A033580 -id:A033580 - cp's OEIS frontend

A049451 Twice second pentagonal numbers.

0, 4, 14, 30, 52, 80, 114, 154, 200, 252, 310, 374, 444, 520, 602, 690, 784, 884, 990, 1102, 1220, 1344, 1474, 1610, 1752, 1900, 2054, 2214, 2380, 2552, 2730, 2914, 3104, 3300, 3502, 3710, 3924, 4144, 4370, 4602, 4840, 5084, 5334, 5590, 5852, 6120, 6394, 6674, 6960, 7252, 7550, 7854

Offset: 0

Joe Keane (jgk(AT)jgk.org)

From Floor van Lamoen, Jul 21 2001: (Start)
Write 1,2,3,4,... in a hexagonal spiral around 0, then a(n) is the sequence found by reading the line from 0 in the direction 0,4,... . The spiral begins:
                            .
                          52
                          . \
            33--32--31--30  51
            /           . \   \
          34  16--15--14  29  50
          /   /       . \   \   \
        35  17   5---4  13  28  49
        /   /   /   . \   \   \   \
      36  18   6   0   3  12  27  48
      /   /   /   /   /   /   /   /
    37  19   7   1---2  11  26  47
      \   \   \         /   /   /
      38  20   8---9--10  25  46
        \   \             /   /
        39  21--22--23--24  45
          \                 /
          40--41--42--43--44
(End)
Number of edges in the join of the complete bipartite graph of order 2n and the cycle graph of order n, K_n,n * C_n. - Roberto E. Martinez II, Jan 07 2002
The average of the first n elements starting from a(1) is equal to (n+1)^2. - Mario Catalani (mario.catalani(AT)unito.it), Apr 10 2003
If Y is a 4-subset of an n-set X then, for n >= 4, a(n-4) is the number of (n-4)-subsets of X having either one element or two elements in common with Y. - Milan Janjic, Dec 28 2007
With offset 1: the maximum possible sum of numbers in an N x N standard Minesweeper grid. - Dmitry Kamenetsky, Dec 14 2008
a(n) = A001399(6*n-2), number of partitions of 6*n-2 into parts < 4. For example a(2)=14 where the partitions of 6*2-2=10 into parts < 4 are [1,1,1,1,1,1,1,1,1,1], [1,1,1,1,1,1,1,1,2], [1,1,1,1,1,1,1,3], [1,1,1,1,1,1,2,2], [1,1,1,1,1,2,3], [1,1,1,1,2,2,2], [1,1,1,1,3,3], [1,1,1,2,2,3], [1,1,2,2,2,2], [1,1,2,3,3], [1,2,2,2,3], [2,2,2,2,2], [1,3,3,3], [2,2,3,3]. - Adi Dani, Jun 07 2011
A003056 is the following array A read by antidiagonals:
  0,  1,  2,  3,  4,  5, ...
  1,  2,  3,  4,  5,  6, ...
  2,  3,  4,  5,  6,  7, ...
  3,  4,  5,  6,  7,  8, ...
  4,  5,  6,  7,  8,  9, ...
  5,  6,  7,  8,  9, 10, ...
and a(n) is the hook sum Sum_{k=0..n} A(n,k) + Sum_{r=0..n-1} A(r,n). - R. J. Mathar, Jun 30 2013
a(n)*Pi is the total length of 3 points circle center spiral after n rotations. The spiral length at each rotation (L(n)) is A016957. The spiral length ratio rounded down [floor(L(n)/L(1))] is A001651. See illustration in links. - Kival Ngaokrajang, Dec 27 2013
Partial sums give A114364. - Leo Tavares, Feb 25 2022
For n >= 1, the continued fraction expansion of sqrt(27*a(n)) is [9n+1; {2, 2n-1, 1, 4, 1, 2n-1, 2, 18n+2}]. - Magus K. Chu, Oct 13 2022

			From _Dmitry Kamenetsky_, Dec 14 2008, with slight rewording by Raymond Martineau (mart0258(AT)yahoo.com), Dec 16 2008: (Start)
For an N x N Minesweeper grid the highest sum of numbers is (N-1)(3*N-2). This is achieved by filling every second row with mines (shown as 'X'). For example, when N=5 the best grids are:
.
  X X X X X
  4 6 6 6 4
  X X X X X
  4 6 6 6 4
  X X X X X
.
  and
.
  2 3 3 3 2
  X X X X X
  4 6 6 6 4
  X X X X X
  2 3 3 3 2
.
each giving a total of 52. (End)

L. B. W. Jolley, Summation of Series, Dover Publications, 1961, p. 12.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Raghavendra N. Bhat, Cristian Cobeli, and Alexandru Zaharescu, A lozenge triangulation of the plane with integers, arXiv:2403.10500 [math.NT], 2024.
Kival Ngaokrajang, Illustration of 3 points circle center spiral.
Leo Tavares, Illustration: Double Hexagonal Trapezoids.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567, A001105, A002378, A005449, A033580, A049450.
Similar sequences are listed in A316466.
Cf. A003215, A005408, A114364.

List([0..55], n-> n*(3*n+1)); # G. C. Greubel, Sep 01 2019

a049451 n = n * (3 * n + 1)  -- Reinhard Zumkeller, Jul 07 2012

[n*(3*n+1): n in [0..55]]; // G. C. Greubel, Sep 01 2019

Table[n(3n+1), {n,0,55}] (* or *) CoefficientList[Series[2x(2+x)/(1-x)^3, {x,0,55}], x] (* Michael De Vlieger, Apr 05 2017 *)

a(n)=n*(3*n+1) \\ Charles R Greathouse IV, Sep 24 2015

[n*(3*n+1) for n in range(60)] # Gennady Eremin, Feb 27 2022

[n*(3*n+1) for n in (0..55)] # G. C. Greubel, Sep 01 2019

a(n) = n*(3*n+1).
G.f.: 2*x*(2+x)/(1-x)^3.
Sum_{i=1..n} a(i) = A045991(n+1). - Gary W. Adamson, Dec 20 2006
a(n) = 2*A005449(n). - Omar E. Pol, Dec 18 2008
a(n) = a(n-1) + 6*n -2, n > 0. - Vincenzo Librandi, Aug 06 2010
a(n) = A100104(n+1) - A100104(n). - Reinhard Zumkeller, Jul 07 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 4, a(2) = 14. - Philippe Deléham, Mar 26 2013
a(n) = A174709(6*n+3). - Philippe Deléham, Mar 26 2013
a(n) = (24/(n+2)!)*Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*j^(n+2). - Bruno Berselli, Jun 04 2013 - after the similar formula of Vladimir Kruchinin in A002411
a(n) = A002061(n+1) + A056220(n). - Bruce J. Nicholson, Sep 21 2017
a(n) = Sum_{i = 2..5} P(i,n), where P(i,m) = m*((i-2)*m-(i-4))/2. - Bruno Berselli, Jul 04 2018
E.g.f.: x*(4 + 3*x)*exp(x). - G. C. Greubel, Sep 01 2019
a(n) = A003215(n) - A005408(n). - Leo Tavares, Feb 25 2022
From Amiram Eldar, Feb 27 2022: (Start)
Sum_{n>=1} 1/a(n) = 3 - Pi/(2*sqrt(3)) - 3*log(3)/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/sqrt(3) + 2*log(2) - 3. (End)
a(n) = A001105(n) + A002378(n). - Torlach Rush, Jul 11 2022

A033579 Four times pentagonal numbers: a(n) = 2n(3*n-1).

Original entry on oeis.org

0, 4, 20, 48, 88, 140, 204, 280, 368, 468, 580, 704, 840, 988, 1148, 1320, 1504, 1700, 1908, 2128, 2360, 2604, 2860, 3128, 3408, 3700, 4004, 4320, 4648, 4988, 5340, 5704, 6080, 6468, 6868, 7280, 7704, 8140, 8588, 9048, 9520, 10004, 10500, 11008, 11528, 12060

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, 4, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - Omar E. Pol, Sep 08 2011

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Pentagonal Number.
Wikipedia, Pentagonal number.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000326, A001318, A014642, A033579, A033580, A033581, A049450, A186423.

List([0..45], n-> 4*Binomial(3*n,2)/3 ); # G. C. Greubel, Oct 09 2019

[4*Binomial(3*n,2)/3: n in [0..45]]; // G. C. Greubel, Oct 09 2019

seq(4*binomial(3*n,2)/3, n=0..45); # G. C. Greubel, Oct 09 2019

4 PolygonalNumber[5, Range[0, 45]] (* Michael De Vlieger, Aug 02 2016, Version 10.4 *)

a(n)=2*n*(3*n-1) \\ Charles R Greathouse IV, Jun 28 2013

[4*binomial(3*n,2)/3 for n in (0..45)] # G. C. Greubel, Oct 09 2019

a(n) = 4*n*(3*n-1)/2 = 6*n^2 - 2*n = 4*A000326(n). - Omar E. Pol, Dec 11 2008
a(n) = 2*A049450(n). - Omar E. Pol, Dec 13 2008
a(n) = a(n-1) + 12*n - 8 for n > 0, a(0)=0. - Vincenzo Librandi, Aug 05 2010
a(n) = A014642(n)/2. - Omar E. Pol, Aug 19 2011
G.f.: x*(4+8*x)/(1-3*x+3*x^2-x^3). - Colin Barker, Jan 06 2012
a(n) = A191967(2*n). - Reinhard Zumkeller, Jul 07 2012
a(n) = A181617(n+1) - A181617(n). - J. M. Bergot, Jun 28 2013
a(n) = (A174371(n) - 1)/6. - Miquel Cerda, Jul 28 2016
From Ilya Gutkovskiy, Jul 28 2016: (Start)
E.g.f.: 2*x*(2 + 3*x)*exp(x).
a(n+1) = Sum_{k=0..n} A017569(k).
Sum_{i>0} 1/a(i) = (9*log(3) - sqrt(3)*Pi)/12 = 0.3705093754425278... (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(2*sqrt(3)) - log(2). - Amiram Eldar, Feb 20 2022

More terms from Michel Marcus, Mar 04 2014

A062717 Numbers m such that 6*m+1 is a perfect square.

Original entry on oeis.org

0, 4, 8, 20, 28, 48, 60, 88, 104, 140, 160, 204, 228, 280, 308, 368, 400, 468, 504, 580, 620, 704, 748, 840, 888, 988, 1040, 1148, 1204, 1320, 1380, 1504, 1568, 1700, 1768, 1908, 1980, 2128, 2204, 2360, 2440, 2604, 2688, 2860, 2948, 3128, 3220, 3408, 3504

Offset: 1

Jason Earls, Jul 14 2001

X values of solutions to the equation 6*X^3 + X^2 = Y^2. - Mohamed Bouhamida, Nov 06 2007
Arithmetic averages of the k triangular numbers 0, 1, 3, 6, ..., (k-1)*k/2 that take integer values. - Vladimir Joseph Stephan Orlovsky, Aug 05 2009 [edited by Jon E. Schoenfield, Jan 10 2015]
Even terms in A186423; union of A033579 and A033580, A010052(6*a(n)+1) = 1. - Reinhard Zumkeller, Feb 21 2011
a(n) are integers produced by Sum_{i = 1..k-1} i*(k-i)/k, for some k > 0. Values for k are given by A007310 = sqrt(6*a(n)+1), the square roots of those perfect squares. - Richard R. Forberg, Feb 16 2015
Equivalently, numbers of the form 2*h*(3*h+1), where h = 0, -1, 1, -2, 2, -3, 3, -4, 4, ... (see also the sixth comment of A152749). - Bruno Berselli, Feb 02 2017

Harry J. Smith, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Equals 4 * A001318.
Cf. A005563, A046092, A001082, A002378, A036666.
Cf. A160757, A000217. - Vladimir Joseph Stephan Orlovsky, Aug 05 2009
Cf. A007310.
Diagonal of array A323674. - Sally Myers Moite, Feb 03 2019

[(6*n*(n-1) + (2*n-1)*(-1)^n + 1)/4: n in [1..70]]; // Wesley Ivan Hurt, Apr 21 2021

seq(n^2+n+2*ceil(n/2)^2,n=0..48); # Gary Detlefs, Feb 23 2010

Select[Range[0, 3999], IntegerQ[Sqrt[6# + 1]] &] (* Harvey P. Dale, Mar 10 2013 *)

je=[]; for(n=0,7000, if(issquare(6*n+1),je=concat(je,n))); je

{ n=0; for (m=0, 10^9, if (issquare(6*m + 1), write("b062717.txt", n++, " ", m); if (n==1000, break)) ) } \\ Harry J. Smith, Aug 09 2009

def A062717(n): return (n*(3*n + 4) + 1 if n&1 else n*(3*n + 2))>>1 # Chai Wah Wu, Jan 31 2023

G.f.: 4*x^2*(1 + x + x^2) / ( (1+x)^2*(1-x)^3 ).
a(2*k) = k*(6*k+2), a(2*k+1) = 6*k^2 + 10*k + 4. - Mohamed Bouhamida, Nov 06 2007
a(n) = n^2 - n + 2*ceiling((n-1)/2)^2. - Gary Detlefs, Feb 23 2010
From Bruno Berselli, Nov 28 2010: (Start)
a(n) = (6*n*(n-1) + (2*n-1)*(-1)^n + 1)/4.
6*a(n) + 1 = A007310(n)^2. (End)
E.g.f.: (3*x^2*exp(x) - x*exp(-x) + sinh(x))/2. - Ilya Gutkovskiy, Jun 18 2016
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Apr 21 2021
From Amiram Eldar, Mar 11 2022: (Start)
Sum_{n>=2} 1/a(n) = (9-sqrt(3)*Pi)/6.
Sum_{n>=2} (-1)^n/a(n) = 3*(log(3)-1)/2. (End)

A254963 a(n) = n(11n + 3)/2.

Original entry on oeis.org

0, 7, 25, 54, 94, 145, 207, 280, 364, 459, 565, 682, 810, 949, 1099, 1260, 1432, 1615, 1809, 2014, 2230, 2457, 2695, 2944, 3204, 3475, 3757, 4050, 4354, 4669, 4995, 5332, 5680, 6039, 6409, 6790, 7182, 7585, 7999, 8424, 8860, 9307, 9765, 10234, 10714, 11205, 11707

Offset: 0

Bruno Berselli, Feb 11 2015

This sequence provides the first differences of A254407 and the partial sums of A017473.
Also:
a(n) - n         = A022269(n);
a(n) + n         = n*(11*n+5)/2: 0, 8, 27, 57, 98, 150, 213, 287, ...;
a(n) - 2*n       = A022268(n);
a(n) + 2*n       = n*(11*n+7)/2: 0, 9, 29, 60, 102, 155, 219, 294, ...;
a(n) - 3*n       = n*(11*n-3)/2: 0, 4, 19, 45, 82, 130, 189, 259, ...;
a(n) + 3*n       = A211013(n);
a(n) - 4*n       = A226492(n);
a(n) + 4*n       = A152740(n);
a(n) - 5*n       = A180223(n);
a(n) + 5*n       = n*(11*n+13)/2: 0, 12, 35, 69, 114, 170, 237, 315, ...;
a(n) - 6*n       = A051865(n);
a(n) + 6*n       = n*(11*n+15)/2: 0, 13, 37, 72, 118, 175, 243, 322, ...;
a(n) - 7*n       = A152740(n-1) with A152740(-1) = 0;
a(n) + 7*n       = n*(11*n+17)/2: 0, 14, 39, 75, 122, 180, 249, 329, ...;
a(n) - n*(n-1)/2 = A168668(n);
a(n) + n*(n-1)/2 = A049453(n);
a(n) - n*(n+1)/2 = A202803(n);
a(n) + n*(n+1)/2 = A033580(n).

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

Cf. A008729 and A218530 (seventh column); A017473, A254407.
Cf. similar sequences of the type 4*n^2 + k*n*(n+1)/2: A055999 (k=-7, n>6), A028552 (k=-6, n>2), A095794 (k=-5, n>1), A046092 (k=-4, n>0), A000566 (k=-3), A049450 (k=-2), A022264 (k=-1), A016742 (k=0), A022267 (k=1), A202803 (k=2), this sequence (k=3), A033580 (k=4).
Cf. A069125: (2*n+1)^2 + 3*n*(n+1)/2; A147875: n^2 + 3*n*(n+1)/2.
Cf. A022268, A022269, A049453, A051865, A152740, A168668, A180223, A211013, A226492.

Magma
```
[n*(11*n+3)/2: n in [0..50]];
```

Table[n (11 n + 3)/2, {n, 0, 50}]
LinearRecurrence[{3,-3,1},{0,7,25},50] (* Harvey P. Dale, Mar 25 2018 *)

Maxima
```
makelist(n*(11*n+3)/2, n, 0, 50);
```
PARI
```
vector(50, n, n--; n*(11*n+3)/2)
```
Sage
```
[n*(11*n+3)/2 for n in (0..50)]
```

G.f.: x*(7 + 4*x)/(1 - x)^3.
From Elmo R. Oliveira, Dec 15 2024: (Start)
E.g.f.: exp(x)*x*(14 + 11*x)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A139267 Twice octagonal numbers: 2n(3*n-2).

Original entry on oeis.org

0, 2, 16, 42, 80, 130, 192, 266, 352, 450, 560, 682, 816, 962, 1120, 1290, 1472, 1666, 1872, 2090, 2320, 2562, 2816, 3082, 3360, 3650, 3952, 4266, 4592, 4930, 5280, 5642, 6016, 6402, 6800, 7210, 7632, 8066, 8512, 8970, 9440, 9922

Offset: 0

Omar E. Pol, May 14 2008, May 19 2008

Sequence found by reading the line from 0, in the direction 0, 2,..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Opposite numbers to the members of A033580 in the same spiral. - Omar E. Pol, Sep 09 2011

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

Cf. A000567, A017545.

Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=12).

List([0..50], n-> 2*n*(3*n-2)); # G. C. Greubel, Sep 18 2019

[2*n*(3*n-2): n in [0..50]]; // G. C. Greubel, Sep 18 2019

seq(2*n*(3*n-2), n=0..50); # G. C. Greubel, Sep 18 2019

Table[2*n*(3*n-2), {n,0,50}] (* G. C. Greubel, Jun 07 2017 *)

a(n)=2*n*(3*n-2) \\ Charles R Greathouse IV, Oct 07 2015

[2*n*(3*n-2) for n in (0..50)] # G. C. Greubel, Sep 18 2019

a(n) = 2*A000567(n) = 6*n^2 - 4*n = 2*n*(3*n - 2).
a(n) = a(n-1) + 12*n - 10, with n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
G.f.: x*(2+10*x)/(1-3*x+3*x^2-x^3). - Colin Barker, Jan 06 2012
After 0, a(n) = Sum_{i=0..n-1} (12*i + 2). - Bruno Berselli, Sep 11 2013
E.g.f.: 2*x*(1 + 3*x)*exp(x). - G. C. Greubel, Sep 18 2019

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)

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)

A136392 a(n) = 6n^2 - 10n + 5.

Original entry on oeis.org

1, 9, 29, 61, 105, 161, 229, 309, 401, 505, 621, 749, 889, 1041, 1205, 1381, 1569, 1769, 1981, 2205, 2441, 2689, 2949, 3221, 3505, 3801, 4109, 4429, 4761, 5105, 5461, 5829, 6209, 6601, 7005, 7421, 7849, 8289, 8741, 9205, 9681, 10169, 10669, 11181, 11705, 12241

Offset: 1

Gary W. Adamson, Dec 28 2007

Binomial transform of [1, 8, 12, 0, 0, 0, ...].
Numbers k such that 6*k - 5 is the square of a number of the form 6*k - 5, contained in A199859. - Eleonora Echeverri-Toro, Nov 29 2011
Central terms of the triangle A033292. - Reinhard Zumkeller, Feb 06 2012
Sequence found by reading the line from 1, in the direction 1, 9, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - Omar E. Pol, Jul 18 2012

Kelvin Voskuijl, Table of n, a(n) for n = 1..10000
John Elias, Illustration: Centered nesting cube frames.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567, A001318, A016777, A033292, A033580, A059722, A199859, A201279, A204675.

a136392 n = 2 * n * (3*n - 5) + 5 -- Reinhard Zumkeller, Feb 06 2012

Table[6n^2-10n+5,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{1,9,29},50] (* Harvey P. Dale, Mar 05 2023 *)

a(n)=6*n^2-10*n+5 \\ Charles R Greathouse IV, Nov 29 2011

a(n) = n*(3*n - 2) + (n-1)*(3*n - 5), n > 1.
a(n) = n*A016777(n-1) + (n-1)*A016777(n-2).
a(n) = a(n-1) + 12*n - 16 (with a(1)=1). - Vincenzo Librandi, Nov 24 2010
G.f.: x*(1+x)*(1+5*x)/(1-x)^3. - Colin Barker, Jan 09 2012
a(n) = 1 + A033580(n-1). - Omar E. Pol, Jul 18 2012
a(n) = A059722(n) - A059722(n-1). - J. M. Bergot, Nov 02 2012
a(n) = A000567(n-1) + A000567(n). - Charlie Marion, May 29 2024
From Elmo R. Oliveira, Oct 31 2024: (Start)
E.g.f.: exp(x)*(2*x*(3*x - 2) + 5) - 5.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A202804 a(n) = n(6n+4).

Original entry on oeis.org

0, 10, 32, 66, 112, 170, 240, 322, 416, 522, 640, 770, 912, 1066, 1232, 1410, 1600, 1802, 2016, 2242, 2480, 2730, 2992, 3266, 3552, 3850, 4160, 4482, 4816, 5162, 5520, 5890, 6272, 6666, 7072, 7490, 7920, 8362, 8816, 9282, 9760, 10250, 10752, 11266, 11792, 12330

Offset: 0

Jeremy Gardiner, Dec 24 2011

Sequence found by reading the line from 0, in the direction 0, 10, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Opposite numbers to the members of A033579 in the same spiral. - Omar E. Pol, Jul 17 2012

Partial sums give A163815. - Leo Tavares, Feb 25 2022

Jeremy Gardiner, Table of n, a(n) for n = 0..3000
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Leo Tavares, Illustration: Star Halves.
Pavlos Vavolas, Cellular automaton model of cardiac arrhythmias, University of Sheffield Department of Computer Science (2005), (see page 43). [broken link, abstract]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A003154, A005408, A033580, A033581, A049453, A163815, A195319.

Cf. A001318, A033579, A045944, A080859.

A202804:=n->n*(6*n+4): seq(A202804(n), n=0..100); # Wesley Ivan Hurt, Apr 09 2017

Table[n(6n+4),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,10,32},50] (* Harvey P. Dale, Dec 28 2015 *)

x='x + O('x^50); concat([0], Vec(-2*x*(5 + x)/(x - 1)^3)) \\ Indranil Ghosh, Apr 10 2017

a(n) = 2*n(3*n+2) = 6*n^2 + 4*n = 2*A045944(n).
a(n) = A080859(n) - 1. - Omar E. Pol, Jul 18 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Dec 28 2015
G.f.: 2*x*(5 + x)/(1 - x)^3. - Indranil Ghosh, Apr 10 2017
a(n) = A003154(n+1) - A005408(n). - Leo Tavares, Feb 25 2022
From Amiram Eldar, Mar 01 2022: (Start)
Sum_{n>=1} 1/a(n) = (Pi/sqrt(3) - 3*log(3) + 3)/8.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(4*sqrt(3)) - 3/8. (End)
E.g.f.: 2*exp(x)*x*(5 + 3*x). - Elmo R. Oliveira, Dec 12 2024

A282513 a(n) = floor((3*n + 2)^2/24 + 1/3).

Original entry on oeis.org

0, 1, 3, 5, 8, 12, 17, 22, 28, 35, 43, 51, 60, 70, 81, 92, 104, 117, 131, 145, 160, 176, 193, 210, 228, 247, 267, 287, 308, 330, 353, 376, 400, 425, 451, 477, 504, 532, 561, 590, 620, 651, 683, 715, 748, 782, 817, 852, 888, 925, 963

Offset: 0

Luce ETIENNE, Feb 17 2017

List of quadruples: 2*n*(3*n+1), (2*n+1)*(3*n+1), 6*n^2+8*n+3, (n+1)*(6*n+5). These terms belong to the sequences A033580, A033570, A126587 and A049452, respectively. See links for all the permutations.
After 0, subsequence of A025767.
It seems that a(n) is the smallest number of cells that need to be painted in a (n+1) X (n+1) grid, such that it has no unpainted hexominoes (see link to Kamenetsky and Pratt). - Rob Pratt, Dmitry Kamenetsky, Aug 30 2020

			Rectangular array with four columns:
.   0,   1,   3,   5;
.   8,  12,  17,  22;
.  28,  35,  43,  51;
.  60,  70,  81,  92;
. 104, 117, 131, 145, etc.
From _Rob Pratt_, Aug 30 2020: (Start)
For n = 3, painting only 2 cells would leave an unpainted hexomino, but painting the following 3 cells avoids all unpainted hexominoes:
    . . .
    . . X
    X X .
(End)

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Luce ETIENNE, Permutations
Dmitry Kamenetsky and Rob Pratt, 10x10 grid with no unpainted hexominoes, Puzzling StackExchange, October 2019.
Rob Pratt, Optimal solution for n=11.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

Cf. A033436: floor((3*n)^2/24 + 1/3).
Cf. A000326, A000567, A025767, A033570, A033580, A049452, A064412, A126587, A222017, A269064, A274221.
Cf. A130519.
Minimum number of painted cells in other n-ominoes: A337501, A337502, A337503.

[(3*n^2+4*n+4) div 8: n in [0..50]]; // Bruno Berselli, Feb 17 2017

Table[Floor[(3 n + 2)^2/24 + 1/3], {n, 0, 50}] (* or *) CoefficientList[Series[x (1 + x + x^3)/((1 + x) (1 + x^2) (1 - x)^3), {x, 0, 50}], x] (* or *) Table[(6 n^2 + 8 n + 3 + Cos[n Pi] - 4 Cos[n Pi/2])/16, {n, 0, 50}] (* or *) Table[(3 n + 2)^2/24 + 1/3 + (-6 + (1 + (-1)^n) (1 + 2 I^((n + 1) (n + 2))))/16, {n, 0, 50}] (* Michael De Vlieger, Feb 17 2017 *)
LinearRecurrence[{2,-1,0,1,-2,1},{0,1,3,5,8,12},60] (* Harvey P. Dale, Aug 10 2024 *)

a(n)=(3*n^2 + 4*n + 4)\8 \\ Charles R Greathouse IV, Feb 17 2017

G.f.: x*(1 + x + x^3)/((1 + x)*(1 + x^2)*(1 - x)^3).
a(n) = 2*a(n-1) - a(n-2) + a(n-4) - 2*a(n-5) + a(n-6) for n>5.
a(n) = floor((3*n + 2)^2/24 + 2/3).
a(n) = (6*n^2 + 8*n + 3 + (-1)^n - 2*((-1)^((2*n - 1 + (-1)^n)/4) + (-1)^((2*n + 1 - (-1)^n)/4)))/16. Therefore:
a(2*k)   = (6*k^2 + 4*k + 1 - (-1)^k)/4,
a(2*k+1) = (k + 1)*(3*k + 2)/2.
a(n) = (6*n^2 + 8*n + 3 + cos(n*Pi) - 4*cos(n*Pi/2))/16.
a(n) = (3*n + 2)^2/24 + 1/3 + (-6 + (1 + (-1)^n)*(1 + 2*i^((n+1)*(n+2))))/16, where i=sqrt(-1).
a(n) = A130519(n+3)+A130519(n+2)+A130519(n). - R. J. Mathar, Jun 23 2021

Corrected and extended by Bruno Berselli, Feb 17 2017

cp's OEIS Frontend

A049451 Twice second pentagonal numbers.

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A033579 Four times pentagonal numbers: a(n) = 2*n*(3*n-1).

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A062717 Numbers m such that 6*m+1 is a perfect square.

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A254963 a(n) = n*(11*n + 3)/2.

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A139267 Twice octagonal numbers: 2*n*(3*n-2).

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

A033579 Four times pentagonal numbers: a(n) = 2n(3*n-1).

A254963 a(n) = n(11n + 3)/2.

A139267 Twice octagonal numbers: 2n(3*n-2).

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

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

A136392 a(n) = 6n^2 - 10n + 5.

A202804 a(n) = n(6n+4).