A003215 -id:A003215 - 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

A069099 Centered heptagonal numbers.

Original entry on oeis.org

1, 8, 22, 43, 71, 106, 148, 197, 253, 316, 386, 463, 547, 638, 736, 841, 953, 1072, 1198, 1331, 1471, 1618, 1772, 1933, 2101, 2276, 2458, 2647, 2843, 3046, 3256, 3473, 3697, 3928, 4166, 4411, 4663, 4922, 5188, 5461, 5741, 6028, 6322, 6623, 6931, 7246

Offset: 1

Terrel Trotter, Jr., Apr 05 2002

Equals the triangular numbers convolved with [ 1, 5, 1, 0, 0, 0, ...]. - Gary W. Adamson and Alexander R. Povolotsky, May 29 2009
Number of ordered pairs of integers (x,y) with abs(x) < n, abs(y) < n and abs(x + y) < n, counting twice pairs of equal numbers. - Reinhard Zumkeller, Jan 23 2012; corrected and extended by Mauro Fiorentini, Jan 01 2018
The number of pairs without repetitions is a(n) - 2n + 3 for n > 1. For example, there are 19 such pairs for n = 3: (-2, 0), (-2, 1), (-2, 2), (-1, -1), (-1, 0), (-1, 1), (-1, 2), (0, -2), (0, -1), (0, 0), (0, 1), (0, 2), (1, -2), (1, -1), (1, 0), (1, 1), (2, -2), (2, -1), (2, 0). - Mauro Fiorentini, Jan 01 2018

			a(5) = 71 because 71 = (7*5^2 - 7*5 + 2)/2 = (175 - 35 + 2)/2 = 142/2.
From _Bruno Berselli_, Oct 27 2017: (Start)
1   =         -(0) + (1).
8   =       -(0+1) + (2+3+4).
22  =     -(0+1+2) + (3+4+5+6+7).
43  =   -(0+1+2+3) + (4+5+6+7+8+9+10).
71  = -(0+1+2+3+4) + (5+6+7+8+9+10+11+12+13). (End)

T. D. Noe, Table of n, a(n) for n = 1..1000
Nicolas Bělohoubek and Viktor Javor, Centered heptagonal numbers appearing in aperiodic heptagonal tiling
Leo Tavares, Illustration: Crystal Numbers
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers.
Index entries for sequences related to centered polygonal numbers
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000566 (heptagonal numbers).
Cf. A001263, A057655, A001106.
Cf. A003215, A000217.

a069099 n = length
   [(x,y) | x <- [-n+1..n-1], y <- [-n+1..n-1], x + y <= n - 1]
-- Reinhard Zumkeller, Jan 23 2012

FoldList[#1 + #2 &, 1, 7 Range@ 50] (* Robert G. Wilson v, Feb 02 2011 *)
LinearRecurrence[{3,-3,1},{1,8,22},50] (* Harvey P. Dale, Jun 04 2011 *)

a(n)=(7*n^2-7*n+2)/2 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = (7*n^2 - 7*n + 2)/2.
a(n) = 1 + Sum_{k=1..n} 7*k. - Xavier Acloque, Oct 26 2003
Binomial transform of [1, 7, 7, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 7, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
a(n) = 7*n + a(n-1) - 7 (with a(1)=1). - Vincenzo Librandi, Aug 08 2010
G.f.: x*(1+5*x+x^2) / (1-x)^3. - R. J. Mathar, Feb 04 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=1, a(1)=8, a(2)=22. - Harvey P. Dale, Jun 04 2011
a(n) = A024966(n-1) + 1. - Omar E. Pol, Oct 03 2011
a(n) = 2*a(n-1) - a(n-2) + 7. - Ant King, Jun 17 2012
From Ant King, Jun 17 2012: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi/sqrt(7)*tanh(Pi/(2*sqrt(7))) = 1.264723171685652...
a(n) == 1 (mod 7) for all n.
The sequence of digital roots of the a(n) is period 9: repeat [1, 8, 4, 7, 8, 7, 4, 8, 1] (the period is a palindrome).
The sequence of a(n) mod 10 is period 20: repeat [1, 8, 2, 3, 1, 6, 8, 7, 3, 6, 6, 3, 7, 8, 6, 1, 3, 2, 8, 1] (the period is a palindrome).
(End)
E.g.f.: -1 +  (2 + 7*x^2)*exp(x)/2. - Ilya Gutkovskiy, Jun 30 2016
a(n) = A101321(7,n-1). - R. J. Mathar, Jul 28 2016
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=1} a(n)/n! = 9*e/2 - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 9/(2*e) - 1. (End)
a(n) = A003215(n-1) + A000217(n-1). - Leo Tavares, Jul 19 2022

A008458 Coordination sequence for hexagonal lattice.

Original entry on oeis.org

1, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 186, 192, 198, 204, 210, 216, 222, 228, 234, 240, 246, 252, 258, 264, 270, 276, 282, 288, 294, 300, 306, 312, 318, 324, 330, 336, 342, 348

Offset: 0

N. J. A. Sloane

The hexagonal lattice is the familiar 2-dimensional lattice in which each point has 6 neighbors. This is sometimes called the triangular lattice. It is also the planar net 3.3.3.3.3.3.
Coordination sequence for 2-dimensional cyclotomic lattice Z[zeta_6].
Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0( 20 ).
Also the Engel expansion of exp^(1/6); cf. A006784 for the Engel expansion definition. - Benoit Cloitre, Mar 03 2002
Numbers k such that k+floor(k/2) | k*floor(k/2). - Wesley Ivan Hurt, Dec 01 2020

			From _Omar E. Pol_, Aug 20 2011: (Start)
Illustration of initial terms:
.                                             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
. 1             o o o       o       o       o           o
.       6                    o o o o         o         o
.                 12                          o o o o o
.                               18
.                                                 24
(End)
G.f. = 1 + 6*x + 12*x^2 + 18*x^3 + 24*x^4 + 30*x^5 + 36*x^6 + 42*x^7 + 48*x^8 + 54*x^9 + ...

T. D. Noe, Table of n, a(n) for n = 0..1000
M. Beck and S. Hosten, Cyclotomic polytopes and growth series of cyclotomic lattices, arXiv:math/0508136 [math.CO], 2005-2006.
J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).
Brian Galebach, k-uniform tilings (k <= 6) and their A-numbers
Chaim Goodman-Strauss and N. J. A. Sloane, A Coloring Book Approach to Finding Coordination Sequences, Acta Cryst. A75 (2019), 121-134, also on NJAS's home page. Also on arXiv, arXiv:1803.08530 [math.CO], 2018-2019.
Rostislav Grigorchuk and Cosmas Kravaris, On the growth of the wallpaper groups, arXiv:2012.13661 [math.GR], 2020. See section 4.1 p. 19.
Branko Grünbaum and Geoffrey C. Shephard, Tilings by regular polygons, Mathematics Magazine, 50 (1977), 227-247.
Tom Karzes, Tiling Coordination Sequences
G. Nebe and N. J. A. Sloane, Home page for hexagonal (or triangular) lattice A2
Reticular Chemistry Structure Resource, hxl
N. J. A. Sloane, The uniform planar nets and their A-numbers [Annotated scanned figure from Gruenbaum and Shephard (1977)]
N. J. A. Sloane, Overview of coordination sequences of Laves tilings [Fig. 2.7.1 of Grünbaum-Shephard 1987 with A-numbers added and in some cases the name in the RCSR database]
William A. Stein, Dimensions of the spaces S_k(Gamma_0(N))
William A. Stein, The modular forms database
Index entries for sequences related to A2 = hexagonal = triangular lattice
Index entries for linear recurrences with constant coefficients, signature (2,-1).
Index entries for coordination sequences

Essentially the same as A008588.
List of coordination sequences for uniform planar nets: A008458 (the planar net 3.3.3.3.3.3), A008486 (6^3), A008574(4.4.4.4 and 3.4.6.4), A008576 (4.8.8), A008579(3.6.3.6), A008706 (3.3.3.4.4), A072154 (4.6.12), A219529(3.3.4.3.4), A250120 (3.3.3.3.6), A250122 (3.12.12).
List of coordination sequences for Laves tilings (or duals of uniform planar nets): [3,3,3,3,3.3] = A008486; [3.3.3.3.6] = A298014, A298015, A298016; [3.3.3.4.4] = A298022, A298024; [3.3.4.3.4] = A008574, A296368; [3.6.3.6] = A298026, A298028; [3.4.6.4] = A298029, A298031, A298033; [3.12.12] = A019557, A298035; [4.4.4.4] = A008574; [4.6.12] = A298036, A298038, A298040; [4.8.8] = A022144, A234275; [6.6.6] = A008458.
Cf. A032528. - Omar E. Pol, Aug 20 2011
Cf. A048477 (binomial Transf.)

[0^n+6*n: n in [0..60] ]; // Vincenzo Librandi, Aug 21 2011

Maple
```
1, seq(6*n, n=1..65);
```

Join[{1},6*Range[60]] (* Harvey P. Dale, Jul 21 2013 *)
a[ n_] := Boole[n == 0] + 6 n; (* Michael Somos, May 21 2015 *)

makelist(if n=0 then 1 else 6*n,n,0,65); /* Martin Ettl, Nov 12 2012 */

PARI
```
{a(n) = 6*n + (!n)};
```

[6*n+int(n==0) for n in range(66)] # G. C. Greubel, May 25 2023

G.f.: (1 + 4*x + x^2)/(1 - x)^2.
a(n) = A003215(n) - A003215(n-1), n > 0.
Equals binomial transform of [1, 5, 1, -1, 1, -1, 1, ...]. - Gary W. Adamson, Jul 08 2008
G.f.: Hypergeometric2F1([3,-2], [1], -x/(1-x)). - Paul Barry, Sep 18 2008
a(n) = 0^n + 6*n. - Vincenzo Librandi, Aug 21 2011
n*a(1) + (n-1)*a(2) + (n-2)*a(3) + ... + 2*a(n-1) + a(n) = n^3. - Warren Breslow, Oct 28 2013
E.g.f.: 1 + 6*x*exp(x). - Stefano Spezia, Jun 26 2022

A054552 a(n) = 4n^2 - 3n + 1.

Original entry on oeis.org

1, 2, 11, 28, 53, 86, 127, 176, 233, 298, 371, 452, 541, 638, 743, 856, 977, 1106, 1243, 1388, 1541, 1702, 1871, 2048, 2233, 2426, 2627, 2836, 3053, 3278, 3511, 3752, 4001, 4258, 4523, 4796, 5077, 5366, 5663, 5968, 6281, 6602, 6931, 7268, 7613, 7966, 8327

Offset: 0

Enoch Haga and G. L. Honaker, Jr., Apr 09 2000

Also indices in any square spiral organized like A054551.
Equals binomial transform of [1, 1, 8, 0, 0, 0, ...]. - Gary W. Adamson, May 11 2008
Ulam's spiral (E spoke). - Robert G. Wilson v, Oct 31 2011
For n > 0: left edge of the triangle A033293. - Reinhard Zumkeller, Jan 18 2012

			The spiral begins:
.
197-196-195-194-193-192-191-190-189-188-187-186-185-184-183
  |                                                       |
198 145-144-143-142-141-140-139-138-137-136-135-134-133 182
  |   |                                               |   |
199 146 101-100--99--98--97--96--95--94--93--92--91 132 181
  |   |   |                                       |   |   |
200 147 102  65--64--63--62--61--60--59--58--57  90 131 180
  |   |   |   |                               |   |   |   |
201 148 103  66  37--36--35--34--33--32--31  56  89 130 179
  |   |   |   |   |                       |   |   |   |   |
202 149 104  67  38  17--16--15--14--13  30  55  88 129 178
  |   |   |   |   |   |               |   |   |   |   |   |
203 150 105  68  39  18   5---4---3  12  29  54  87 128 177
  |   |   |   |   |   |   |       |   |   |   |   |   |   |
204 151 106  69  40  19   6   1---2  11  28  53  86 127 176
  |   |   |   |   |   |   |           |   |   |   |   |   |
205 152 107  70  41  20   7---8---9--10  27  52  85 126 175
  |   |   |   |   |   |                   |   |   |   |   |
206 153 108  71  42  21--22--23--24--25--26  51  84 125 174
  |   |   |   |   |                           |   |   |   |
207 154 109  72  43--44--45--46--47--48--49--50  83 124 173
  |   |   |   |                                   |   |   |
208 155 110  73--74--75--76--77--78--79--80--81--82 123 172
  |   |   |                                           |   |
209 156 111-112-113-114-115-116-117-118-119-120-121-122 171
  |   |                                                   |
210 157-158-159-160-161-162-163-164-165-166-167-168-169-170
  |
211-212-213-214-215-216-217-218-219-220-221-222-223-224-225
.
- _Robert G. Wilson v_, Jul 04 2014

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Scientific American, Cover of the March 1964 issue
Leo Tavares, Illustration: Hexagon/Square Pairs
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A033293, A054551, A108781.
Spokes of square spiral: A054552, A054554, A054556, A053755, A054567, A054569, A033951, A016754.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Cf. A003215.

[4*n^2-3*n+1 : n in [0..50]]; // Wesley Ivan Hurt, Jul 11 2014

A054552:=n->4*n^2-3*n+1: seq(A054552(n), n=0..50); # Wesley Ivan Hurt, Jul 11 2014

f[n_] := 4*n^2 - 3*n + 1; Array[f, 50, 0] (* Vladimir Joseph Stephan Orlovsky, Sep 01 2008 *)
LinearRecurrence[{3,-3,1},{1,2,11},50] (* Harvey P. Dale, Jun 01 2024 *)

a(n)= 4*n^2-3*n+1 \\ Charles R Greathouse IV, Jan 15 2012

G.f.: (1 - x + 8*x^2)/(1-x)^3.
a(n) = 8*n + a(n-1) - 7 (with a(0)=1). - Vincenzo Librandi, Aug 07 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=1, a(1)=2, a(2)=11. - Harvey P. Dale, Oct 10 2011
E.g.f.: exp(x)*(1 + x + 4*x^2). - Stefano Spezia, May 14 2021
a(n) = A003215(n-1) + A000290(n). - Leo Tavares, Jul 21 2022

A108625 Square array, read by antidiagonals, where row n equals the crystal ball sequence for the A_n lattice.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 13, 19, 7, 1, 1, 21, 55, 37, 9, 1, 1, 31, 131, 147, 61, 11, 1, 1, 43, 271, 471, 309, 91, 13, 1, 1, 57, 505, 1281, 1251, 561, 127, 15, 1, 1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1, 1, 91, 1405, 6637, 12559, 11253, 5321, 1415, 217, 19, 1

Offset: 0

Paul D. Hanna, Jun 12 2005

Compare to the corresponding array A108553 of crystal ball sequences for D_n lattice.
From Peter Bala, Jul 18 2008: (Start)
Row reverse of A099608.
This array has a remarkable relationship with the constant zeta(2). The row, column and diagonal entries of the array occur in series acceleration formulas for zeta(2).
For the entries in row n we have zeta(2) = 2*(1 - 1/2^2 + 1/3^2 - ... + (-1)^(n+1)/n^2) + (-1)^n*Sum_{k >= 1} 1/(k^2*T(n,k-1)*T(n,k)). For example, n = 4 gives zeta(2) = 2*(1 - 1/4 + 1/9 - 1/16) + 1/(1*21) + 1/(4*21*131) + 1/(9*131*471) + ... . See A142995 for further details.
For the entries in column k we have zeta(2) = (1 + 1/4 + 1/9 + ... + 1/k^2) + 2*Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,k)*T(n,k)). For example, k = 4 gives zeta(2) = (1 + 1/4 + 1/9 + 1/16) + 2*(1/(1*9) - 1/(4*9*61) + 1/(9*61*309) - ... ). See A142999 for further details.
Also, as consequence of Apery's proof of the irrationality of zeta(2), we have a series acceleration formula along the main diagonal of the table: zeta(2) = 5 * Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n,n)*T(n-1,n-1)) = 5*(1/3 - 1/(2^2*3*19) + 1/(3^2*19*147) - ...).
There also appear to be series acceleration results along other diagonals. For example, for the main subdiagonal, calculation supports the result zeta(2) = 2 - Sum_{n >= 1} (-1)^(n+1)*(n^2+(2*n+1)^2)/(n^2*(n+1)^2*T(n,n-1)*T(n+1,n)) = 2 - 10/(2^2*7) + 29/(6^2*7*55) - 58/(12^2*55*471) + ..., while for the main superdiagonal we appear to have zeta(2) = 1 + Sum_{n >= 1} (-1)^(n+1)*((n+1)^2 + (2*n+1)^2)/(n^2*(n+1)^2*T(n-1,n)*T(n,n+1)) = 1 + 13/(2^2*5) - 34/(6^2*5*37) + 65/(12^2*37*309) - ... .
Similar series acceleration results hold for Apery's constant zeta(3) involving the crystal ball sequences for the product lattices A_n x A_n; see A143007 for further details. Similar results also hold between the constant log(2) and the crystal ball sequences of the hypercubic lattices A_1 x...x A_1 and between log(2) and the crystal ball sequences for lattices of type C_n ; see A008288 and A142992 respectively for further details. (End)
This array is the Hilbert transform of triangle A008459 (see A145905 for the definition of the Hilbert transform). - Peter Bala, Oct 28 2008

			Square array begins:
  1,   1,    1,     1,      1,       1,       1, ... A000012;
  1,   3,    5,     7,      9,      11,      13, ... A005408;
  1,   7,   19,    37,     61,      91,     127, ... A003215;
  1,  13,   55,   147,    309,     561,     923, ... A005902;
  1,  21,  131,   471,   1251,    2751,    5321, ... A008384;
  1,  31,  271,  1281,   4251,   11253,   25493, ... A008386;
  1,  43,  505,  3067,  12559,   39733,  104959, ... A008388;
  1,  57,  869,  6637,  33111,  124223,  380731, ... A008390;
  1,  73, 1405, 13237,  79459,  350683, 1240399, ... A008392;
  1,  91, 2161, 24691, 176251,  907753, 3685123, ... A008394;
  1, 111, 3191, 43561, 365751, 2181257, ...      ... A008396;
  ...
As a triangle:
  [0]  1
  [1]  1,  1
  [2]  1,  3,   1
  [3]  1,  7,   5,    1
  [4]  1, 13,  19,    7,    1
  [5]  1, 21,  55,   37,    9,    1
  [6]  1, 31, 131,  147,   61,   11,   1
  [7]  1, 43, 271,  471,  309,   91,  13,   1
  [8]  1, 57, 505, 1281, 1251,  561, 127,  15,  1
  [9]  1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1
       ...
Inverse binomial transform of rows yield rows of triangle A063007:
  1;
  1,  2;
  1,  6,   6;
  1, 12,  30,  20;
  1, 20,  90, 140,  70;
  1, 30, 210, 560, 630, 252; ...
Product of the g.f. of row n and (1-x)^(n+1) generates the symmetric triangle A008459:
  1;
  1,  1;
  1,  4,   1;
  1,  9,   9,   1;
  1, 16,  36,  16,  1;
  1, 25, 100, 100, 25, 1;
  ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
R. Bacher, P. de la Harpe, and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
Joseph T. Iosue, T. C. Mooney, Adam Ehrenberg, and Alexey V. Gorshkov, Projective toric designs, difference sets, and quantum state designs, arXiv:2311.13479 [quant-ph], 2023. See page 6.
Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; arXiv preprint, arXiv:1401.0854 [math.NT], 2014.
A. van der Poorten, A proof that Euler missed ... Apery's proof of the irrationality of zeta(3). An informal report. Math. Intelligencer 1 (1978/79), no 4, 195-203.
Eric Weisstein's World of Mathematics, Apéry number.

Rows include: A003215 (row 2), A005902 (row 3), A008384 (row 4), A008386 (row 5), A008388 (row 6), A008390 (row 7), A008392 (row 8), A008394 (row 9), A008396 (row 10).
Cf. A063007, A099601 (n-th term of A_{2n} lattice), A108553.
Cf. A008459 (h-vectors type B associahedra), A145904, A145905.
Cf. A005258 (main diagonal), A108626 (antidiagonal sums).
Cf. A108628, A208675, A363867, A363868, A363869, A363870, A363871.

T:= func< n,k | (&+[Binomial(n,j)^2*Binomial(n+k-j,k-j): j in [0..k]]) >; // array
A108625:= func< n,k | T(n-k,k) >; // antidiagonals
[A108625(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023

T := (n,k) -> binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1):
seq(seq(simplify(T(n,k)),k=0..n),n=0..10); # Peter Luschny, Feb 10 2018

T[n_, k_]:= HypergeometricPFQ[{-n, -k, n+1}, {1, 1}, 1] (* Michael Somos, Jun 03 2012 *)

T(n,k)=sum(i=0,k,binomial(n,i)^2*binomial(n+k-i,k-i))

def T(n,k): return sum(binomial(n,j)^2*binomial(n+k-j, k-j) for j in range(k+1)) # array
def A108625(n,k): return T(n-k, k) # antidiagonals
flatten([[A108625(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

T(n, k) = Sum_{i=0..k} C(n, i)^2 * C(n+k-i, k-i).
G.f. for row n: (Sum_{i=0..n} C(n, i)^2 * x^i)/(1-x)^(n+1).
Sum_{k=0..n} T(n-k, k) = A108626(n) (antidiagonal sums).
From Peter Bala, Jul 23 2008 (Start):
O.g.f. row n: 1/(1 - x)*Legendre_P(n,(1 + x)/(1 - x)).
G.f. for square array: 1/sqrt((1 - x)*((1 - t)^2 - x*(1 + t)^2)) = (1 + x + x^2 + x^3 + ...) + (1 + 3*x + 5*x^2 + 7*x^3 + ...)*t + (1 + 7*x + 19*x^2 + 37*x^3 + ...)*t^2 + ... . Cf. A142977.
Main diagonal is A005258.
Recurrence relations:
Row n entries: (k+1)^2*T(n,k+1) = (2*k^2+2*k+n^2+n+1)*T(n,k) - k^2*T(n,k-1), k = 1,2,3,... ;
Column k entries: (n+1)^2*T(n+1,k) = (2*k+1)*(2*n+1)*T(n,k) + n^2*T(n-1,k), n = 1,2,3,... ;
Main diagonal entries: (n+1)^2*T(n+1,n+1) = (11*n^2+11*n+3)*T(n,n) + n^2*T(n-1,n-1), n = 1,2,3,... .
Series acceleration formulas for zeta(2):
Row n: zeta(2) = 2*(1 - 1/2^2 + 1/3^2 - ... + (-1)^(n+1)/n^2) + (-1)^n*Sum_{k >= 1} 1/(k^2*T(n,k-1)*T(n,k));
Column k: zeta(2) = 1 + 1/2^2 + 1/3^2 + ... + 1/k^2 + 2*Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,k)*T(n,k));
Main diagonal: zeta(2) = 5 * Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,n-1)*T(n,n)).
Conjectural result for superdiagonals: zeta(2) = 1 + 1/2^2 + ... + 1/k^2 + Sum_{n >= 1} (-1)^(n+1) * (5*n^2 + 6*k*n + 2*k^2)/(n^2*(n+k)^2*T(n-1,n+k-1)*T(n,n+k)), k = 0,1,2... .
Conjectural result for subdiagonals: zeta(2) = 2*(1 - 1/2^2 + ... + (-1)^(k+1)/k^2) + (-1)^k*Sum_{n >= 1} (-1)^(n+1)*(5*n^2 + 4*k*n + k^2)/(n^2*(n+k)^2*T(n+k-1,n-1)*T(n+k,n)), k = 0,1,2... .
Conjectural congruences: the main superdiagonal numbers S(n) := T(n,n+1) appear to satisfy the supercongruences S(m*p^r - 1) = S(m*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers m and r. If p is prime of the form 4*n + 1 we can write p = a^2 + b^2 with a an odd number. Then calculation suggests the congruence S((p-1)/2) == 2*a^2 (mod p). (End)
From Michael Somos, Jun 03 2012: (Start)
T(n, k) = hypergeom([-n, -k, n + 1], [1, 1], 1).
T(n, n-1) = A208675(n).
T(n+1, n) = A108628(n). (End)
T(n, k) = binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1). - Peter Luschny, Feb 10 2018
From Peter Bala, Jun 23 2023: (Start)
T(n, k) = Sum_{i = 0..k} (-1)^i * binomial(n, i)*binomial(n+k-i, k-i)^2.
T(n, k) = binomial(n+k, k)^2 * hypergeom([-n, -k, -k], [-n - k, -n - k], 1). (End)
From Peter Bala, Jun 28 2023; (Start)
T(n,k) = the coefficient of (x^n)*(y^k)*(z^n) in the expansion of 1/( (1 - x - y)*(1 - z ) - x*y*z ).
T(n,k) = B(n, k, n) in the notation of Straub, equation 24.
The supercongruences T(n*p^r, k*p^r) == T(n*p^(r-1), k*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and k.
The formula T(n,k) = hypergeom([n+1, -n, -k], [1, 1], 1) allows the table indexing to be extended to negative values of n and k; clearly, we find that T(-n,k) = T(n-1,k) for all n and k. It appears that T(n,-k) = (-1)^n*T(n,k-1) for n >= 0, while T(n,-k) = (-1)^(n+1)*T(n,k-1) for n <= -1 [added Sep 10 2023: these follow from the identities immediately below]. (End)
T(n,k) = Sum_{i = 0..n} (-1)^(n+i) * binomial(n, i)*binomial(n+i, i)*binomial(k+i, i) = (-1)^n * hypergeom([n + 1, -n, k + 1], [1, 1], 1). - Peter Bala, Sep 10 2023
From G. C. Greubel, Oct 05 2023: (Start)
Let t(n,k) = T(n-k, k) (antidiagonals).
t(n, k) = Hypergeometric3F2([k-n, -k, n-k+1], [1,1], 1).
T(n, 2*n) = A363867(n).
T(3*n, n) = A363868(n).
T(2*n, 2*n) = A363869(n).
T(n, 3*n) = A363870(n).
T(2*n, 3*n) = A363871(n). (End)
T(n, k) = Sum_{i = 0..n} binomial(n, i)*binomial(n+i, i)*binomial(k, i). - Peter Bala, Feb 26 2024
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n, k) = A005259(n), the Apéry numbers associated with zeta(3). - Peter Bala, Jul 18 2024
From Peter Bala, Sep 21 2024: (Start)
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*T(n, k) = binomial(2*n, n) = A000984(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n-1, n-k) = A376458(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(i, k) = A143007(n, i). (End)
From Peter Bala, Oct 12 2024: (Start)
The square array = A063007 * transpose(A007318).
Conjecture: for positive integer m, Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * T(m*n, k) = ((m+1)*n)!/( ((m-1)*n)!*n!^2) (verified up to m = 10 using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). (End)

A024166 a(n) = Sum_{1 <= i < j <= n} (j-i)^3.

Original entry on oeis.org

0, 1, 10, 46, 146, 371, 812, 1596, 2892, 4917, 7942, 12298, 18382, 26663, 37688, 52088, 70584, 93993, 123234, 159334, 203434, 256795, 320804, 396980, 486980, 592605, 715806, 858690, 1023526, 1212751, 1428976, 1674992, 1953776, 2268497, 2622522, 3019422

Offset: 0

Clark Kimberling

Convolution of the cubes (A000578) with the positive integers a(n)=n+1, where all sequences have offset zero. - Graeme McRae, Jun 06 2006
a(n) gives the n-th antidiagonal sum of the convolution array A212891. - Clark Kimberling, Jun 16 2012
In general, the r-th successive summation of the cubes from 1 to n is (6*n^2 + 6*n*r + r^2 - r)*(n+r)!/((r+3)!*(n-1)!), n>0. Here r = 2. - Gary Detlefs, Mar 01 2013
The inverse binomial transform is (essentially) row n=2 of A087127. - R. J. Mathar, Aug 31 2022

			4*a(7) = 6384 = (0*1)^2 + (1*2)^2 + (2*3)^2 + (3*4)^2 + (4*5)^2 + (5*6)^2 + (6*7)^2 + (7*8)^2. - _Bruno Berselli_, Feb 05 2014

Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

T. D. Noe, Table of n, a(n) for n = 0..1000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See pp. 13, 15.
C. P. Neuman and D. I. Schonbach, Evaluation of sums of convolved powers using Bernoulli numbers, SIAM Rev. 19 (1977), no. 1, 90--99. MR0428678 (55 #1698). See Table 1. - _N. J. A. Sloane_, Mar 23 2014
Claudio de J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq. 16 (2013) #13.5.7.
Alexander R. Povolotsky, Problem 1147, Pi Mu Epsilon Fall 2006 Problems.
Alexander R. Povolotsky, Problem, Pi Mu Epsilon Spring 2007 Problems.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1)

Cf. A000292, A000332, A000389, A000579, A000580, A024166, A027555, A085438, A085439, A085440, A085441, A085442, A086020, A086021, A086022, A086023, A086024, A086025, A086026, A086027, A086028, A086029, A086030, A087127.

Cf. A000330, A000537 (first diffs), A001286, A003215, A100536, A101094 (partial sums), A101097, A101102, A222716.

a024166 n = sum $ zipWith (*) [n+1,n..0] a000578_list
-- Reinhard Zumkeller, Oct 14 2001

[n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: n in [0..30]]; // G. C. Greubel, Nov 21 2017

A024166:=n->n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: seq(A024166(n), n=0..50); # Wesley Ivan Hurt, Nov 21 2017

Nest[Accumulate,Range[0,40]^3,2] (* Harvey P. Dale, Jan 10 2016 *)
Table[n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, {n,0,30}] (* G. C. Greubel, Nov 21 2017 *)

a(n)=sum(j=1,n, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2, (2*i-1)))) \\ Alexander R. Povolotsky, May 17 2008

for(n=0,30, print1(n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, ", ")) \\ G. C. Greubel, Nov 21 2017

From Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999: (Start)
a(n) = Sum_{i=0..n} A000537(i), partial sums of A000537.
a(n) = n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60. (End)
a(A004772(n)) mod 2 = 0; a(A016813(n)) mod 2 = 1. - Reinhard Zumkeller, Oct 14 2001
a(n) = Sum_{k=0..n} k^3*(n+1-k). - Paul Barry, Sep 14 2003; edited by Jon E. Schoenfield, Dec 29 2014
a(n) = 2*n*(n+1)*(n+2)*((n+1)^2 + 2*n*(n+2))/5!. This sequence could be obtained from the general formula a(n) = n*(n+1)*(n+2)*(n+3)* ...* (n+k) *(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=2. - Alexander R. Povolotsky, May 17 2008
O.g.f.: x*(1 + 4*x + x^2)/(-1 + x)^6. - R. J. Mathar, Jun 06 2008
a(n) = (6*n^2 + 12*n + 2)*(n+2)!/(120*(n-1)!), n > 0. - Gary Detlefs, Mar 01 2013
a(n) = A222716(n+1)/10 = A000292(n)*A100536(n+1)/10. - Jonathan Sondow, Mar 04 2013
4*a(n) = Sum_{i=0..n} A000290(i)*A000290(i+1). - Bruno Berselli, Feb 05 2014
a(n) = Sum_{i=1..n} Sum_{j=1..n} i*j*(n - max(i, j) + 1). - Melvin Peralta, May 12 2016
a(n) = n*binomial(n+3, 4) + binomial(n+2, 5). - Tony Foster III, Nov 14 2017
a(n) = Sum_{i=1..n} i*A143037(n,n-i+1). - J. M. Bergot, Aug 30 2022

A048766 Integer part of cube root of n. Or, number of cubes <= n. Or, n appears 3n^2 + 3n + 1 times.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 0

Charles T. Le (charlestle(AT)yahoo.com)

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
K. Atanassov, On the 100th, 101st and 102nd Smarandache Problems, On Some of Smarandache's Problems, American Research Press, 1999, pp. 57-61. Reprinted in Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 3, 94-96.
F. Smarandache, Only Problems, Not Solutions!.

Cf. A000196, A003215, A007412.

a048766 = round . (** (1/3)) . fromIntegral
a048766_list = concatMap (\x -> take (a003215 x) $ repeat x) [0..]
-- Reinhard Zumkeller, Sep 15 2013, Oct 22 2011

[n eq 0 select 0 else Iroot(n,3): n in [0..110]]; // Bruno Berselli, Feb 20 2015

A048766 := proc(n)
    floor(root[3](n)) ;
end proc:
seq(A048766(n),n=0..80) ; # R. J. Mathar, Dec 20 2020

a[n_]:=IntegerPart[n^(1/3)];lst={};Do[AppendTo[lst, a[n]], {n, 0, 6!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 02 2008 *)

a(n)=floor(n^(1/3)) \\ Charles R Greathouse IV, Mar 20 2012

a(n) = sqrtnint(n, 3); \\ Michel Marcus, Nov 10 2015

from sympy import integer_nthroot
def a(n): return integer_nthroot(n, 3)[0]
print([a(n) for n in range(105)]) # Michael S. Branicky, Oct 19 2021

G.f.: Sum_{k>=1} x^(k^3)/(1-x). - Geoffrey Critzer, Feb 05 2014

a(n) = Sum_{i=1..n} A210826(i)*floor(n/i). - Ridouane Oudra, Jan 21 2021

Additional comments from Reinhard Zumkeller, Oct 07 2001

More terms from Benoit Cloitre, Jan 30 2003

A049450 Pentagonal numbers multiplied by 2: a(n) = n(3n-1).

Original entry on oeis.org

0, 2, 10, 24, 44, 70, 102, 140, 184, 234, 290, 352, 420, 494, 574, 660, 752, 850, 954, 1064, 1180, 1302, 1430, 1564, 1704, 1850, 2002, 2160, 2324, 2494, 2670, 2852, 3040, 3234, 3434, 3640, 3852, 4070, 4294, 4524, 4760, 5002, 5250, 5504, 5764

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 0,2,.... The spiral begins:
.
                   56--55--54--53--52
                   /                 \
                 57  33--32--31--30  51
                 /   /             \   \
               58  34  16--15--14  29  50
               /   /   /         \   \   \
             59  35  17   5---4  13  28  49
             /   /   /   /     \   \   \   \
           60  36  18   6   0   3  12  27  48
           /   /   /   /   / . /   /   /   /
         61  37  19   7   1---2  11  26  47
           \   \   \   \       . /   /   /
           62  38  20   8---9--10  25  46
             \   \   \           . /   /
             63  39  21--22--23--24  45
               \   \               . /
               64  40--41--42--43--44
                 \                   .
                 65--66--67--68--69--70
(End)
Starting with offset 1 = binomial transform of [2, 8, 6, 0, 0, 0, ...]. - Gary W. Adamson, Jan 09 2009
Number of possible pawn moves on an (n+1) X (n+1) chessboard (n=>3). - Johannes W. Meijer, Feb 04 2010
a(n) = A069905(6n-1): Number of partitions of 6*n-1 into 3 parts. - Adi Dani, Jun 04 2011
Even octagonal numbers divided by 4. - Omar E. Pol, Aug 19 2011
Partial sums give A011379. - Omar E. Pol, Jan 12 2013
First differences are A016933; second differences equal 6. - Bob Selcoe, Apr 02 2015
For n >= 1, the continued fraction expansion of sqrt(27*a(n)) is [9n-2; {2, 2n-1, 6, 2n-1, 2, 18n-4}]. - Magus K. Chu, Oct 13 2022

			On a 4 X 4 chessboard pawns at the second row have (3+4+4+3) moves and pawns at the third row have (2+3+3+2) moves so a(3) = 24. - _Johannes W. Meijer_, Feb 04 2010
From _Adi Dani_, Jun 04 2011: (Start)
a(1)=2: the partitions of 6*1-1=5 into 3 parts are [1,1,3] and[1,2,2].
a(2)=10: the partitions of 6*2-1=11 into 3 parts are [1,1,9], [1,2,8], [1,3,7], [1,4,6], [1,5,5], [2,2,7], [2,3,6], [2,4,5], [3,3,5], and [3,4,4].
(End)
.
.                                                         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    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
.    2      10         24             44                 70
- _Philippe Deléham_, Mar 30 2013

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16)
Leo Tavares, Illustration: X Hexagons
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567.
Bisection of A001859. Cf. A045944, A000326, A033579, A027599, A049451.
Cf. A033586 (King), A035005 (Queen), A035006 (Rook), A035008 (Knight) and A002492 (Bishop).
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488. [Bruno Berselli, Jun 10 2013]
Cf. sequences listed in A254963.
Cf. A003215, A016813, A000290, A000217.

List([0..50], n-> n*(3*n-1)); # G. C. Greubel, Aug 31 2019

[n*(3*n-1) : n in [0..50]]; // Wesley Ivan Hurt, Sep 24 2017

seq(n*(3*n-1),n=0..44); # Zerinvary Lajos, Jun 12 2007

Table[n(3n-1),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,2,10},50] (* Harvey P. Dale, Jun 21 2014 *)
2*PolygonalNumber[5,Range[0,50]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jun 01 2018 *)

a(n)=n*(3*n-1) \\ Charles R Greathouse IV, Nov 20 2012

[n*(3*n-1) for n in (0..50)] # G. C. Greubel, Aug 31 2019

O.g.f.: A(x) = 2*x*(1+2*x)/(1-x)^3.
a(n) = A049452(n) - A033428(n). - Zerinvary Lajos, Jun 12 2007
a(n) = 2*A000326(n), twice pentagonal numbers. - Omar E. Pol, May 14 2008
a(n) = A022264(n) - A000217(n). - Reinhard Zumkeller, Oct 09 2008
a(n) = a(n-1) + 6*n - 4 (with a(0)=0). - Vincenzo Librandi, Aug 06 2010
a(n) = A014642(n)/4 = A033579(n)/2. - Omar E. Pol, Aug 19 2011
a(n) = A000290(n) + A000384(n) = A000217(n) + A000566(n). - Omar E. Pol, Jan 11 2013
a(n+1) = A014107(n+2) + A000290(n). - Philippe Deléham, Mar 30 2013
E.g.f.: x*(2 + 3*x)*exp(x). - Vincenzo Librandi, Apr 28 2016
a(n) = (2/3)*A000217(3*n-1). - Bruno Berselli, Feb 13 2017
a(n) = A002061(n) + A056220(n). - Bruce J. Nicholson, Sep 21 2017
From Amiram Eldar, Feb 20 2022: (Start)
Sum_{n>=1} 1/a(n) = 3*log(3)/2 - Pi/(2*sqrt(3)).
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/sqrt(3) - 2*log(2). (End)
From Leo Tavares, Feb 23 2022: (Start)
a(n) = A003215(n) - A016813(n).
a(n) = 2*A000290(n) + 2*A000217(n-1). (End)

A063496 a(n) = (2n - 1)(8n^2 - 8n + 3)/3.

Original entry on oeis.org

1, 19, 85, 231, 489, 891, 1469, 2255, 3281, 4579, 6181, 8119, 10425, 13131, 16269, 19871, 23969, 28595, 33781, 39559, 45961, 53019, 60765, 69231, 78449, 88451, 99269, 110935, 123481, 136939, 151341, 166719, 183105, 200531, 219029, 238631, 259369, 281275, 304381

Offset: 1

N. J. A. Sloane, Aug 01 2001

Number of potential flows in a 2 X 2 matrix with integer velocities in -n..n, i.e., number of 2 X 2 matrices with adjacent elements differing by no more than n, counting matrices differing by a constant only once. - R. H. Hardin, Feb 27 2002
Number of ordered quadruples (a,b,c,d), -(n-1) <= a,b,c,d <= n-1, such that a+b+c+d = 0. - Benoit Cloitre, Jun 14 2003
If Y and Z are 2-blocks of a (2n+1)-set X then a(n-1) is the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Oct 28 2007
Equals binomial transform of [1, 18, 48, 32, 0, 0, 0, ...]. - Gary W. Adamson, Jul 19 2008

Harry J. Smith, Table of n, a(n) for n = 1..1000
R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
Milan Janjic, Two Enumerative Functions
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

(1/12)*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

Cf. A003215, A010006, A142992, A142993, A142994.

[(2*n-1)*(8*n^2-8*n+3)/3: n in [1..40]]; // Wesley Ivan Hurt, May 09 2014

A063496:=n->(2*n-1)*(8*n^2-8*n+3)/3; seq(A063496(n), n=1..40); # Wesley Ivan Hurt, May 09 2014

Table[(2*n - 1)*(8*n^2 - 8*n + 3)/3, {n, 40}] (* Wesley Ivan Hurt, May 09 2014 *)
LinearRecurrence[{4,-6,4,-1}, {1,19,85,231}, 30] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(8*n^2 - 8*n + 3)/3 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-3+6*x+24*x^2+16*x^3)*exp(x)/3 + 1)) \\ G. C. Greubel, Dec 01 2017

From Peter Bala, Jul 18 2008: (Start)
The following remarks about the C_3 lattice assume the sequence offset is 0.
Partial sums of A010006. So this sequence is the crystal ball sequence for the C_3 lattice - row 3 of A142992. The lattice C_3 consists of all integer lattice points v = (a,b,c) in Z^3 such that a + b + c is even, equipped with the taxicab type norm ||v|| = (1/2) * (|a| + |b| + |c|).
The crystal ball sequence of C_3 gives the number of lattice points v in C_3 with ||v|| <= n for n = 0,1,2,3,... [Bacher et al.].
For example, a(1) = 19 because the origin has norm 0 and the 18 lattice points in Z^3 of norm 1 (as defined above) are +-(2,0,0), +-(0,2,0), +-(0,0,2), +-(1,1,0), +-(1,0,1), +-(0,1,1), +-(1,-1,0), +-(1,0,-1) and +-(0,1,-1). These 18 vectors form a root system of type C_3.
O.g.f.: x*(1 + 15*x + 15*x^2 + x^3)/(1 - x)^4 = x/(1 - x) * T(3, (1 + x)/(1 - x)), where T(n, x) denotes the Chebyshev polynomial of the first kind.
2*log(2) = 4/3 + Sum_{n >= 1} 1/(n*a(n)*a(n+1)). (End)
a(n+1) = (1/Pi) * Integral_{x=0..Pi} (sin((n+1/2)*x)/sin(x/2))^4. - Yalcin Aktar, Nov 02 2011, corrected by R. J. Mathar, Dec 01 2011
From G. C. Greubel, Dec 01 2017: (Start)
G.f.: x*(1 + 15*x + 15*x^2 + x^3)/(1 - x)^4.
E.g.f.: (-3 + 6*x + 24*x^2 + 16*x^3)*exp(x)/3 + 1. (End)
a(n) = A005900(2n-1). - Ivan N. Ianakiev, Mar 27 2022
From Peter Bala, Mar 11 2024: (Start)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k+1)) = 1/(19 - 3/(27 - 60/(43 - 315/(67 - ... -n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*3^2))))).
E.g.f.: exp(x)*(1 + 18*x + 48*x^2/2! + 32*x^3/3!). Note that -T(6, i*sqrt(x)) = 1 + 18*x + 48*x^2 + 32*x^3, where T(n, x) denotes the n-th Chebyshev polynomial of the first kind. See A008310. (End)

A015441 Generalized Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 7, 13, 55, 133, 463, 1261, 4039, 11605, 35839, 105469, 320503, 953317, 2876335, 8596237, 25854247, 77431669, 232557151, 697147165, 2092490071, 6275373061, 18830313487, 56482551853, 169464432775, 508359743893, 1525146340543, 4575304803901, 13726182847159

Offset: 0

Olivier Gérard

a(n) is the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x,y) = xF(n-1)(x,y) + yF(n-2)(x,y), F(0)(x,y)=0, F(1)(x,y)=1, when y=6x^2. - Mario Catalani (mario.catalani(AT)unito.it), Dec 06 2002
For n>=1: number of length-(n-1) words with letters {0,1,2,3,4,5,6,7} where no two consecutive letters are nonzero, see fxtbook link below. - Joerg Arndt, Apr 08 2011
Starting with offset 1 and convolved with (1, 3, 3, 3, ...) = A003462: (1, 4, 13, 40, ...). - Gary W. Adamson, May 28 2009
a(n) is identical to its inverse binomial transform signed. Differences: A102901. - Paul Curtz, Feb 23 2010
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=2, 7*a(n-2) equals the number of 7-colored compositions of n with all parts >=2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
Pisano period lengths: 1, 1, 1, 2, 20, 1, 6, 2, 3, 20, 5, 2, 12, 6, 20, 4, 16, 3, 18, 20, ... - R. J. Mathar, Aug 10 2012
A015441 and A015518 are the only integer sequences (from the family of homogeneous linear recurrence relation of order 2 with positive integer coefficients with initial values a(0)=0 and a(1)=1) whose ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 14 2014
This is an autosequence of the first kind: the array of successive differences shows a main diagonal of zeros and the inverse binomial transform is identical to the sequence (with alternating signs). - Pointed out by Paul Curtz, Dec 05 2016
First two upper diagonals: A000400(n).
This is a variation on the Starhex honeycomb configuration A332243, see illustration in links. It is an alternating pattern of the 2nd iteration of the centered hexagonal numbers A003215 and centered 12-gonal 'Star' numbers A003154. - John Elias, Oct 06 2021

			G.f. = x + x^2 + 7*x^3 + 13*x^4 + 55*x^5 + 133*x^6 + 463*x^7 + 1261*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
Joerg Arndt, Matters Computational (The Fxtbook), p. 317-318
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
John Elias, Illustration: A Starhex honeycomb variation
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
H. Li and T. MacHenry, Permanents and Determinants, Weighted Isobaric Polynomials, and Integer Sequences, J. Int. Seq. 16 (2013) #13.3.5, example 48.
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.
A. G. Shannon and J. V. Leyendekkers, The Golden Ratio family and the Binet equation, Notes on Number Theory and Discrete Mathematics, Vol. 21, No. 2, (2015), 35-42.
Index entries for linear recurrences with constant coefficients, signature (1,6).

Cf. A000079, A003462, A016153, A109466, A122117. Cf. A001656, A087451. Cf. A000400.

I:=[0,1]; [n le 2 select I[n] else Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

A015441:=n->(1/5)*((3^n)-((-2)^n)); seq(A015441(n), n=0..30); # Wesley Ivan Hurt, Mar 14 2014

a[n_]:=(MatrixPower[{{1,4},{1,-2}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]], {n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
LinearRecurrence[{1,6},{0,1},30] (* Harvey P. Dale, Apr 26 2011 *)
CoefficientList[Series[x/((1 + 2 x) (1 - 3 x)), {x, 0, 29}], x] (* Michael De Vlieger, Dec 05 2016 *)

PARI
```
{a(n) = (3^n - (-2)^n) / 5};
```

[lucas_number1(n,1,-6) for n in range(0, 27)] # Zerinvary Lajos, Apr 22 2009

G.f.: x/((1+2*x)*(1-3*x)).
a(n) = a(n-1) + 6*a(n-2).
a(n) = (1/5)*((3^n)-((-2)^n)). - henryk.wicke(AT)stud.uni-hannover.de
E.g.f.: (exp(3*x) - exp(-2*x))/5. - Paul Barry, Apr 20 2003
a(n+1) = Sum_{k=0..ceiling(n/2)} 6^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004
a(n) = (A000244(n) - A001045(n+1)(-1)^n - A001045(n)(-1)^n)/5. - Paul Barry, Apr 27 2004
The binomial transform of [1,1,7,13,55,133,463,...] is A122117. - Philippe Deléham, Oct 19 2006
a(n+1) = Sum_{k=0..n} A109466(n,k)*(-6)^(n-k). - Philippe Deléham, Oct 26 2008
a(n) = 3a(n-1) + (-1)^(n+1)*A000079(n-1). - Paul Curtz, Feb 23 2010
G.f.: Q(0) -1, where Q(k) = 1 + 6*x^2 + (k+2)*x - x*(k+1 + 6*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013
a(n) = (Sum_{1<=k<=n, k odd} binomial(n,k)*5^(k-1))/2^(n-1). - Vladimir Shevelev, Feb 05 2014
a(-n) = -(-1)^n * a(n) / 6^n for all n in Z. - Michael Somos, Mar 18 2014
From Peter Bala, Apr 01 2015: (Start)
Sum_{n >= 0} a(n+1)*x^n = exp( Sum_{n >= 1} A087451(n)*x^n/n ).
For k = 0, 1, 2, ... and for n >= 1, (5^k)*a(n) | a((5^k)*n).
The expansion of exp( Sum_{n >= 1} a(5*n)/(5*a(n))*x^n/n ) has integral coefficients. Cf. A001656. (End)
From  Peter Bala, Jun 27 2025: (Start)
Sum_{n >= 1} (-6)^n/(a(n)*a(n+1)) = -2, since (-6)^n/(a(n)*a(n+1)) = (-2)^n/a(n) - (-2)^(n+1)/a(n+1) for n >= 1.
The following are examples of telescoping infinite products:
Product_{n >= 0} (1 + 6^n/a(2*n+2)) = 6, since (1 + 6^(2*n-1)/a(4*n))*(1 + 6^(2*n)/a(4*n+2)) = (6 - 4^(n+1)/b(n)) / (6 - 4^n/b(n-1)), where b(n) = (2*4^n + 3*9^n)/5 = A096951(n). Similarly,
Product_{n >= 1} (1 - 6^n/a(2*n+2)) = 3/13.
Product_{n >= 0} (1 + (-6)^n/a(2*n+2)) = 6/5.
Product_{n >= 1} (1 - (-6)^n/a(2*n+2)) = 15/13.
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n. (End)

cp's OEIS Frontend

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

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A069099 Centered heptagonal numbers.

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A008458 Coordination sequence for hexagonal lattice.

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A054552 a(n) = 4*n^2 - 3*n + 1.

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A108625 Square array, read by antidiagonals, where row n equals the crystal ball sequence for the A_n lattice.

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A024166 a(n) = Sum_{1 <= i < j <= n} (j-i)^3.

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A054552 a(n) = 4n^2 - 3n + 1.

A049450 Pentagonal numbers multiplied by 2: a(n) = n(3n-1).

A063496 a(n) = (2n - 1)(8n^2 - 8n + 3)/3.