A046092 -id:A046092 - cp's OEIS frontend

A028896 6 times triangular numbers: a(n) = 3n(n+1).

0, 6, 18, 36, 60, 90, 126, 168, 216, 270, 330, 396, 468, 546, 630, 720, 816, 918, 1026, 1140, 1260, 1386, 1518, 1656, 1800, 1950, 2106, 2268, 2436, 2610, 2790, 2976, 3168, 3366, 3570, 3780, 3996, 4218, 4446, 4680, 4920, 5166, 5418, 5676

Offset: 0

Joe Keane (jgk(AT)jgk.org), Dec 11 1999

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, 6, ...
The spiral begins:
                 85--84--83--82--81--80
                 /                     \
               86  56--55--54--53--52  79
               /   /                 \   \
             87  57  33--32--31--30  51  78
             /   /   /             \   \   \
           88  58  34  16--15--14  29  50  77
           /   /   /   /         \   \   \   \
         89  59  35  17   5---4  13  28  49  76
         /   /   /   /   /     \   \   \   \   \
    <==90==60==36==18===6===0   3  12  27  48  75
           /   /   /   /   /   /   /   /   /   /
         61  37  19   7   1---2  11  26  47  74
           \   \   \   \         /   /   /   /
           62  38  20   8---9--10  25  46  73
             \   \   \             /   /   /
             63  39  21--22--23--24  45  72
               \   \                 /   /
               64  40--41--42--43--44  71
                 \                     /
                 65--66--67--68--69--70
(End)
If Y is a 4-subset of an n-set X then, for n >= 5, a(n-5) is the number of (n-4)-subsets of X having exactly two elements in common with Y. - Milan Janjic, Dec 28 2007
a(n) is the maximal number of points of intersection of n+1 distinct triangles drawn in the plane. For example, two triangles can intersect in at most a(1) = 6 points (as illustrated in the Star of David configuration). - Terry Stickels (Terrystickels(AT)aol.com), Jul 12 2008
Also sequence found by reading the line from 0, in the direction 0, 6, ... and the same line from 0, in the direction 0, 18, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. Axis perpendicular to A195143 in the same spiral. - Omar E. Pol, Sep 18 2011
Partial sums of A008588. - R. J. Mathar, Aug 28 2014
Also the number of 5-cycles in the (n+5)-triangular honeycomb acute knight graph. - Eric W. Weisstein, Jul 27 2017
a(n-4) is the maximum irregularity over all maximal 3-degenerate graphs with n vertices.  The extremal graphs are 3-stars (K_3 joined to n-3 independent vertices).  (The irregularity of a graph is the sum of the differences between the degrees over all edges of the graph.) - Allan Bickle, May 29 2023

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Allan Bickle and Zhongyuan Che, Irregularities of Maximal k-degenerate Graphs, Discrete Applied Math. 331 (2023) 70-87.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
Enrique Navarrete and Daniel Orellana, Finding Prime Numbers as Fixed Points of Sequences, arXiv:1907.10023 [math.NT], 2019.
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
Leo Tavares, Illustration: Centroid Hexagons.
Eric Weisstein's World of Mathematics, Graph Cycle.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000567, A003215, A008588, A024966, A028895, A033996, A046092, A049598, A084939, A084940, A084941, A084942, A084943, A084944, A124080.
Cf. A002378 (3-cycles in triangular honeycomb acute knight graph), A045943 (4-cycles), A152773 (6-cycles).
Cf. A007531.
The partial sums give A007531. - Leo Tavares, Jan 22 2022
Cf. A002378, A046092, A028896 (irregularities of maximal k-degenerate graphs).

List([0..44],n->3*n*(n+1)); # Muniru A Asiru, Mar 15 2019

[3*n*(n+1): n in [0..50]]; // Wesley Ivan Hurt, Jun 09 2014

[seq(6*binomial(n,2),n=1..44)]; # Zerinvary Lajos, Nov 24 2006

6 Accumulate[Range[0, 50]] (* Harvey P. Dale, Mar 05 2012 *)
6 PolygonalNumber[Range[0, 20]] (* Eric W. Weisstein, Jul 27 2017 *)
LinearRecurrence[{3, -3, 1}, {0, 6, 18}, 20] (* Eric W. Weisstein, Jul 27 2017 *)

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

first(n) = Vec(6*x/(1 - x)^3 + O(x^n), -n) \\ Iain Fox, Feb 14 2018

def A028896(n): return 3*n*(n+1) # Chai Wah Wu, Aug 07 2025

O.g.f.: 6*x/(1 - x)^3.
E.g.f.: 3*x*(x + 2)*exp(x). - G. C. Greubel, Aug 19 2017
a(n) = 6*A000217(n).
a(n) = polygorial(3, n+1). - Daniel Dockery (peritus(AT)gmail.com), Jun 16 2003
From Zerinvary Lajos, Mar 06 2007: (Start)
a(n) = A049598(n)/2.
a(n) = A124080(n) - A046092(n).
a(n) = A033996(n) - A002378(n). (End)
a(n) = A002378(n)*3 = A045943(n)*2. - Omar E. Pol, Dec 12 2008
a(n) = a(n-1) + 6*n for n>0, a(0)=0. - Vincenzo Librandi, Aug 05 2010
a(n) = A003215(n) - 1. - Omar E. Pol, Oct 03 2011
From Philippe Deléham, Mar 26 2013: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2, a(0)=0, a(1)=6, a(2)=18.
a(n) = A174709(6*n + 5). (End)
a(n) = A049450(n) + 4*n. - Lear Young, Apr 24 2014
a(n) = Sum_{i = n..2*n} 2*i. - Bruno Berselli, Feb 14 2018
a(n) = A320047(1, n, 1). - Kolosov Petro, Oct 04 2018
a(n) = T(3*n) - T(2*n-2) + T(n-2), where T(n) = A000217(n). In general, T(k)*T(n) = Sum_{i=0..k-1} (-1)^i*T((k-i)*(n-i)). - Charlie Marion, Dec 04 2020
From Amiram Eldar, Feb 15 2022: (Start)
Sum_{n>=1} 1/a(n) = 1/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2)/3 - 1/3. (End)
From Amiram Eldar, Feb 21 2023: (Start)
Product_{n>=1} (1 - 1/a(n)) = -(3/Pi)*cos(sqrt(7/3)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (3/Pi)*cosh(Pi/(2*sqrt(3))). (End)

A028895 5 times triangular numbers: a(n) = 5n(n+1)/2.

Original entry on oeis.org

0, 5, 15, 30, 50, 75, 105, 140, 180, 225, 275, 330, 390, 455, 525, 600, 680, 765, 855, 950, 1050, 1155, 1265, 1380, 1500, 1625, 1755, 1890, 2030, 2175, 2325, 2480, 2640, 2805, 2975, 3150, 3330, 3515, 3705, 3900, 4100, 4305, 4515, 4730, 4950, 5175, 5405, 5640

Offset: 0

Joe Keane (jgk(AT)jgk.org), Dec 11 1999

Sequence found by reading the line from 0, in the direction 0, 5, ... and the same line from 0, in the direction 0, 15, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. Axis perpendicular to A195142 in the same spiral. - Omar E. Pol, Sep 18 2011
Bisection of A195014. Sequence found by reading the line from 0, in the direction 0, 5, ..., and the same line from 0, in the direction 0, 15, ..., in the square spiral whose edges have length A195013 and whose vertices are the numbers A195014. This is the main diagonal of the spiral. - Omar E. Pol, Sep 25 2011
a(n) = the Wiener index of the graph obtained by applying Mycielski's construction to the complete graph K(n) (n>=2). - Emeric Deutsch, Aug 29 2013
Sum of the numbers from 2*n to 3*n for n=0,1,2,... - Wesley Ivan Hurt, Nov 27 2015
Numbers k such that the concatenation k625 is a square, where also 625 is a square. - Bruno Berselli, Nov 07 2018
From Paul Curtz, Nov 29 2019: (Start)
Main column of the pentagonal spiral for n (A001477):
                         50
                      49 30 31
                   48 29 15 16 32
                47 28 14  5  6 17 33
             46 27 13  4  0  1  7 18 34
               45 26 12  3  2  8 19 35
                 44 25 11 10  9 20 36
                   43 24 23 22 21 37
                    42 41 40 39 38
(End)

D. B. West, Introduction to Graph Theory, 2nd ed., Prentice-Hall, NJ, 2001, p. 205.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Rangaswami Balakrishnan and S. Francis Raj, The Wiener number of powers of the Mycielskian, Discussiones Math. Graph Theory, Vol. 30, No. 3 (2010), pp. 489-498 (see Theorem 2.1).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A001068, A005891, A028896, A046092, A055998, A085787, A130520, A195013, A195014, A195142.
Cf. index to numbers of the form n*(d*n+10-d)/2 in A140090.
Cf. A000566, A005475, A005476, A033583, A085787, A147875, A192136, A326725 (all in the spiral).

[5*n*(n+1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Jun 09 2014

[seq(5*binomial(n,2), n=1..45)]; # Zerinvary Lajos, Nov 24 2006

Table[Sum[i + 2*n - 1, {i, 2, n}], {n, 45}] (* Zerinvary Lajos, Jul 11 2009 *)
Table[5 n (n + 1)/2, {n, 0, 50}] (* Bruno Berselli, Sep 23 2016 *)
5 Accumulate[Range[0,60]] (* Harvey P. Dale, Jun 17 2025 *)

a(n)=5*n*(n+1)/2 \\ Charles R Greathouse IV, Sep 24 2015

def A028895(n): return 5*n*(n+1)>>1 # Chai Wah Wu, Aug 07 2025

G.f.: 5*x/(1-x)^3.
a(n) = 5*n*(n+1)/2 = 5*A000217(n).
a(n+1) = 5*n+a(n). - Vincenzo Librandi, Aug 05 2010
a(n) = A005891(n) - 1. - Omar E. Pol, Oct 03 2011
a(n) = A130520(5n+4). - Philippe Deléham, Mar 26 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. - Wesley Ivan Hurt, Nov 27 2015
a(n) = Sum_{i=0..n} A001068(4i). - Wesley Ivan Hurt, May 06 2016
E.g.f.: 5*x*(2 + x)*exp(x)/2. - Ilya Gutkovskiy, May 06 2016
a(n) = A055998(3*n) - A055998(2*n). - Bruno Berselli, Sep 23 2016
From Amiram Eldar, Feb 26 2022: (Start)
Sum_{n>=1} 1/a(n) = 2/5.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2/5)*(2*log(2) - 1). (End)
Product_{n>=1} (1 - 1/a(n)) = -(5/(2*Pi))*cos(sqrt(13/5)*Pi/2). - Amiram Eldar, Feb 21 2023

A077221 a(0) = 0 and then alternately even and odd numbers in increasing order such that the sum of any two successive terms is a square.

Original entry on oeis.org

0, 1, 8, 17, 32, 49, 72, 97, 128, 161, 200, 241, 288, 337, 392, 449, 512, 577, 648, 721, 800, 881, 968, 1057, 1152, 1249, 1352, 1457, 1568, 1681, 1800, 1921, 2048, 2177, 2312, 2449, 2592, 2737, 2888, 3041, 3200, 3361, 3528, 3697, 3872, 4049, 4232

Offset: 0

Amarnath Murthy, Nov 03 2002

This sequence arises from reading the line from 0, in the direction 0, 1, ... and the same line from 0, in the direction 0, 8, ..., in the square spiral whose vertices are the triangular numbers A000217. Cf. A139591, etc. - Omar E. Pol, May 03 2008
The general formula for alternating sums of powers of odd integers is in terms of the Swiss-Knife polynomials P(n,x) A153641 (P(n,0)-(-1)^k*P(n,2*k))/2. Here n=2, thus a(k) = |(P(2,0)-(-1)^k*P(2,2*k))/2|. - Peter Luschny, Jul 12 2009
Axis perpendicular to A046092 in the square spiral whose vertices are the triangular numbers A000217. See the comment above. - Omar E. Pol, Sep 14 2011
Column 8 of A195040. - Omar E. Pol, Sep 28 2011
Concentric octagonal numbers. A139098 and A069129 interleaved. - Omar E. Pol, Sep 17 2011
Subsequence of A194274. - Bruno Berselli, Sep 22 2011
Partial sums of A047522. - Reinhard Zumkeller, Jan 07 2012
Alternating sum of the first n odd squares in decreasing order, n >= 1. Also number of "ON" cells at n-th stage in simple 2-dimensional cellular automaton. The rules are: on the infinite square grid, start with all cells OFF, so a(0) = 0. Turn a single cell to the ON state, so a(1) = 1. At each subsequent step, the neighbor cells of each cell of the old generation are turned ON, and the cells of the old generation are turned OFF. Here "neighbor" refers to the eight adjacent cells of each ON cell. See example. - Omar E. Pol, Feb 16 2014

			From _Omar E. Pol_, Feb 16 2014: (Start)
Illustration of initial terms as a cellular automaton:
.
.                                   O O O O O O O
.                     O O O O O     O           O
.           O O O     O       O     O   O O O   O
.     O     O   O     O   O   O     O   O   O   O
.           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       8           17              32
.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Bruno Berselli, An origin of A077221 (illustration) (see Pol's comment).
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
Index entries for sequences related to cellular automata.

Cf. A032528, A069129, A077222, A131925, A139098, A194274, A195040, A195041, A195042, A195142.

a077221 n = a077221_list !! n
a077221_list = scanl (+) 0 a047522_list
-- Reinhard Zumkeller, Jan 07 2012

[2*n^2 - (n mod 2): n in [0..50]]; // Vincenzo Librandi, Sep 22 2011

a := n -> 2*n^2 - (n mod 2); # Peter Luschny, Jul 12 2009

a=1;lst={a};Do[b=n^2-a;AppendTo[lst,b];a=b,{n,3,6!,2}];lst (* Vladimir Joseph Stephan Orlovsky, May 18 2009 *)

a(2n) = 8*n^2, a(2n+1) = 8*n(n+1) + 1.
From Ralf Stephan, Mar 31 2003: (Start)
a(n) = 2*n^2 + 4*n + 1 [+1 if n is odd] with a(0)=1.
G.f.: x*(x^2+6*x+1)/(1-x)^3/(1+x). (End)
Row sums of triangle A131925; binomial transform of (1, 7, 2, 4, -8, 16, -32, ...). - Gary W. Adamson, Jul 29 2007
a(n) = a(-n); a(n+1) = A195605(n) - (-1)^n. - Bruno Berselli, Sep 22 2011
a(n) = 2*n^2 + ((-1)^n-1)/2. - Omar E. Pol, Sep 28 2011
Sum_{n>=1} 1/a(n) = Pi^2/48 + tan(Pi/(2*sqrt(2)))*Pi /(4*sqrt(2)). - Amiram Eldar, Jan 16 2023

Extended by Ralf Stephan, Mar 31 2003

A142463 a(n) = 2n^2 + 2n - 1.

Original entry on oeis.org

-1, 3, 11, 23, 39, 59, 83, 111, 143, 179, 219, 263, 311, 363, 419, 479, 543, 611, 683, 759, 839, 923, 1011, 1103, 1199, 1299, 1403, 1511, 1623, 1739, 1859, 1983, 2111, 2243, 2379, 2519, 2663, 2811, 2963, 3119, 3279, 3443, 3611, 3783, 3959, 4139, 4323, 4511, 4703, 4899, 5099

Offset: 0

Roger L. Bagula, Sep 19 2008

Essentially the same as A132209.
From Vincenzo Librandi, Nov 25 2010: (Start)
Numbers k such that 2*k + 3 is a square.
First diagonal of A144562. (End)
The terms a(n) give the values for c of indefinite binary quadratic forms [a, b, c] = [2, 4n+2, a(n)] of discriminant D = 12, where a and c can be switched. The positive numbers represented by these forms are given in A084917. - Klaus Purath, Aug 31 2023

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Leo Tavares, Illustration: Hexagonic Diamonds.
Leo Tavares, Illustration: Hexagonic Rectangles.
Leo Tavares, Illustration: Hexagonic Crosses.
Leo Tavares, Illustration: Hexagonic Columns.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A005408, A132209, A144562, A188653.

Cf. A005563, A016945, A001105, A000290, A004767, A046092, A002378, A028387.

Magma
```
[2*n^2+2*n-1: n in [0..100]];
```

A142463:= n-> 2*n^2 +2*n -1; seq(A142463(n), n=0..50); # G. C. Greubel, Mar 01 2021

Array[ -#*(2-#*2)-1&,5!,1] (* Vladimir Joseph Stephan Orlovsky, Dec 21 2008 *)
Table[2n^2+2n-1,{n,0,50}] (* Harvey P. Dale, Feb 29 2024 *)

a(n)=2*n^2+2*n-1 \\ Charles R Greathouse IV, Sep 24 2015

[2*n^2 +2*n -1 for n in (0..50)] # G. C. Greubel, Mar 01 2021

a(n) = a(n-1) + 4*n.
From Paul Barry, Nov 03 2009: (Start)
G.f.: (1 - 6*x + x^2)/(1-x)^3.
a(n) = 4*C(n+1,2) - 1. (End)
a(n) = -A188653(2*n+1). - Reinhard Zumkeller, Apr 13 2011
a(n) = 3*( Sum_{k=1..n} k^5 )/( Sum_{k=1..n} k^3 ), n > 0. - Gary Detlefs, Oct 18 2011
a(n) = (A005408(n)^2 - 3)/2. - Zhandos Mambetaliyev, Feb 11 2017
E.g.f.: (-1 + 4*x + 2*x^2)*exp(x). - G. C. Greubel, Mar 01 2021
From Leo Tavares, Nov 22 2021: (Start)
a(n) = 2*A005563(n) - A005408(n). See Hexagonic Diamonds illustration.
a(n) = A016945(n-1) + A001105(n-1). See Hexagonic Rectangles illustration.
a(n) = A004767(n-1) + A046092(n-1). See Hexagonic Crosses illustration.
a(n) = A002378(n) + A028387(n-1). See Hexagonic Columns illustration. (End)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Wesley Ivan Hurt, Dec 03 2021
Sum_{n>=0} 1/a(n) = tan(sqrt(3)*Pi/2)*Pi/(2*sqrt(3)). - Amiram Eldar, Sep 16 2022

Edited by the Associate Editors of the OEIS, Sep 02 2009

A122141 Array: T(d,n) = number of ways of writing n as a sum of d squares, read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 4, 0, 1, 6, 4, 0, 1, 8, 12, 0, 2, 1, 10, 24, 8, 4, 0, 1, 12, 40, 32, 6, 8, 0, 1, 14, 60, 80, 24, 24, 0, 0, 1, 16, 84, 160, 90, 48, 24, 0, 0, 1, 18, 112, 280, 252, 112, 96, 0, 4, 2, 1, 20, 144, 448, 574, 312, 240, 64, 12, 4, 0, 1, 22, 180, 672, 1136, 840, 544, 320, 24, 30, 8, 0

Offset: 1

R. J. Mathar, Oct 29 2006

This is the transpose of the array in A286815.

T(d,n) is divisible by 2d for any n != 0 iff d is a power of 2. - Jianing Song, Sep 05 2018

			Array T(d,n) with rows d = 1,2,3,... and columns n = 0,1,2,3,... reads
  1  2   0   0    2    0     0     0     0     2      0 ...
  1  4   4   0    4    8     0     0     4     4      8 ...
  1  6  12   8    6   24    24     0    12    30     24 ...
  1  8  24  32   24   48    96    64    24   104    144 ...
  1 10  40  80   90  112   240   320   200   250    560 ...
  1 12  60 160  252  312   544   960  1020   876   1560 ...
  1 14  84 280  574  840  1288  2368  3444  3542   4424 ...
  1 16 112 448 1136 2016  3136  5504  9328 12112  14112 ...
  1 18 144 672 2034 4320  7392 12672 22608 34802  44640 ...
  1 20 180 960 3380 8424 16320 28800 52020 88660 129064 ...

Alois P. Heinz, Antidiagonals d = 1..141, flattened
Wikipedia, Sum of squares function (transposed)
Index entries for sequences related to sums of squares

Cf. A066535, A286815.
Cf. A000122 (1st row), A004018 (2nd row), A005875 (3rd row), A000118 (4th row), A000132 (5th row), A000141 (6th row), A008451 (7th row), A000143 (8th row), A008452 (9th row), A000144 (10th row), A008453 (11th row), A000145 (12th row), A276285 (13th row), A276286 (14th row), A276287 (15th row), A000152 (16th row).
Cf. A005843 (2nd column), A046092 (3rd column), A130809 (4th column).
Cf. A010052 (1st row divides 2), A002654 (2nd row divides 4), A046897 (4th row divides 8), A008457 (8th row divides 16), A302855 (16th row divides 32), A302857 (32nd row divides 64).

A122141 := proc(d,n) local i,cnts ; cnts := 0 ; for i from -trunc(sqrt(n)) to trunc(sqrt(n)) do if n-i^2 >= 0 then if d > 1 then cnts := cnts+procname(d-1,n-i^2) ; elif n-i^2 = 0 then cnts := cnts+1 ; fi ; fi ; od ; cnts ;
end:
for diag from 1 to 14 do for n from 0 to diag-1 do d := diag-n ; printf("%d,",A122141(d,n)) ; od ; od;
# second Maple program:
A:= proc(d, n) option remember; `if`(n=0, 1, `if`(n<0 or d<1, 0,
      A(d-1, n) +2*add(A(d-1, n-j^2), j=1..isqrt(n))))
    end:
seq(seq(A(h-n, n), n=0..h-1), h=1..14); # Alois P. Heinz, Jul 16 2014

Table[ SquaresR[d - n, n], {d, 1, 12}, {n, 0, d - 1}] // Flatten (* Jean-François Alcover, Jun 13 2013 *)
A[d_, n_] := A[d, n] = If[n==0, 1, If[n<0 || d<1, 0, A[d-1, n] + 2*Sum[A[d-1, n-j^2], {j, 1, Sqrt[n]}]]]; Table[A[h-n, n], {h, 1, 14}, {n, 0, h-1}] // Flatten (* Jean-François Alcover, Feb 28 2018, after Alois P. Heinz *)

from sympy.core.power import isqrt
from functools import cache
@cache
def T(d, n):
  if n == 0: return 1
  if n < 0 or d < 1: return 0
  return T(d-1, n) + sum(T(d-1, n-(j**2)) for j in range(1, isqrt(n)+1)) * 2  # Darío Clavijo, Feb 06 2024

T(n,n) = A066535(n). - Alois P. Heinz, Jul 16 2014

A027468 9 times the triangular numbers A000217.

Original entry on oeis.org

0, 9, 27, 54, 90, 135, 189, 252, 324, 405, 495, 594, 702, 819, 945, 1080, 1224, 1377, 1539, 1710, 1890, 2079, 2277, 2484, 2700, 2925, 3159, 3402, 3654, 3915, 4185, 4464, 4752, 5049, 5355, 5670, 5994, 6327, 6669, 7020, 7380, 7749, 8127, 8514, 8910, 9315

Offset: 0

Olivier Gérard

Staggered diagonal of triangular spiral in A051682, between (0,1,11) spoke and (0,8,25) spoke. - Paul Barry, Mar 15 2003
Number of permutations of n distinct letters (ABCD...) each of which appears thrice with n-2 fixed points. - Zerinvary Lajos, Oct 15 2006
Number of n permutations (n>=2) of 4 objects u, v, z, x with repetition allowed, containing n-2=0 u's. Example: if n=2 then n-2 =zero (0) u, a(1)=9 because we have vv, zz, xx, vx, xv, zx, xz, vz, zv. A027465 formatted as a triangular array: diagonal: 9, 27, 54, 90, 135, 189, 252, 324, ... . - Zerinvary Lajos, Aug 06 2008
a(n) is also the least weight of self-conjugate partitions having n different parts such that each part is a multiple of 3. - Augustine O. Munagi, Dec 18 2008
Also sequence found by reading the line from 0, in the direction 0, 9, ..., and the same line from 0, in the direction 0, 27, ..., in the square spiral whose vertices are the generalized hendecagonal numbers A195160. Axis perpendicular to A195147 in the same spiral. - Omar E. Pol, Sep 18 2011
Sum of the numbers from 4*n to 5*n. - Wesley Ivan Hurt, Nov 01 2014

			The first such self-conjugate partitions, corresponding to a(n)=1,2,3,4 are 3+3+3, 6+6+6+3+3+3, 9+9+9+6+6+6+3+3+3, 12+12+12+9+9+9+6+6+6+3+3+3. - _Augustine O. Munagi_, Dec 18 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Augustine O. Munagi, Pairing conjugate partitions by residue classes, Discrete Math., Vol. 308, No. 12 (2008), pp. 2492-2501.
Enrique Navarrete and Daniel Orellana, Finding Prime Numbers as Fixed Points of Sequences, arXiv:1907.10023 [math.NT], 2019.
Leo Tavares, Illustration: Centroid Triangles.
D. Zvonkine, Counting ramified coverings and intersection theory on Hurwitz spaces II (local structure of Hurwitz spaces and combinatorial results), Moscow Mathematical Journal, Vol. 7, No. 1 (2007), pp. 135-162.
D. Zvonkine, Home Page.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Index entries for two-way infinite sequences.

Third diagonal of A027465.
Cf. A000459, A002378, A008585, A024966, A028895, A028896, A038764, A033996, A045943, A046092, A049598, A059073, A080855, A134171, A283394.
Cf. A060544.

[9*n*(n+1)/2: n in [0..50]]; // Vincenzo Librandi, Dec 29 2012

[seq(9*binomial(n+1,2), n=0..50)]; # Zerinvary Lajos, Nov 24 2006

Table[(9/2)*n*(n+1), {n,0,50}] (* G. C. Greubel, Aug 22 2017 *)

PARI
```
a(n)=9*n*(n+1)/2
```

[9*binomial(n+1, 2) for n in (0..50)] # G. C. Greubel, May 20 2021

Numerators of sequence a[n, n-2] in (a[i, j])^2 where a[i, j] = binomial(i-1, j-1)/2^(i-1) if j<=i, 0 if j>i.
a(n) = (9/2)*n*(n+1).
a(n) = 9*C(n, 1) + 9*C(n, 2) (binomial transform of (0, 9, 9, 0, 0, ...)). - Paul Barry, Mar 15 2003
G.f.: 9*x/(1-x)^3.
a(-1-n) = a(n).
a(n) = 9*C(n+1,2), n>=0. - Zerinvary Lajos, Aug 06 2008
a(n) = a(n-1) + 9*n (with a(0)=0). - Vincenzo Librandi, Nov 19 2010
a(n) = A060544(n+1) - 1. - Omar E. Pol, Oct 03 2011
a(n) = A218470(9*n+8). - Philippe Deléham, Mar 27 2013
E.g.f.: (9/2)*x*(x+2)*exp(x). - G. C. Greubel, Aug 22 2017
a(n) = A060544(n+1) - 1. See Centroid Triangles illustration. - Leo Tavares, Dec 27 2021
From Amiram Eldar, Feb 15 2022: (Start)
Sum_{n>=1} 1/a(n) = 2/9.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/9 - 2/9. (End)
From Amiram Eldar, Feb 21 2023: (Start)
Product_{n>=1} (1 - 1/a(n)) = -(9/(2*Pi))*cos(sqrt(17)*Pi/6).
Product_{n>=1} (1 + 1/a(n)) = 9*sqrt(3)/(4*Pi). (End)

More terms from Patrick De Geest, Oct 15 1999

A182368 Triangle T(n,k), n>=1, 0<=k<=n^2, read by rows: row n gives the coefficients of the chromatic polynomial of the square grid graph G_(n,n), highest powers first.

Original entry on oeis.org

1, 0, 1, -4, 6, -3, 0, 1, -12, 66, -216, 459, -648, 594, -323, 79, 0, 1, -24, 276, -2015, 10437, -40614, 122662, -292883, 557782, -848056, 1022204, -960627, 682349, -346274, 112275, -17493, 0, 1, -40, 780, -9864, 90798, -647352, 3714180, -17590911, 69997383

Offset: 1

Alois P. Heinz, Apr 26 2012

The square grid graph G_(n,n) has n^2 = A000290(n) vertices and 2*n*(n-1) = A046092(n-1) edges. The chromatic polynomial of G_(n,n) has n^2+1 = A002522(n) coefficients.

			3 example graphs:                          o---o---o
.                                          |   |   |
.                             o---o        o---o---o
.                             |   |        |   |   |
.                o            o---o        o---o---o
Graph:        G_(1,1)        G_(2,2)        G_(3,3)
Vertices:        1              4              9
Edges:           0              4             12
The square grid graph G_(2,2) is the cycle graph C_4 with chromatic polynomial q^4 -4*q^3 +6*q^2 -3*q => row 2 = [1, -4, 6, -3, 0].
Triangle T(n,k) begins:
  1,    0;
  1,   -4,     6,      -3,        0;
  1,  -12,    66,    -216,      459,       -648,         594, ...
  1,  -24,   276,   -2015,    10437,     -40614,      122662, ...
  1,  -40,   780,   -9864,    90798,    -647352,     3714180, ...
  1,  -60,  1770,  -34195,   486210,   -5421612,    49332660, ...
  1,  -84,  3486,  -95248,  1926585,  -30755376,   403410654, ...
  1, -112,  6216, -227871,  6205479, -133865298,  2382122274, ...
  1, -144, 10296, -487280, 17169852, -480376848, 11114098408, ...
  ...

Alois P. Heinz, Rows n = 1..9, flattened
Eric Weisstein's World of Mathematics, Chromatic Polynomial
Eric Weisstein's World of Mathematics, Grid Graph
Wikipedia, Chromatic Polynomial

Columns 0, 1 give: A000012, (-1)*A046092(n-1).
Sums of absolute values of row elements give: A080690(n).
Cf. A000290, A002522, A182406, A185442, A193233, A193277, A193283.

Reverse /@ CoefficientList[Table[ChromaticPolynomial[GridGraph[{n, n}], x], {n, 5}], x] // Flatten (* Eric W. Weisstein, May 01 2017 *)

A182406 Square array A(n,k), n>=1, k>=1, read by antidiagonals: A(n,k) is the number of n-colorings of the square grid graph G_(k,k).

Original entry on oeis.org

1, 0, 2, 0, 2, 3, 0, 2, 18, 4, 0, 2, 246, 84, 5, 0, 2, 7812, 9612, 260, 6, 0, 2, 580986, 6000732, 142820, 630, 7, 0, 2, 101596896, 20442892764, 828850160, 1166910, 1302, 8, 0, 2, 41869995708, 380053267505964, 50820390410180, 38128724910, 6464682, 2408, 9

Offset: 1

Alois P. Heinz, Apr 27 2012

The square grid graph G_(n,n) has n^2 = A000290(n) vertices and 2*n*(n-1) = A046092(n-1) edges. The chromatic polynomial of G_(n,n) has n^2+1 = A002522(n) coefficients.

			Square array A(n,k) begins:
  1,   0,       0,           0,                 0, ...
  2,   2,       2,           2,                 2, ...
  3,  18,     246,        7812,            580986, ...
  4,  84,    9612,     6000732,       20442892764, ...
  5, 260,  142820,   828850160,    50820390410180, ...
  6, 630, 1166910, 38128724910, 21977869327169310, ...

Eric Weisstein's World of Mathematics, Grid Graph
Wikipedia, Chromatic Polynomial

Columns k=1-7 give: A000027, A091940, A068239*2, A068240*2, A068241*2, A068242*2, A068243*2.
Rows n=1-20 give: A000007, A007395, A068253*3, A068254*4, A068255*5, A068256*6, A068257*7, A068258*8, A068259*9, A068260*10, A068261*11, A068262*12, A068263*13, A068264*14, A068265*15, A068266*16, A068267*17, A068268*18, A068269*19, A068270*20.
Cf. A182368.

A217843 Numbers which are the sum of one or more consecutive nonnegative cubes.

Original entry on oeis.org

0, 1, 8, 9, 27, 35, 36, 64, 91, 99, 100, 125, 189, 216, 224, 225, 341, 343, 405, 432, 440, 441, 512, 559, 684, 729, 748, 775, 783, 784, 855, 1000, 1071, 1196, 1241, 1260, 1287, 1295, 1296, 1331, 1584, 1728, 1729, 1800, 1925, 1989, 2016, 2024, 2025, 2197

Offset: 1

T. D. Noe, Oct 23 2012

nonn

Contains A000578 (cubes), A005898 (two consecutive cubes), A027602 (three consecutive cubes), A027603 (four consecutive cubes) etc. - R. J. Mathar, Nov 04 2012
See A265845 for sums of consecutive positive cubes in more than one way. - Reinhard Zumkeller, Dec 17 2015
From Lamine Ngom, Apr 15 2021: (Start)
a(n) can always be expressed as the difference of the squares of two triangular numbers (A000217).
A168566 is the subsequence A000217(n)^2 - 1.
a(n) is also the product of two nonnegative integers whose sum and difference are both promic.
See example and formula sections for details. (End)

			From _Lamine Ngom_, Apr 15 2021: (Start)
Arrange the positive terms in a triangle as follows:
n\k |   1    2    3    4    5    6    7
----+-----------------------------------
  0 |   1;
  1 |   8,   9;
  2 |  27,  35,  36;
  3 |  64,  91,  99, 100;
  4 | 125, 189, 216, 224, 225;
  5 | 216, 341, 405, 432, 440, 441;
  6 | 343, 559, 684, 748, 775, 783, 784;
Column 1: cubes = A000217(n+1)^2 - A000217(n)^2.
The difference of the squares of two consecutive triangular numbers (A000217) is a cube (A000578).
Column 2: sums of 2 consecutive cubes (A027602).
Column 3: sums of 3 consecutive cubes (A027603).
etc.
Column k: sums of k consecutive cubes.
Row n: A000217(n)^2 - A000217(m)^2, m < n.
T(n,n) = A000217(n)^2 (main diagonal).
T(n,n-1) = A000217(n)^2 - 1 (A168566) (2nd diagonal).
Now rectangularize this triangle as follows:
n\k |   1    2     3     4    5     6   ...
----+--------------------------------------
  0 |   1,   9,   36,  100,  225,  441, ...
  1 |   8,  35,   99,  224,  440,  783, ...
  2 |  27,  91,  216,  432,  775, 1287, ...
  3 |  64, 189,  405,  748, 1260, 1989, ...
  4 | 125, 341,  684, 1196, 1925, 2925, ...
  5 | 216, 559, 1071, 1800, 2800, 4131, ...
  6 | 343, 855, 1584, 2584, 3915, 5643, ...
The general form of terms is:
T(n,k) = [n^4 + A016825(k)*n^3 + A003154(k)*n^2 + A300758(k)*n]/4, sum of n consecutive cubes after k^3.
This expression can be factorized into [n*(n + A005408(k))*(n*(n + A005408(k)) + 4*A000217(k))]/4.
For k = 1, the sequence provides all cubes: T(n,1) = A000578(k).
For k = 2, T(n,2) = A005898(k), centered cube numbers, sum of two consecutive cubes.
For k = 3, T(n,3) = A027602(k), sum of three consecutive cubes.
For k = 4, T(n,4) = A027603(k), sum of four consecutive cubes.
For k = 5, T(n,5) = A027604(k), sum of five consecutive cubes.
T(n,n) = A116149(n), sum of n consecutive cubes after n^3 (main diagonal).
For n = 0, we obtain the subsequence T(0,k) = A000217(n)^2, product of two numbers whose difference is 0*1 (promic) and sum is promic too.
For n = 1, we obtain the subsequence T(1,k) = A168566(x), product of two numbers whose difference is 1*2 (promic) and sum is promic too.
For n = 2, we obtain the subsequence T(2,k) = product of two numbers whose difference is 2*3 (promic) and sum is promic too.
etc.
For n = x, we obtain the subsequence formed by products of two numbers whose difference is the promic x*(x+1) and sum is promic too.
Consequently, if m is in the sequence, then m can be expressed as the product of two nonnegative integers whose sum and difference are both promic. (End)

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A034705, A217844-A217850, A062682, A131643, A240137.
Cf. A000578, A005898, A027602, A027603, A027604.
Cf. A265845 (subsequence).
Cf. A000217 (triangular numbers), A046092 (4*A000217).
Cf. A168566 (A000217^2 - 1).
Cf. A002378 (promics), A016825 (singly even numbers), A003154 (stars numbers).
Cf. A000330 (square pyramidal numbers), A300758 (12*A000330).
Cf. A005408 (odd numbers).

import Data.Set (singleton, deleteFindMin, insert, Set)
a217843 n = a217843_list !! (n-1)
a217843_list = f (singleton (0, (0,0))) (-1) where
   f s z = if y /= z then y : f s'' y else f s'' y
              where s'' = (insert (y', (i, j')) $
                           insert (y' - i ^ 3 , (i + 1, j')) s')
                    y' = y + j' ^ 3; j' = j + 1
                    ((y, (i, j)), s') = deleteFindMin s
-- Reinhard Zumkeller, Dec 17 2015, May 12 2015

nMax = 3000; t = {0}; Do[k = n; s = 0; While[s = s + k^3; s <= nMax, AppendTo[t, s]; k++], {n, nMax^(1/3)}]; t = Union[t]

lista(nn) = {my(list = List([0])); for (i=1, nn, my(s = 0); forstep(j=i, 1, -1, s += j^3; if (s > nn^3, break); listput(list, s););); Set(list);} \\ Michel Marcus, Nov 13 2020

a(n) >> n^2. Probably a(n) ~ kn^2 for some k but I cannot prove this. - Charles R Greathouse IV, Aug 07 2013

a(n) is of the form [x*(x+2*k+1)*(x*(x+2*k+1)+2*k*(k+1))]/4, sum of n consecutive cubes starting from (k+1)^3. - Lamine Ngom, Apr 15 2021

Name edited by N. J. A. Sloane, May 24 2021

A130519 a(n) = Sum_{k=0..n} floor(k/4). (Partial sums of A002265.)

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 3, 4, 6, 8, 10, 12, 15, 18, 21, 24, 28, 32, 36, 40, 45, 50, 55, 60, 66, 72, 78, 84, 91, 98, 105, 112, 120, 128, 136, 144, 153, 162, 171, 180, 190, 200, 210, 220, 231, 242, 253, 264, 276, 288, 300, 312, 325, 338, 351, 364, 378, 392, 406, 420, 435, 450

Offset: 0

Hieronymus Fischer, Jun 01 2007

Complementary to A130482 with respect to triangular numbers, in that A130482(n) + 4*a(n) = n(n+1)/2 = A000217(n).
Disregarding the first three 0's the resulting sequence a'(n) is the sum of the positive integers <= n that have the same residue modulo 4 as n. This is the additive counterpart of the quadruple factorial numbers. - Peter Luschny, Jul 06 2011
From Heinrich Ludwig, Dec 23 2017: (Start)
Column sums of (shift of rows = 4):
  1 2 3 4 5 6  7  8  9 10 11 12 13 14 ...
          1 2  3  4  5  6  7  8  9 10 ...
                     1  2  3  4  5  6 ...
                                 1  2 ...
  .......................................
  ---------------------------------------
  1 2 3 4 6 8 10 12 15 18 21 24 28 32 ...
  shift of rows = 1 see A000217
  shift of rows = 2 see A002620
  shift of rows = 3 see A001840
  shift of rows = 5 see A130520
(End)
Conjecture: a(n+2) is the maximum effective weight of a numerical semigroup S of genus n (see Nathan Pflueger). - Stefano Spezia, Jan 04 2019

			G.f. = x^4 + 2*x^5 + 3*x^6 + 4*x^7 + 6*x^8 + 8*x^9 + 10*x^10 + 12*x^11 + ...
[ n] a(n)
---------
[ 4] 1
[ 5] 2
[ 6] 3
[ 7] 4
[ 8] 1 + 5
[ 9] 2 + 6
[10] 3 + 7
[11] 4 + 8

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Bakir Farhi, On the Representation of the Natural Numbers as the Sum of Three Terms of the Sequence floor(n^2/a), Journal of Integer Sequences, Vol. 16 (2013), Article 13.6.4.
Darren Glass and Joshua Wagner, Arithmetical Structures on Paths With a Doubled Edge, arXiv:1903.01398 [math.CO], 2019.
Nathan Pflueger, On non-primitive Weierstrass points, Alg. Number Th. 12 (2018) 1923-1947 and arXiv:1608.0566 [math.AG], 2016.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

Cf. A000217, A001840, A002264, A002265, A002266, A002620, A004526, A010872, A010873, A010874, A130481, A130483, A130520.

Cf. A000290, A007590, A000212, A118015, A056827, A056834, A056838, A056865.

a:=List([0..65],n->Sum([0..n],k->Int(k/4)));; Print(a); # Muniru A Asiru, Jan 04 2019

[Round(n*(n-2)/8): n in [0..70]]; // Vincenzo Librandi, Jun 25 2011

quadsum := n -> add(k, k = select(k -> k mod 4 = n mod 4, [$1 .. n])):
A130519 := n ->`if`(n<3,0,quadsum(n-3)); seq(A130519(n),n=0..58); # Peter Luschny, Jul 06 2011

a[ n_] := Quotient[ (n - 1)^2, 8]; (* Michael Somos, Oct 14 2011 *)

makelist(floor((n-1)^2/8), n, 0, 70); /* Stefano Spezia, Jan 04 2019 */

{a(n) = (n - 1)^2 \ 8}; /* Michael Somos, Oct 14 2011 */

def A130519(n): return (n-1)**2>>3  # Chai Wah Wu, Jul 30 2022

G.f.: x^4/((1-x^4)*(1-x)^2) = x^4/((1+x)*(1+x^2)*(1-x)^3).
a(n) = +2*a(n-1) -1*a(n-2) +1*a(n-4) -2*a(n-5) +1*a(n-6).
a(n) = floor(n/4)*(n - 1 - 2*floor(n/4)) = A002265(n)*(n - 1 - 2*A002265(n)).
a(n) = (1/2)*A002265(n)*(n - 2 + A010873(n)).
a(n) = floor((n-1)^2/8). - Mitch Harris, Sep 08 2008
a(n) = round(n*(n-2)/8) = round((n^2-2*n-1)/8) = ceiling((n+1)*(n-3)/8). - Mircea Merca, Nov 28 2010
a(n) = A001972(n-4), n>3. - Franklin T. Adams-Watters, Jul 10 2009
a(n) = a(n-4)+n-3, n>3. - Mircea Merca, Nov 28 2010
Euler transform of length 4 sequence [ 2, 0, 0, 1]. - Michael Somos, Oct 14 2011
a(n) = a(2-n) for all n in Z. - Michael Somos, Oct 14 2011
a(n) = A214734(n, 1, 4). - Renzo Benedetti, Aug 27 2012
a(4n) = A000384(n), a(4n+1) = A001105(n), a(4n+2) = A014105(n), a(4n+3) = A046092(n). - Philippe Deléham, Mar 26 2013
a(n) = Sum_{i=1..ceiling(n/2)-1} (i mod 2) * (n - 2*i - 1). - Wesley Ivan Hurt, Jan 23 2014
a(n) = ( 2*n^2-4*n-1+(-1)^n+2*((-1)^((2*n-1+(-1)^n)/4)-(-1)^((6*n-1+(-1)^n)/4)) )/16 = ( 2*n*(n-2) - (1-(-1)^n)*(1-2*i^(n*(n-1))) )/16, where i=sqrt(-1). - Luce ETIENNE, Aug 29 2014
E.g.f.: (1/8)*((- 1 + x)*x*cosh(x) + 2*sin(x) + (- 1 - x + x^2)*sinh(x)). - Stefano Spezia, Jan 15 2019
a(n) = (A002620(n-1) - A011765(n+1)) / 2, for n > 0. - Yuchun Ji, Feb 05 2021
Sum_{n>=4} 1/a(n) = Pi^2/12 + 5/2. - Amiram Eldar, Aug 13 2022

Partially edited by R. J. Mathar, Jul 11 2009

cp's OEIS Frontend

A028896 6 times triangular numbers: a(n) = 3*n*(n+1).

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A028895 5 times triangular numbers: a(n) = 5*n*(n+1)/2.

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A077221 a(0) = 0 and then alternately even and odd numbers in increasing order such that the sum of any two successive terms is a square.

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A142463 a(n) = 2*n^2 + 2*n - 1.

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A122141 Array: T(d,n) = number of ways of writing n as a sum of d squares, read by ascending antidiagonals.

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A027468 9 times the triangular numbers A000217.

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A028896 6 times triangular numbers: a(n) = 3n(n+1).

A028895 5 times triangular numbers: a(n) = 5n(n+1)/2.

A142463 a(n) = 2n^2 + 2n - 1.