A069894 -id:A069894 - cp's OEIS frontend

A166911 a(n) = (9 + 14n + 12n^2 + 4*n^3)/3.

3, 13, 39, 89, 171, 293, 463, 689, 979, 1341, 1783, 2313, 2939, 3669, 4511, 5473, 6563, 7789, 9159, 10681, 12363, 14213, 16239, 18449, 20851, 23453, 26263, 29289, 32539, 36021, 39743, 43713, 47939, 52429, 57191, 62233, 67563, 73189, 79119, 85361, 91923

Offset: 0

Paul Curtz, Oct 23 2009

The inverse binomial transform yields the quasi-finite sequence 3,10,16,8,0,.. (0 continued).
These are the bottom-left numbers in the blocks (each with 2 rows) shown in A172002, the
atomic number of the leftmost element in the 2nd, 4th, 6th etc. row of the Janet table.

Charles Janet, La structure du noyau de l'atome .., Nov 1927, page 15.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

[(9+14*n+12*n^2+4*n^3)/3: n in [0..40] ]; // Vincenzo Librandi, Aug 06 2011

LinearRecurrence[{4,-6,4,-1}, {3, 13, 39, 89}, 100] (* G. C. Greubel, May 28 2016 *)

a(n)=n*(4*n^2+12*n+14)/3+3 \\ Charles R Greathouse IV, Dec 21 2011

First differences: a(n)-a(n-1) = 2+4*n+4*n^2 = 1+(1+2n)^2 = 1 + A016754(n+1) = A069894(n+1).
Second differences: a(n) - 2*a(n-1) + a(n-2) = 8*n = A008590(n+2).
Third differences: a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 8.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: (3 + x + 5*x^2 - x^3)/(1-x)^4.
a(n) = A166464(n) + 2*(n+1)^2 = A166464(n) + A001105(n+1).
E.g.f.: (1/3)*(9 + 30*x + 24*x^2 + 4*x^3)*exp(x). - G. C. Greubel, May 28 2016

Edited and extended by R. J. Mathar, Mar 02 2010

A248800 a(n) = (2*n^2 + 3 + (-1)^n)/2.

Original entry on oeis.org

2, 2, 6, 10, 18, 26, 38, 50, 66, 82, 102, 122, 146, 170, 198, 226, 258, 290, 326, 362, 402, 442, 486, 530, 578, 626, 678, 730, 786, 842, 902, 962, 1026, 1090, 1158, 1226, 1298, 1370, 1446, 1522, 1602, 1682, 1766, 1850, 1938, 2026, 2118

Offset: 0

Paul Curtz, Oct 14 2014

Numbers belonging to A016825: a(n) = A016825( A002620(n) ). - Bruno Berselli, Oct 15 2014

For n>1, a(n) is the number of row vectors of length 2n with entries in [1,n], first entry 1, with maximum inner distance. That is, vectors where the modular distance between adjacent entries is at least (n-2)/2. Modular distance is the minimum of remainders of (x - y) and (y - x) when dividing by n. Geometrically, this metric counts how far the integers mod n are from each other if 1 and n are adjacent as on a circle. - Omar Aceval Garcia, Jan 30 2021

G. C. Greubel, Table of n, a(n) for n = 0..1000
Omar Aceval Garcia, On the Number of Maximum Inner Distance Latin Squares, arXiv:2112.13912 [math.CO], 2021.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A000034, A000290, A002620, A016825, A042964, A080827, A168273, A005899, A069894.

[n^2+3/2+(-1)^n/2: n in [0..50]]; // Vincenzo Librandi, Oct 15 2014

Table[n^2 + 3/2 + (-1)^n/2, {n, 0, 50}] (* Bruno Berselli, Oct 15 2014 *)
CoefficientList[Series[2(x^3+x^2-x+1)/((1-x)^3 (x+1)), {x, 0, 50}], x] (* Vincenzo Librandi, Oct 15 2014 *)
LinearRecurrence[{2,0,-2,1},{2,2,6,10},60] (* Harvey P. Dale, Apr 08 2019 *)

Vec(-2*(x^3+x^2-x+1)/((x-1)^3*(x+1)) + O(x^100)) \\ Colin Barker, Oct 15 2014

[(2*n^2 +3 +(-1)^n)/2 for n in (0..50)] # G. C. Greubel, Dec 14 2021

a(n) = A000290(n) + A000034(n+1) = 4*A002620(n) + 2.
a(n+1) = 2*A080827(n+1) = (n+2)^2 - A042964(n+1) = a(n) + 2*n + 1 -(-1)^n.
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - Colin Barker, Oct 15 2014
G.f.: 2*(1-x+x^2+x^3) / ((1-x)^3*(x+1)). - Colin Barker, Oct 15 2014
E.g.f.: cosh(x) + (1 + x + x^2)*exp(x). - G. C. Greubel, Dec 14 2021
a(2n) = A005899(n) for n > 0, a(2n+1) = A069894(n). - Omar Aceval Garcia, Dec 30 2021

Typo in data fixed by Colin Barker, Oct 15 2014

A069477 a(n) = 60n^2 + 180n + 150.

Original entry on oeis.org

390, 750, 1230, 1830, 2550, 3390, 4350, 5430, 6630, 7950, 9390, 10950, 12630, 14430, 16350, 18390, 20550, 22830, 25230, 27750, 30390, 33150, 36030, 39030, 42150, 45390, 48750, 52230, 55830, 59550, 63390, 67350, 71430, 75630, 79950, 84390, 88950, 93630, 98430, 103350

Offset: 1

Eli McGowan (ejmcgowa(AT)mail.lakeheadu.ca), Apr 11 2002

First differences of A068236, successive differences of (n+1)^5 - n^5 (A022521).

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001844, A022521, A068236, A069477, A069894, A101096.

[30*(2*n^2 + 6*n + 5): n in [1..40]]; // Vincenzo Librandi, Nov 23 2011

Table[30 (2 n^2 + 6 n + 5), {n, 1, 100}] (* Vladimir Joseph Stephan Orlovsky, Jun 19 2011 *)
LinearRecurrence[{3,-3,1},{390,750,1230},40] (* Harvey P. Dale, Apr 06 2012 *)

a(n)=60*n^2+180*n+150 \\ Charles R Greathouse IV, Nov 23 2011

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(1)=390, a(2)=750, a(3)=1230. - Harvey P. Dale, Apr 06 2012
Sum_{n>=1} 1/a(n) = (Pi/60)*tanh(Pi/2) - 1/25. - Amiram Eldar, Jan 27 2022
From Elmo R. Oliveira, Feb 08 2025: (Start)
G.f.: 30*x*(5*x^2 - 14*x + 13)/(1-x)^3.
E.g.f.: 30*(exp(x)*(2*x^2 + 8*x + 5) - 5).
a(n) = 30*A001844(n+1) = 15*A069894(n+1). (End)

A293958 Smallest odd prime divisor of (2n+1)^2 + 1.

Original entry on oeis.org

5, 13, 5, 41, 61, 5, 113, 5, 181, 13, 5, 313, 5, 421, 13, 5, 613, 5, 761, 29, 5, 1013, 5, 1201, 1301, 5, 17, 5, 1741, 1861, 5, 2113, 5, 2381, 2521, 5, 29, 5, 3121, 17, 5, 3613, 5, 17, 41, 5, 4513, 5, 13, 5101, 5, 37, 5, 13, 61, 5, 17, 5, 73, 7321, 5, 13, 5, 53, 8581, 5, 13, 5, 9661, 9941, 5

Offset: 1

N. J. A. Sloane, Nov 04 2017, following a suggestion from Zoran Sunic

nonn

If the map "x -> smallest odd prime divisor of n^2+1" is iterated, does it always terminate in the 2-cycle (5 <-> 13)? - Zoran Sunic, Oct 25 2017

A027862 is a subsequence. - David A. Corneth, Nov 04 2017

David A. Corneth, Table of n, a(n) for n = 1..10000

A bisection of A125256. Cf. A027862, A069894, A078701, A256970.

sod[n_]:=With[{fi=FactorInteger[n]},If[fi[[1,1]]==2,fi[[2,1]],fi[1,1]]]; sod/@(Range[3,151,2]^2+1) (* Harvey P. Dale, Dec 23 2023 *)

a(n) = factor((2*n+1)^2 + 1)[2,1]; \\ Michel Marcus, Nov 04 2017

a(n) = A078701(A069894(n)). - Michel Marcus, Nov 04 2017

A364361 Table read by rows. T(n, k) = Sum_{j=0..n-k} kbinomial(k, j)binomial(n - j, k).

Original entry on oeis.org

0, 0, 1, 0, 3, 2, 0, 5, 10, 3, 0, 7, 26, 21, 4, 0, 9, 50, 75, 36, 5, 0, 11, 82, 189, 164, 55, 6, 0, 13, 122, 387, 516, 305, 78, 7, 0, 15, 170, 693, 1284, 1155, 510, 105, 8, 0, 17, 226, 1131, 2724, 3405, 2262, 791, 136, 9, 0, 19, 290, 1725, 5156, 8415, 7734, 4025, 1160, 171, 10

Offset: 0

Peter Luschny, Jul 30 2023

			The triangle begins:
  [0] 0;
  [1] 0,  1;
  [2] 0,  3,   2;
  [3] 0,  5,  10,   3;
  [4] 0,  7,  26,   21,    4;
  [5] 0,  9,  50,   75,   36,    5;
  [6] 0, 11,  82,  189,  164,   55,    6;
  [7] 0, 13, 122,  387,  516,  305,   78,   7;
  [8] 0, 15, 170,  693, 1284, 1155,  510, 105,   8;
  [9] 0, 17, 226, 1131, 2724, 3405, 2262, 791, 136, 9;
Seen as an array:
  [0] 0,  1,   2,   3,     4,     5,      6,      7, ...  A001477
  [1] 0,  3,  10,   21,   36,    55,     78,    105, ...  A014105
  [2] 0,  5,  26,   75,  164,   305,    510,    791, ...  A048395
  [3] 0,  7,  50,  189,  516,  1155,   2262,   4025, ...
  [4] 0,  9,  82,  387, 1284,  3405,   7734,  15687, ...
  [5] 0, 11, 122,  693, 2724,  8415,  21918,  50281, ...
  [6] 0, 13, 170, 1131, 5156, 18265,  53934, 138775, ...
  [7] 0, 15, 226, 1725, 8964, 35915, 118950, 340473, ...
    A005408|A069894

Cf. A364553 (row sums),  A364634 (main diagonal).
Rows: A001477, A014105, A048395.
Columns: A005408, A069894.

T := (n, k) -> local j; add(k*binomial(k, j)*binomial(n-j, k), j = 0..n-k):
seq(seq(T(n, k), k = 0..n), n = 0..10);

T(2*n, n) = n * LegendreP(n, 3).

A383466 a(0) = 1; thereafter a(n) = 10n^2 - 5n + 2.

Original entry on oeis.org

1, 7, 32, 77, 142, 227, 332, 457, 602, 767, 952, 1157, 1382, 1627, 1892, 2177, 2482, 2807, 3152, 3517, 3902, 4307, 4732, 5177, 5642, 6127, 6632, 7157, 7702, 8267, 8852, 9457, 10082, 10727, 11392, 12077, 12782, 13507, 14252, 15017, 15802, 16607, 17432, 18277, 19142, 20027, 20932, 21857, 22802, 23767, 24752, 25757, 26782, 27827

Offset: 0

Scott R. Shannon and N. J. A. Sloane, Jul 22 2025

Definition: A regular pentagram of radius R is formed by placing five equally-spaced points P_0 .. P_4 around the boundary of a circle of radius R, and drawing line segments P_0 - P_2 - P_4 - P_1 - P_3 - P_0.
Theorem 1: a(n) is the maximum number of regions that can be formed in the plane by drawing n regular pentagrams with the same radius and the same center.
Conjecture 2: a(n) is the maximum number of regions that can be formed in the plane by drawing n regular pentagrams with any radii and any centers.
The following construction works for any n >= 1. Take 5*n equally-spaced points P_i around a circle, and draw a pentagram through P_i, P_{i+n}, P_{i+2*n}, P_{i+3*n}, P_{i+4*n} for i = 0, ..., n-1.
The resulting planar graph decomposes into 5*n triangular regions each with 2*n-1 cells (see the red triangle in "Illustration for a(n)..."), plus the interior and exterior regions, for a total of 10*n^2 - 5*n + 2 regions. There are 10*n^2 vertices (10 for n=1, 40 for n=2, and so on).

Paolo Xausa, Table of n, a(n) for n = 0..10000
Scott R. Shannon, Illustration for a(1) = 7. [Note that the cell counts shown on these four figures do not include the black exterior region, so the totals are off by 1]
Scott R. Shannon, Illustration for a(2) = 32.
Scott R. Shannon, Illustration for a(3) = 77.
Scott R. Shannon, Illustration for a(8) = 602.
N. J. A. Sloane, Illustration for a(1) = 7.
N. J. A. Sloane, Illustration for a(2) = 32.
N. J. A. Sloane, Illustration for a(n), n >= 1, showing a(3) = 77.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

See A077588, A069894, and A386477 for analogous sequences based on triangles, squares, and hexagrams.

Without the "+2" in the definition, the sequence is A152745.

A383466[n_] := If[n == 0, 1, 5*n*(2*n - 1) + 2]; Array[A383466, 50, 0] (* or *)
Join[{1}, 5*PolygonalNumber[6, Range[49]] + 2] (* or *)
LinearRecurrence[{3, -3, 1}, {1, 7, 32, 77}, 50] (* Paolo Xausa, Jul 22 2025 *)

From Elmo R. Oliveira, Sep 03 2025: (Start)
G.f.: (1 + 4*x + 14*x^2 + x^3)/(1 - x)^3.
E.g.f.: exp(x)*(2 + 5*x + 10*x^2) - 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A386477 a(0) = 1; thereafter a(n) = 2(6n^2 - 3*n + 1).

Original entry on oeis.org

1, 8, 38, 92, 170, 272, 398, 548, 722, 920, 1142, 1388, 1658, 1952, 2270, 2612, 2978, 3368, 3782, 4220, 4682, 5168, 5678, 6212, 6770, 7352, 7958, 8588, 9242, 9920, 10622, 11348, 12098, 12872, 13670, 14492, 15338, 16208, 17102, 18020, 18962, 19928, 20918, 21932, 22970, 24032, 25118, 26228, 27362, 28520, 29702, 30908, 32138, 33392, 34670

Offset: 0

Scott R. Shannon and N. J. A. Sloane, Jul 22 2025

Definition: A regular hexagram of radius R is formed by placing six equally-spaced points P_0 .. P_5 around the boundary of a circle of radius R, and drawing line segments P_0 - P_2 - P_4 - P_0 and P_1 - P_3 - P_5 - P_1.
Theorem 1: a(n) is the maximum number of regions that can be formed in the plane by drawing n regular hexagrams with the same radius and the same center.
Conjecture 2: a(n) is the maximum number of regions that can be formed in the plane by drawing n regular hexagrams with any radii and any centers.
The following construction works for any n >= 1. Take 6*n equally-spaced points P_i around a circle, and draw hexagrams starting at P_i for i = 0, ..., n-1.
The resulting planar graph decomposes into 6*n triangular cells each with 2*n-1 cells (see the red triangle in the "Three pentagons" illustration), plus the interior and exterior regions, for a total of 12*n^2 - 6*n + 2 regions. There are 12*n^2 vertices, for n>0.

Paolo Xausa, Table of n, a(n) for n = 0..10000
Scott R. Shannon, Illustration for a(1) = 8 [Note that the cell counts shown on these five figures do not include the black exterior region, so the totals are off by 1]
Scott R. Shannon, Illustration for a(2) = 38
Scott R. Shannon, Illustration for a(3) = 92
Scott R. Shannon, Illustration for a(8) = 722
Scott R. Shannon, Illustration for a(10) = 1142
N. J. A. Sloane, Sketch to illustrate a(2) = 38. The two hexagrams are colored red and black, respectively.
N. J. A. Sloane, Sketch to illustrate a(3) = 92. The three hexagrams are colored red, blue, and black, respectively.
N. J. A. Sloane, Analogous illustration for three pentagrams
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A094159, A152746.

See A077588, A069894, and A383466 for analogous sequences based on triangles, squares, and pentagrams.

A386477[n_] := If[n == 0, 1, 6*n*(2*n - 1) + 2]; Array[A386477, 50, 0] (* or *)
Join[{1}, 6*PolygonalNumber[6, Range[49]] + 2] (* or *)
LinearRecurrence[{3, -3, 1}, {1, 8, 38, 92}, 50] (* Paolo Xausa, Jul 24 2025 *)

From Stefano Spezia, Jul 23 2025: (Start)
G.f.: (1 + 5*x + 17*x^2 + x^3)/(1 - x)^3.
E.g.f.: 2*exp(x)*(1 + 3*x + 6*x^2) - 1. (End)
a(n) = A152746(n) + 2, for n >= 1. - Paolo Xausa, Jul 24 2025

A185669 a(n) = 4n^2 + 3n + 2.

Original entry on oeis.org

2, 9, 24, 47, 78, 117, 164, 219, 282, 353, 432, 519, 614, 717, 828, 947, 1074, 1209, 1352, 1503, 1662, 1829, 2004, 2187, 2378, 2577, 2784, 2999, 3222, 3453, 3692, 3939, 4194, 4457, 4728, 5007, 5294, 5589, 5892, 6203, 6522, 6849, 7184, 7527, 7878, 8237, 8604, 8979, 9362, 9753, 10152, 10559, 10974, 11397, 11828

Offset: 0

Paul Curtz, Feb 09 2011

Natural numbers A000027 written clockwise as a square spiral:
.
  43--44--45--46--47--48--49
   |
  42  21--22--23--24--25--26
   |   |                   |
  41  20   7---8---9--10  27
   |   |   |           |   |
  40  19   6   1---2  11  28
   |   |   |       |   |   |
  39  18   5---4---3  12  29
   |   |               |   |
  38  17--16--15--14--13  30
   |                       |
  37--36--35--34--33--32--31
.
Walking in straight lines away from the center:
1,  2, 11, ... = A054552(n)     = 1 -3*n+4*n^2,
1,  8, 23, ... = A033951(n)     = 1 +3*n+4*n^2,
1,  3, 13, ... = A054554(n+1)   = 1 -2*n-4*n^2,
1,  7, 21, ... = A054559(n+1)   = 1 +2*n+4*n^2,
1,  4, 15, ... = A054556(n+1)   = 1   -n+4*n^2,
1,  6, 19, ... = A054567(n+1)   = 1   +n+4*n^2,
1,  5, 17, ... = A053755(n)     = 1     +4*n^2,
1,  9, 25, ... = A016754(n)     = 1 +4*n+4*n^2 = (1+2*n)^2,
2,  8, 22, ... = 2*A084849(n)   = 2 +2*n+4*n^2,
2, 12, 30, ... = A002939(n+1)   = 2 +6*n+4*n^2,
2,  9, 24, ... = a(n)           = 2 +3*n+4*n^2,
2, 10, 26, ... = A069894(n)     = 2 +4*n+4*n^2,
3, 11, 27, ... = A164897(n)     = 3 +4*n+4*n^2,
3, 12, 29, ... = A054552(n+1)+1 = 3 +5*n+4*n^2,
3, 14, 33, ... = A033991(n+1)   = 3 +7*n+4*n^2,
3, 15, 35, ... = A000466(n+1)   = 3 +8*n+4*n^2,
4, 14, 32, ... = 2*A130883(n+1) = 4 +6*n+4*n^2,
4, 16, 36, ... = A016742(n+1)   = 4 +8*n+4*n^2 = (2+2*n)^2,
5, 18, 39, ... = A007742(n+1)   = 5 +9*n+4*n^2,
5, 19, 41, ... = A125202(n+2)   = 5+10*n+4*n^2.

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

[2+3*n+4*n^2: n in [0..80]];  // Vincenzo Librandi, Feb 09 2011

Table[4n^2 + 3n + 2, {n,0,50}] (* G. C. Greubel, Jul 09 2017 *)
LinearRecurrence[{3,-3,1},{2,9,24},60] (* Harvey P. Dale, Aug 11 2021 *)

a(n)=4*n^2+3*n+2 \\ Charles R Greathouse IV, Apr 14 2014

a(n) = a(n-1) + 8*n - 1.
a(n) = 2*a(n-1) - a(n-2) + 8.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: (2 +3*x +3*x^2)/(1-x)^3 . - R. J. Mathar, Feb 11 2011
a(n) = A033954(n) + 2. - Bruno Berselli, Apr 10 2011
E.g.f.: (4*x^2 + 7*x + 2)*exp(x). - G. C. Greubel, Jul 09 2017

A265129 Triangle read by rows, formed as the sum of the two versions of the natural numbers filling an equilateral triangle.

Original entry on oeis.org

2, 5, 5, 10, 10, 10, 17, 17, 17, 17, 26, 26, 26, 26, 26, 37, 37, 37, 37, 37, 37, 50, 50, 50, 50, 50, 50, 50, 65, 65, 65, 65, 65, 65, 65, 65, 82, 82, 82, 82, 82, 82, 82, 82, 82, 101, 101, 101, 101, 101, 101, 101, 101, 101, 101

Offset: 1

Craig Knecht, Dec 02 2015

The natural numbers can sequentially fill a right- or left-handed equilateral triangle. Componentwise addition of the values of these two triangles produces the present triangle.
The row sums for this triangle give A034262.
The difference between the right- and left-handed triangles produces A049581.

			Displayed as a triangle:
   2;
   5  5;
  10 10 10;
  17 17 17 17;
  26 26 26 26 26;
  37 37 37 37 37 37;
  ...

Craig Knecht, Sum of right and left triangles.
Craig Knecht, Difference between right and left triangles A049581.

Column k=1 gives A002522.
Cf. A049581 (difference of triangles), A034262 (row sum of triangle), A069894 (center column).
Cf. A071237.

seq(seq(n^2+1,k=1..n),n=1..10); # Georg Fischer, Oct 01 2021

T(n,k) = n^2 + 1 for k = 1..n and n >= 1. - Georg Fischer, Oct 01 2021

Sum_{k=1..n} k * T(n,k) = A071237(n). - Alois P. Heinz, Oct 01 2021

Row 6 with T(6,k)=37 inserted by Georg Fischer, Oct 01 2021

A267654 Irregular triangle of palindromic subsequences. Every row has 2n+1 terms. From the second row, there are only two alternated numbers: 2n+4 and 2*n+2.

Original entry on oeis.org

2, 4, 2, 4, 6, 4, 6, 4, 6, 8, 6, 8, 6, 8, 6, 8, 10, 8, 10, 8, 10, 8, 10, 8, 10, 12, 10, 12, 10, 12, 10, 12, 10, 12, 10, 12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 16, 14, 16, 14, 16, 14, 16, 14, 16, 14, 16, 14, 16, 14, 16

Offset: 0

Paul Curtz, Jan 19 2016

Row sums = 2, 10, 26, 50, ... = A069894(n).
Starting from A053186(n) =
0,                            for b(n)
0, 1, 2,                              for c(n)
0, 1, 2, 3, 4,                                for d(n)
0, 1, 2, 3, 4, 5, 6,
etc,
a(n) is used for
1) b(n+1) = b(n) + (a(0)=2) i.e. 0, 2, 4, 6, ... = A005843(n).
2) c(n+3) = c(n) + (period 3:repeat 4, 2, 4) i.e. 0, 1, 2, 4, 3, 6, 8, ... =  A265667(n).
3) d(n+5) = d(n) + (period 5:repeat 6, 4, 6, 4, 6) i.e. 0, 1, 2, 3, 4, 6, 5, 8, 7, 10, ... = A265734(n).
Etc.
a(n) has a companion with the same terms,differently distributed,yielding permutations of the nonnegative numbers. See A265672.
a(n) other writing (by pairs):
2,   4,  2,  4,
6,   4,  6,  4,
6,   8,  6,  8,  6,  8,  6,  8,
10   8, 10,  8, 10,  8, 10,  8,
10, 12, 10, 12, 10, 12, 10, 12, 10, 12, 10, 12,
14, 12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 12,
etc.
First column: A168276(n+2). Second column: A168273(n+2).
Row sums: 12, 20, 56, 72, ... = 4*A074378(n+1).
The last term of the successive rows is the number of their terms.
Main diagonal: A005843(n+1).

			The triangle is
2,
4, 2, 4,
6, 4, 6, 4, 6,
8, 6, 8, 6, 8, 6, 8,
etc.

Cf. A007395, A005843, A008586, A016825, A053186, A069894, A074378, A086520, A168273, A168276, A265667, A265672, A265734.

Table[2 (n - 1) + 2 (Boole@ OddQ@ k + 1), {n, 0, 7}, {k, 2 n + 1}] // Flatten (* Michael De Vlieger, Jan 19 2016 *)

a(n) = 2 * A086520(n+2).
a(2n) = 4*n + 2 times 4*n + 2 = 2, 2, 6, 6, 6, 6, 6, 6, 10,....
a(2n+1) = 4*(n+1) times 4*(n+1) = 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, 12, ....

cp's OEIS Frontend

A166911 a(n) = (9 + 14*n + 12*n^2 + 4*n^3)/3.

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A248800 a(n) = (2*n^2 + 3 + (-1)^n)/2.

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A069477 a(n) = 60*n^2 + 180*n + 150.

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A293958 Smallest odd prime divisor of (2n+1)^2 + 1.

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A364361 Table read by rows. T(n, k) = Sum_{j=0..n-k} k*binomial(k, j)*binomial(n - j, k).

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A383466 a(0) = 1; thereafter a(n) = 10*n^2 - 5*n + 2.

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A386477 a(0) = 1; thereafter a(n) = 2*(6*n^2 - 3*n + 1).

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A166911 a(n) = (9 + 14n + 12n^2 + 4*n^3)/3.

A069477 a(n) = 60n^2 + 180n + 150.

A364361 Table read by rows. T(n, k) = Sum_{j=0..n-k} kbinomial(k, j)binomial(n - j, k).

A383466 a(0) = 1; thereafter a(n) = 10n^2 - 5n + 2.

A386477 a(0) = 1; thereafter a(n) = 2(6n^2 - 3*n + 1).

A185669 a(n) = 4n^2 + 3n + 2.

A267654 Irregular triangle of palindromic subsequences. Every row has 2n+1 terms. From the second row, there are only two alternated numbers: 2n+4 and 2*n+2.