, May 14 2014

a(n) = 3*n*(n+1)^2. - Neven Juric (neven.juric(AT)apis-it.hr), Sep 28 2005
From Geoffrey Critzer, May 17 2009: (Start)
a(n) = a(n-1) + 9*n^2 + 3*n.
O.g.f.: 6*x*(2 + x)/(1 - x)^4.
E.g.f.: 3*x*exp(x)*(x^2 + 5*x + 4). (End)
a(n) = A117227(n^3). - Michel Marcus, Jun 19 2013
For n > 0, a(n) = Sum_{k=1..n} 2*(n+1)(k+n+1), which is the sum of all perimeters of Pythagorean triangles with arms 2*k*(n+1) and (n+1)^2 - k^2 with hypotenuse k^2 + (n+1)^2. - J. M. Bergot, May 12 2014
a(n) = a(n+1) - A045945(n+1). - Peter M. Chema, Mar 14 2017
a(n) = (n-1)*t(n+1) + n*(t(n)+t(n+1)) + (n+1)*(t(n-1)+t(n)+t(n+1)), where t = A000217. - J. M. Bergot, May 30 2017
From Amiram Eldar, Jan 14 2021: (Start)
Sum_{n>=1} 1/a(n) = 2/3 - Pi^2/18.
Sum_{n>=1} (-1)^(n+1)/a(n) = -2/3 + Pi^2/36 + 2*log(2)/3. (End)

More terms from Neven Juric (neven.juric(AT)apis-it.hr), Sep 28 2005

A194767 Denominator of the fourth increasing diagonal of the autosequence of second kind from (-1)^n / (n+1).

Original entry on oeis.org

2, 2, 12, 20, 10, 42, 56, 24, 90, 110, 44, 156, 182, 70, 240, 272, 102, 342, 380, 140, 462, 506, 184, 600, 650, 234, 756, 812, 290, 930, 992, 352, 1122, 1190, 420, 1332, 1406, 494, 1560, 1640, 574, 1806, 1892, 660, 2070, 2162, 752, 2352, 2450, 850, 2652, 2756, 954, 2970, 3080, 1064, 3306, 3422, 1180, 3660

Offset: 0

Paul Curtz, Sep 02 2011

The autosequence of first kind from (-1)^n/(n+1) is A189733.
For the second kind (the second increasing diagonal is (-1)^n/(n+1), half of the main one):
      2,     1,     0,  -1/2,  -1/3,   1/6,   1/2,  5/12,
     -1,    -1,  -1/2,   1/6,   1/2,   1/3, -1/12, -7/20,
      0,   1/2,   2/3,   1/3,  -1/6, -5/12, -4/15,  1/12,
    1/2,   1/6,  -1/3,  -1/2,  -1/4,  3/20,  7/20, 13/60,
   -1/3,  -1/2,  -1/6,   1/4,   2/5,   1/5, -2/15, -3/10,
   -1/6,   1/3,  5/12,  3/20,  -1/5,  -1/3,  -1/6,  5/42,
    1/2,  1/12, -4/15, -7/20, -2/15,   1/6,   2/7,   1/7,
  -5/12, -7/20, -1/12, 13/60,  3/10,  5/42,  -1/7,  -1/4.
Main diagonal: (period 2:repeat 2, -1)/A026741(n+1).
Second (increasing) diagonal: (-1)^n / (n+1).
Third (increasing) diagonal: (-1)^(n+1)*A026741(n) / A045896(n).
Fourth (increasing) diagonal: (-1)^(n+1)*A146535(n)/ a(n).

OEIS Wiki, Autosequence.
Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1).

Cf. A001504 = 2*A060544, A049450 = 2*A000326, A045945 = 6*A005449.

Cf. A026741, A045896, A051176, A146535, A189733, A306368.

c = Table[1/9 (7 n + 7 n^2 + 2 n Cos[2 n *Pi/3] + 2 n^2 Cos[2 n *Pi/3] + 2 Sqrt[3] n Sin[2 n *Pi/3] + 2 Sqrt[3] n^2 Sin[2 n *Pi/3]), {n, 1, 50}] (* Roger Bagula, Mar 25 2012 *)
a[n_] := (n+1) * Numerator[(n+2)/3]; Array[a, 60, 0] (* Amiram Eldar, Sep 17 2023 *)
LinearRecurrence[{0,0,3,0,0,-3,0,0,1},{2,2,12,20,10,42,56,24,90},60] (* Harvey P. Dale, May 15 2025 *)

a(3*n) = (3*n+1)*(3*n+2), a(3*n+1) = (n+1)*(3*n+2), a(3*n+2) = 3*(n+1)*(3*n+4).
G.f.: 2*(1+x+6*x^2+7*x^3+2*x^4+3*x^5+x^6)/(1-x^3)^3. - Jean-François Alcover, Nov 11 2016
a(n+2) = 2 * A306368(n) for n >= 0. - Joerg Arndt, Aug 25 2023
a(n) = (n+1) * A051176(n+2) for n >= 0. - Paul Curtz, Sep 13 2023
Sum_{n>=0} 1/a(n) = 1 + log(3) - Pi/(3*sqrt(3)). - Amiram Eldar, Sep 17 2023

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

A270700 Triangular Star of David numbers (the figurate number of triangles framing a hexagram: a(0) = 12; thereafter a(n) = 36*n + 6).

Original entry on oeis.org

12, 42, 78, 114, 150, 186, 222, 258, 294, 330, 366, 402, 438, 474, 510, 546, 582, 618, 654, 690, 726, 762, 798, 834, 870, 906, 942, 978, 1014, 1050, 1086, 1122, 1158, 1194, 1230, 1266, 1302, 1338, 1374, 1410, 1446, 1482, 1518, 1554, 1590, 1626, 1662, 1698, 1734

Offset: 0

Peter M. Chema, Mar 21 2016

Also known as unitary triangular hexagram numbers, according to the author.
After a(0), the sum of inner and outer perimeters of triangle edges forming each hexagram is [36n - 6], always 12 less than the number of triangles framing the hexagram.  Where a(0)=12, the perimeter is also 12.
Compare with A270545, the number of equilateral triangle units forming perimeters of equilateral triangle, which follows the same application.

			Illustration of initial terms are found in the three above links.

Colin Barker, Table of n, a(n) for n = 0..1000
Peter M. Chema, Illustration of a(2)=78.
Peter M. Chema, Illustration of initial terms [0 through 5].
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A045943, A045945, A045946, A270545, A271114.

[12] cat [36*n + 6: n in [1..50]]; // Vincenzo Librandi, Mar 28 2016

CoefficientList[Series[6 (1 + x) (2 + x)/(1 - x)^2, {x, 0, 40}], x] (* Michael De Vlieger, Mar 23 2016 *)
Join[{12},36*Range[50]+6] (* or *) LinearRecurrence[{2,-1},{12,42,78},50] (* Harvey P. Dale, Nov 03 2016 *)

a(n) = if (!n, 12, 36*n + 6); \\ Michel Marcus, Mar 22 2016

Vec(6*(1+x)*(2+x)/(1-x)^2 + O(x^50)) \\ Colin Barker, Mar 22 2016

a(0) = 12; thereafter, a(n) = 36*n + 6.
From Colin Barker, Mar 22 2016: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 2.
G.f.: 6*(1+x)*(2+x)/(1-x)^2. (End)
From Elmo R. Oliveira, Apr 04 2025: (Start)
E.g.f.: 6*(exp(x)*(6*x + 1) + 1).
a(n) = 6*A271114(n). (End)

More terms from Vincenzo Librandi, Mar 28 2016

A274221 List of quadruples: 3n(3n-1), 3n(3n+1), (3n+1)^2, (3n+2)^2.

Original entry on oeis.org

0, 0, 1, 4, 6, 12, 16, 25, 30, 42, 49, 64, 72, 90, 100, 121, 132, 156, 169, 196, 210, 240, 256, 289, 306, 342, 361, 400, 420, 462, 484, 529, 552, 600, 625, 676, 702, 756, 784, 841, 870, 930, 961, 1024, 1056, 1122, 1156, 1225, 1260, 1332, 1369, 1444, 1482

Offset: 0

Luce ETIENNE, Sep 14 2016

For the formulae of the permutations of A152743, A045945, A016778 and A016790, see the link.

Luce ETIENNE, Permutations
Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,-1,-1,1).

Cf. A152743, A045945, A016778, A016790, A064412, A269064.

&cat [[3*n*(3*n-1), 3*n*(3*n+1), (3*n+1)^2, (3*n+2)^2]: n in [0..15]]; // Bruno Berselli, Sep 15 2016

Flatten[Table[{3 n (3 n - 1), 3 n (3 n + 1), (3 n + 1)^2, (3 n + 2)^2}, {n, 0, 15}]] (* Bruno Berselli, Sep 15 2016 *)

G.f.: x^2*(1+3*x+x^2+3*x^3+x^4)/((1-x)^3*(1+x)^2*(1+x^2)). - Robert Israel, Sep 15 2016
a(n) = (18*n^2-18*n+1-3*(2*n-1)*(-1)^n-4*(-1)^((2*n-1+(-1)^n)/4))/32. Therefore: a(2k) = (18*k^2-12*k+1-(-1)^k)/8, a(2k+1) = (18*k^2+12*k+1-(-1)^k)/8.
a(n) = A064412(n) - A269064(n) for n>0.
E.g.f.: ((9*x^2 - 3*x - 1)*sinh(x) + (9*x^2 + 3*x + 2)*cosh(x) - 2*(sin(x) + cos(x)))/16. - Stefano Spezia, Nov 07 2022

cp's OEIS Frontend

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

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A033580 Four times second pentagonal numbers: a(n) = 2*n*(3*n+1).

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A082040 a(n) = 9*n^2 + 3*n + 1.

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A045946 Star of David matchstick numbers: a(n) = 6*n*(3*n+1).

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A045949 Number of equilateral triangles formed out of matches that can be found in a hexagonal chunk of side length n in hexagonal array of matchsticks.

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A059986 Number of rods required to make a 3-D cube of side length n.

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A033580 Four times second pentagonal numbers: a(n) = 2n(3*n+1).

A082040 a(n) = 9n^2 + 3n + 1.

A045946 Star of David matchstick numbers: a(n) = 6n(3*n+1).

A274221 List of quadruples: 3n(3n-1), 3n(3n+1), (3n+1)^2, (3n+2)^2.