A135703 -id:A135703 - cp's OEIS frontend

A045944 Rhombic matchstick numbers: a(n) = n(3n+2).

0, 5, 16, 33, 56, 85, 120, 161, 208, 261, 320, 385, 456, 533, 616, 705, 800, 901, 1008, 1121, 1240, 1365, 1496, 1633, 1776, 1925, 2080, 2241, 2408, 2581, 2760, 2945, 3136, 3333, 3536, 3745, 3960, 4181, 4408, 4641, 4880, 5125, 5376, 5633, 5896, 6165, 6440

Offset: 0

R. K. Guy

From Floor van Lamoen, Jul 21 2001: (Start)
Write 1,2,3,4,... in a hexagonal spiral around 0, then a(n) is the n-th term of the sequence found by reading the line from 0 in the direction 0,5,.... The spiral begins:
                 .
                 85--84--83--82--81--80
                   .                   \
                   56--55--54--53--52  79
                   / .               \   \
                 57  33--32--31--30  51  78
                 /   / .           \   \   \
               58  34  16--15--14  29  50  77
               /   /   / .       \   \   \   \
             59  35  17   5---4  13  28  49  76
             /   /   /   / .   \   \   \   \   \
           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)
Connection to triangular numbers: a(n) = 4*T_n + S_n where T_n is the n-th triangular number and S_n is the n-th square. - William A. Tedeschi, Sep 12 2010
Also, second octagonal numbers. - Bruno Berselli, Jan 13 2011
Sequence found by reading the line from 0, in the direction 0, 16, ... and the line from 5, in the direction 5, 33, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012
Let P denote the points from the n X n grid. A(n-1) also coincides with the minimum number of points Q needed to "block" P, that is, every line segment spanned by two points from P must contain one point from Q. - Manfred Scheucher, Aug 30 2018
Also the number of internal edges of an (n+1)*(n+1) "square" of hexagons; i.e., n+1 rows, each of n+1 edge-adjacent hexagons, stacked with minimal overhang. - Jon Hart, Sep 29 2019
For n >= 1, the continued fraction expansion of sqrt(27*a(n)) is [9n+2; {1, 2n-1, 1, 1, 1, 2n-1, 1, 18n+4}]. - Magus K. Chu, Oct 13 2022

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Ghislain R. Franssens, On a Number Pyramid Related to the Binomial, Deleham, Eulerian, MacMahon and Stirling number triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.4.1.
M. Janjic and B. Petkovic, A Counting Function, arXiv:1301.4550 [math.CO], 2013.
Leo Tavares, Illustration: Square Stars
Leo Tavares, Illustration: Split Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A001859. See Comments of A135713.
Cf. A000217, A000567, A001082, A002378, A016969, A049450, A174709.
Cf. second n-gonal numbers: A005449, A014105, A147875, A179986, A033954, A062728, A135705.
Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734, A139273.
Cf. A000290, A022144, A005563, A001105.
Cf. A056109.
Cf. A003154.

[n*(3*n+2) : n in [0..100]]; // Wesley Ivan Hurt, Sep 24 2017

Table[n*(3n+2), {n,0,60}] (* Harvey P. Dale, May 05 2011 *)
LinearRecurrence[{3,-3,1},{0,5,16},60] (* Harvey P. Dale, Jan 19 2016 *)
CoefficientList[Series[x*(5 + x)/(1 - x)^3,{x, 0, 60}], x] (* Stefano Spezia, Sep 01 2018 *)

a(n)=n*(3*n+2) \\ Charles R Greathouse IV, Nov 20 2012

O.g.f.: x*(5+x)/(1-x)^3. - R. J. Mathar, Jan 07 2008
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=0, a(1)=5, a(2)=16. - Harvey P. Dale, May 06 2011
a(n) = a(n-1) + 6*n - 1 (with a(0)=0). - Vincenzo Librandi, Nov 18 2010
For n > 0, a(n)^3 + (a(n)+1)^3 + ... + (a(n)+n)^3 + 2*A000217(n)^2 = (a(n) + n + 1)^3 + ... + (a(n) + 2n)^3; see also A033954. - Charlie Marion, Dec 08 2007
a(n) = Sum_{i=0..n-1} A016969(i) for n > 0. - Bruno Berselli, Jan 13 2011
a(n) = A174709(6*n+4). - Philippe Deléham, Mar 26 2013
a(n) = A001082(2*n). - Michael Turniansky, Aug 24 2013
Sum_{n>=1} 1/a(n) = (9 + sqrt(3)*Pi - 9*log(3))/12 = 0.3794906245574721941... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A002378(n) + A014105(n). - J. M. Bergot, Apr 24 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/sqrt(12) - 3/4. - Amiram Eldar, Jul 03 2020
E.g.f.: exp(x)*x*(5 + 3*x). - Stefano Spezia, Jun 08 2021
From Leo Tavares, Oct 14 2021: (Start)
a(n) = A000290(n) + 4*A000217(n). See Square Stars illustration.
a(n) = A000567(n+2) - A022144(n+1)
a(n) = A005563(n) + A001105(n).
a(n) = A056109(n) - 1. (End)
From Leo Tavares, Oct 06 2022: (Start)
a(n) = A003154(n+1) - A000567(n+1). See Split Stars illustration.
a(n) = A014105(n) + 2*A000217(n). (End)

A014106 a(n) = n(2n + 3).

Original entry on oeis.org

0, 5, 14, 27, 44, 65, 90, 119, 152, 189, 230, 275, 324, 377, 434, 495, 560, 629, 702, 779, 860, 945, 1034, 1127, 1224, 1325, 1430, 1539, 1652, 1769, 1890, 2015, 2144, 2277, 2414, 2555, 2700, 2849, 3002, 3159, 3320, 3485, 3654, 3827, 4004, 4185, 4370

Offset: 0

N. J. A. Sloane

If Y is a 2-subset of a 2n-set X then, for n >= 1, a(n-1) is the number of (2n-2)-subsets of X intersecting Y. - Milan Janjic, Nov 18 2007
This sequence can also be derived from 1*(2+3)=5, 2*(3+4)=14, 3*(4+5)=27, and so forth. - J. M. Bergot, May 30 2011
Consider the partitions of 2n into exactly two parts.  Then a(n) is the sum of all the parts in the partitions of 2n + the number of partitions of 2n + the total number of partition parts of 2n. - Wesley Ivan Hurt, Jul 02 2013
a(n) is the number of self-intersecting points of star polygon {(2*n+3)/(n+1)}. - Bui Quang Tuan, Mar 25 2015
Bisection of A000096. - Omar E. Pol, Dec 16 2016
a(n+1) is the number of function calls required to compute Ackermann's function ack(2,n). - Olivier Gérard, May 11 2018
a(n-1) is the least denominator d > n of the best rational approximation of sqrt(n^2-2) by x/d (see example and PARI code). - Hugo Pfoertner, Apr 30 2019
The number of cells in a loose n X n+1 rectangular spiral where n is even. See loose rectangular spiral image. - Jeff Bowermaster, Aug 05 2019
a(n-1) is the dimension of the second cohomology group of 2n+1-dimensional Heisenberg Lie algebra h_{2n+1}. - Rafik Khalfi, Jan 27 2025

			a(5-1) = 44: The best approximation of sqrt(5^2-2) = sqrt(23) by x/d with d <= k is 24/5 for all k < 44, but sqrt(23) ~= 211/44 is the first improvement. - _Hugo Pfoertner_, Apr 30 2019

Jolley, Summation of Series, Dover (1961).

Vincenzo Librandi, Table of n, a(n) for n = 0..920
Jeff Bowermaster, Loose Rectangular Spiral
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Milan Janjic, Two Enumerative Functions
L. J. Santharoubane, Cohomology of Heisenberg Lie algebras, Proc. Amer. Math. Soc. 87 (1983), 23-28.
Leo Tavares, Illustration: Hex-tangles
Leo Tavares, Illustration: Second Hex-tangles
Leo Tavares, Illustration: Ob-tangles
Leo Tavares, Illustration: Trap-tangles
Eric Weisstein's World of Mathematics, Star Polygon
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A091823. See A110325 for another version.
Cf. numbers of the form  n*(d*n+10-d)/2:  A008587, A056000, A028347, A140090, A028895, A045944, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734, A139273.
Cf. A000384, A014105, A002378, A005563, A000217.

[n*(2*n+3): n in [0..50]]; // Vincenzo Librandi, Apr 25 2011

A014106 := proc(n) n*(2*n+3) ; end proc: # R. J. Mathar, Feb 13 2011
seq(k*(2*k+3), k=1..100); # Wesley Ivan Hurt, Jul 02 2013

Table[n (2 n + 3), {n, 0, 120}] (* Michael De Vlieger, Apr 02 2015 *)
LinearRecurrence[{3,-3,1},{0,5,14},50] (* Harvey P. Dale, Jul 21 2023 *)

PARI
```
a(n)=2*n^2+3*n
```

\\ least denominator > n in best rational approximation of sqrt(n^2-2)
for(n=2,47,for(k=n,oo,my(m=denominator(bestappr(sqrt(n^2-2),k)));if(m>n,print1(k,", ");break(1)))) \\ Hugo Pfoertner, Apr 30 2019

a(n) - 1 = A091823(n). - Howard A. Landman, Mar 28 2004
A014107(-n) = a(n), A000384(n+1) = a(n)+1. - Michael Somos, Nov 06 2005
G.f.: x*(5 - x)/(1 - x)^3. - Paul Barry, Feb 27 2003
E.g.f: x*(5 + 2*x)*exp(x). - Michael Somos, Nov 06 2005
a(n) = a(n-1) + 4*n + 1, n > 0. - Vincenzo Librandi, Nov 19 2010
a(n) = 4*A000217(n) + n. - Bruno Berselli, Feb 11 2011
Sum_{n>=1} 1/a(n) = 8/9 -2*log(2)/3 = 0.4267907685155920.. [Jolley eq. 265]
Sum_{n>=1} (-1)^(n+1)/a(n) = 4/9 + log(2)/3 - Pi/6. - Amiram Eldar, Jul 03 2020
From Leo Tavares, Jan 27 2022: (Start)
a(n) = A000384(n+1) - 1. See Hex-tangles illustration.
a(n) = A014105(n) + n*2. See Second Hex-tangles illustration.
a(n) = 2*A002378(n) + n. See Ob-tangles illustration.
a(n) = A005563(n) + 2*A000217(n). See Trap-tangles illustration. (End)

A033429 a(n) = 5*n^2.

Original entry on oeis.org

0, 5, 20, 45, 80, 125, 180, 245, 320, 405, 500, 605, 720, 845, 980, 1125, 1280, 1445, 1620, 1805, 2000, 2205, 2420, 2645, 2880, 3125, 3380, 3645, 3920, 4205, 4500, 4805, 5120, 5445, 5780, 6125, 6480, 6845, 7220, 7605, 8000, 8405, 8820, 9245, 9680, 10125, 10580, 11045, 11520, 12005, 12500

Offset: 0

Jeff Burch

Number of edges of the complete bipartite graph of order 6n, K_n,5n. - Roberto E. Martinez II, Jan 07 2002
Number of edges of the complete tripartite graph of order 4n, K_n,n,2n. - Roberto E. Martinez II, Jan 07 2002
a(n+1)-a(n) : 5, 15, 25, 35, 45, ... (see A017329). - Philippe Deléham, Dec 08 2011
From Larry J Zimmermann, Feb 21 2013: (Start)
The sum of the areas of 2 squares that equals the area of a rectangle with whole number sides using the formula x^2 + y^2 = (x+y+sqrt(2*x*y))(x+y-sqrt(2*x*y)), where the substitution y=2*x obtains the whole number sides of the rectangle. So x^2+(2*x)^2=5x(x).
  x     squares sum      rectangle (l,w)   area
  1     1,4     5                   5,1    5
  2     4,16    20                  10,2   20  (End)

G. C. Greubel, Table of n, a(n) for n = 0..5000
Lancelot Hogben, Choice and Chance by Cardpack and Chessboard, Vol. 1, Max Parrish and Co, London, 1950, p. 36.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Central column of A055096.
Cf. A000290.
Cf. numbers of the form  n*(d*n+10-d)/2:  A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A022267, A022268, A049452, A186030, A135703, A152734, A139273.
Cf. A185019.
Cf. A001105, A033428.
Similar sequences are listed in A316466.

5*Range[50]^2 (* Alonso del Arte, May 23 2012 *)

PARI
```
a(n)=5*n^2
```

a(n) = 5*A000290(n). - Omar E. Pol, Dec 11 2008
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: 5*x*(1+x)/(1-x)^3.
a(n) = 4*A000217(n) + A000567(n). (End)
a(n) = a(n-1)+5*(2*n-1) (with a(0)=0). - Vincenzo Librandi, Nov 17 2010
a(n) = A131242(10*n+4). - Philippe Deléham, Mar 27 2013
a(n) = a(n-1) + 10*n - 5, with a(0)=0. - Jean-Bernard François, Oct 04 2013
a(n) = A001105(n) + A033428(n). - Altug Alkan, Sep 28 2015
E.g.f.: 5*x*(x+1)*exp(x). - G. C. Greubel, Jul 17 2017
a(n) = Sum_{i = 2..6} P(i,n), where P(i,m) = m*((i-2)*m-(i-4))/2. - Bruno Berselli, Jul 04 2018
From Amiram Eldar, Feb 03 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/30.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/60.
Product_{n>=1} (1 + 1/a(n)) = sqrt(5)*sinh(Pi/sqrt(5))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(5)*sin(Pi/sqrt(5))/Pi. (End)

Better description from N. J. A. Sloane, May 15 1998

A140090 a(n) = n(3n + 7)/2.

Original entry on oeis.org

0, 5, 13, 24, 38, 55, 75, 98, 124, 153, 185, 220, 258, 299, 343, 390, 440, 493, 549, 608, 670, 735, 803, 874, 948, 1025, 1105, 1188, 1274, 1363, 1455, 1550, 1648, 1749, 1853, 1960, 2070, 2183, 2299, 2418, 2540, 2665, 2793, 2924

Offset: 0

Omar E. Pol, May 22 2008

This sequence is mentioned in the Guo-Niu Han's paper, chapter 6: Dictionary of the standard puzzle sequences, p. 19 (see link). - Omar E. Pol, Oct 28 2011

Number of cards needed to build an n-tier house of cards with a flat, one-card-wide roof. - Tyler Busby, Dec 28 2022

G. C. Greubel, Table of n, a(n) for n = 0..5000
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Sela Fried, Counting r X s rectangles in nondecreasing and Smirnov words, arXiv:2406.18923 [math.CO], 2024. See p. 5.
Julie Zhang, Noah A. Rosenberg, and Julia A. Palacios, The space of multifurcating ranked tree shapes: enumeration, lattice structure, and Markov chains, arXiv:2506.10856 [math.PR], 2025. See p. 33.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

The generalized pentagonal numbers b*n+3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, A005449, A045943, A115067, this sequence, A140091, A059845, A140672, A140673, A140674, A140675, A151542.

Cf. numbers of the form n*(d*n + 10 - d)/2: A008587, A056000, A028347, A014106, A028895, A045944, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734, A139273.

LinearRecurrence[{3,-3,1},{0,5,13},50] (* Harvey P. Dale, Jan 17 2022 *)

a(n)=n*(3*n+7)/2 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: x*(5 - 2*x)/(1 - x)^3. - Bruno Berselli, Feb 11 2011
a(n) = (3*n^2 + 7*n)/2.
a(n) = a(n-1) + 3*n + 2 (with a(0)=0). - Vincenzo Librandi, Nov 24 2010
E.g.f.: (1/2)*(3*x^2 + 10*x)*exp(x). - G. C. Greubel, Jul 17 2017
From Amiram Eldar, Feb 22 2022: (Start)
Sum_{n>=1} 1/a(n) = 117/98 - Pi/(7*sqrt(3)) - 3*log(3)/7.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/(7*sqrt(3)) + 4*log(2)/7 - 75/98. (End)

A022267 a(n) = n(9n + 1)/2.

Original entry on oeis.org

0, 5, 19, 42, 74, 115, 165, 224, 292, 369, 455, 550, 654, 767, 889, 1020, 1160, 1309, 1467, 1634, 1810, 1995, 2189, 2392, 2604, 2825, 3055, 3294, 3542, 3799, 4065, 4340, 4624, 4917, 5219, 5530, 5850, 6179

Offset: 0

N. J. A. Sloane

From Floor van Lamoen, Jul 21 2001: (Start)
Write 0, 1, 2, 3, 4, ... in a triangular spiral; then a(n) is the sequence found by reading the line from 0 in the direction 0, 5, ... . The spiral begins:
.
            15
            / \
          16  14
          /     \
        17   3  13
        /   / \   \
      18   4   2  12
      /   /     \   \
    19   5   0---1  11
    /   /             \
  20   6---7---8---9--10
.
(End)
a(n) is the sum of n consecutive integers starting from 4*n+1: (5), (9+10), (13+14+15), ... - Klaus Purath, Jul 07 2020
a(n) with n>0 are the numbers with the periodic length 3 in the Bulgarian and Mancala solitaire. - Paul Weisenhorn, Jan 29 2022

Lei Zhou, Table of n, a(n) for n = 0..10000
Milan Janjic, Two Enumerative Functions
Leo Tavares, Illustration: X Triangles
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A051682, A110449, A235332.
Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A033429, A022268, A049452, A186030, A135703, A152734, A139273.
Cf. similar sequences listed in A254963.
Cf. similar sequences listed in A022289.
Cf. A060544, A016813.

seq(binomial(9*n+1,2)/9, n=0..37); # Zerinvary Lajos, Jan 21 2007

Table[ n (9 n + 1)/2, {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 19}, 40] (* Harvey P. Dale, Jul 01 2013 *)

vector(100,n,(n-1)*(9*n-8)/2) \\ Derek Orr, Feb 06 2015

a(n) = A110449(n, 4) for n>3.
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 4*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A000566(n). (End)
a(n) = 9*n + a(n-1) - 4 with n>0, a(0)=0. - Vincenzo Librandi, Aug 04 2010
a(n) = A218470(9*n+4). - Philippe Deléham, Mar 27 2013
a(n) = A000217(5*n) - A000217(4*n). - Bruno Berselli, Oct 13 2016
E.g.f.: (1/2)*(9*x^2 + 10*x)*exp(x). - G. C. Greubel, Jul 17 2017
a(n) = A060544(n+1) - A016813(n). - Leo Tavares, Mar 20 2022

A185019 a(n) = n(14n-3).

Original entry on oeis.org

0, 11, 50, 117, 212, 335, 486, 665, 872, 1107, 1370, 1661, 1980, 2327, 2702, 3105, 3536, 3995, 4482, 4997, 5540, 6111, 6710, 7337, 7992, 8675, 9386, 10125, 10892, 11687, 12510, 13361, 14240, 15147, 16082, 17045, 18036, 19055, 20102, 21177, 22280, 23411, 24570

Offset: 0

Bruno Berselli, Oct 14 2011 - based on remarks and sequences by Omar E. Pol

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of initial terms: An origin of A185019.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A013656, A033429, A049452, A049453, A147296, A195025.
Cf. A000384, A195023, A195024, A135703.
Cf. A195020 (vertices of the numerical spiral in link).

Magma
```
[n*(14*n-3): n in [0..42]];
```

I:=[0,11,50]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013

CoefficientList[Series[x (11 + 17 x)/(1 - x)^3, {x, 0, 45}], x] (* Vincenzo Librandi, Aug 18 2013 *)

PARI
```
for(n=0, 42, print1(n*(14*n-3)", "));
```

G.f.: x*(11+17*x)/(1-x)^3.
a(n) = A195025(-n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. - Wesley Ivan Hurt, Dec 18 2020
From Elmo R. Oliveira, Dec 30 2024: (Start)
E.g.f.: exp(x)*x*(11 + 14*x).
a(n) = n + A195023(n). (End)

A139273 a(n) = n(8n - 3).

Original entry on oeis.org

0, 5, 26, 63, 116, 185, 270, 371, 488, 621, 770, 935, 1116, 1313, 1526, 1755, 2000, 2261, 2538, 2831, 3140, 3465, 3806, 4163, 4536, 4925, 5330, 5751, 6188, 6641, 7110, 7595, 8096, 8613, 9146, 9695, 10260, 10841, 11438, 12051, 12680

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 5, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139277 in the same spiral.
Also, sequence of numbers of the form d*A000217(n-1) + 5*n with generating functions x*(5+(d-5)*x)/(1-x)^3; the inverse binomial transform is 0,5,d,0,0,.. (0 continued). See Crossrefs. - Bruno Berselli, Feb 11 2011
Even decagonal numbers divided by 2. - Omar E. Pol, Aug 19 2011

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734.

[ n*(8*n-3) : n in [0..40] ];  // Bruno Berselli, Feb 11 2011

Table[n (8 n - 3), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 26}, 40] (* Harvey P. Dale, Feb 02 2012 *)

a(n)=n*(8*n-3) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 8*n^2 - 3*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 11 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 11*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A051866(n). (End)
a(n) = A028994(n)/2. - Omar E. Pol, Aug 19 2011
a(0)=0, a(1)=5, a(2)=26; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Feb 02 2012
E.g.f.: (8*x^2 + 5*x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = 4*log(2)/3 - (sqrt(2)-1)*Pi/6 - sqrt(2)*arccoth(sqrt(2))/3. - Amiram Eldar, Jul 03 2020

A195023 a(n) = 14n^2 - 4n.

Original entry on oeis.org

0, 10, 48, 114, 208, 330, 480, 658, 864, 1098, 1360, 1650, 1968, 2314, 2688, 3090, 3520, 3978, 4464, 4978, 5520, 6090, 6688, 7314, 7968, 8650, 9360, 10098, 10864, 11658, 12480, 13330, 14208, 15114, 16048, 17010, 18000, 19018, 20064, 21138, 22240, 23370, 24528, 25714

Offset: 0

Omar E. Pol, Oct 13 2011

Sequence found by reading the line from 0, in the direction 0, 10, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the one of the semi-axis of the square spiral, which is related to the primitive Pythagorean triple [3, 4, 5].

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A135703, A144555, A152760, A193053, A195019, A195020, A195024, A195320, A198017.

[14*n^2 - 4*n: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

Table[14n^2-4n,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,10,48},50] (* Harvey P. Dale, Sep 05 2012 *)

a(n)=14*n^2-4*n \\ Charles R Greathouse IV, Apr 10 2012

a(n) = 2*A135703(n). - Bruno Berselli, Oct 13 2011
From Colin Barker, Apr 09 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: 2*x*(5+9*x)/(1-x)^3. (End)
E.g.f.: 2*exp(x)*x*(5 + 7*x). - Elmo R. Oliveira, Dec 30 2024

Corrected by Vincenzo Librandi, Oct 14 2011

A271937 a(n) = (7/4)n^2 + (5/2)n + (7 + (-1)^n)/8.

Original entry on oeis.org

1, 5, 13, 24, 39, 57, 79, 104, 133, 165, 201, 240, 283, 329, 379, 432, 489, 549, 613, 680, 751, 825, 903, 984, 1069, 1157, 1249, 1344, 1443, 1545, 1651, 1760, 1873, 1989, 2109, 2232, 2359, 2489, 2623, 2760, 2901, 3045, 3193, 3344, 3499, 3657, 3819, 3984, 4153

Offset: 0

Vincenzo Librandi, Apr 20 2016

Let P be a polygon with vertices (0,0), (0,2), (1,1) and (0,3/2). The number of integer points in nP is counted by this quasi-polynomial (nP is the n-fold dilation of P). See Wikipedia in Links section.
From Bob Selcoe, Sep 10 2016: (Start)
a(n) = the number of partitions in reverse lexicographic order starting with n 3's followed by n 2's; i.e., the number of partitions summing to 5n such that no part > 3 and the number of 3's digits <= the number of 2's digits.
First differences are A047346(n+1); second differences are 4 when n is even and 3 when n is odd (i.e., A010702(n+1)); third differences are 1 when n is even and -1 when n is odd. (End)

			a(1) = 5; the 5 partitions are: {3,2}; {3,1,1}; {2,2,1}; {2,1,1,1}; {1,1,1,1,1}.
a(3) = 24: floor(8/2) + floor(11/2) + floor(14/2) + floor(17/2) = 4+5+7+8 = 24.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Examples of Ehrhart Quasi-Polynomials.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

First bisection (after 1) is A168235.
Second bisection is A135703 (without 0).
Cf. A006578, A047346.

[(7/4)*n^2+(5/2)*n+(7+(-1)^n)/8: n in [0..50]];

Table[(7/4) n^2 + (5/2) n + (7 + (-1)^n)/8, {n, 0, 50}]
LinearRecurrence[{2,0,-2,1},{1,5,13,24},50] (* Harvey P. Dale, Mar 23 2025 *)

Vec((1+3*x+3*x^2)/((1-x)^3*(1+x)) + O(x^99)) \\ Altug Alkan, Sep 10 2016

O.g.f.: (1 + 3*x + 3*x^2)/((1 - x)^3*(1 + x)).
E.g.f.: (7 + 34*x + 14*x^2)*exp(x)/8 + exp(-x)/8.
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
a(2*k) = k*(7*k + 5) + 1, a(2*k+1) = (k + 1)*(7*k + 5).
From Bob Selcoe, Sep 10 2016 (Start):
a(n) = (n+1)^2 + A006578(n).
a(n) = a(n-1) + A047346(n+1).
a(n) = Sum_{j=0..n} floor((2n+3j+2)/2).
(End)

Edited and extended by Bruno Berselli, Apr 20 2016

cp's OEIS Frontend

A045944 Rhombic matchstick numbers: a(n) = n*(3*n+2).

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A014106 a(n) = n*(2*n + 3).

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A033429 a(n) = 5*n^2.

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A140090 a(n) = n*(3*n + 7)/2.

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A022267 a(n) = n*(9*n + 1)/2.

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A185019 a(n) = n*(14*n-3).

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A139273 a(n) = n*(8*n - 3).

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A045944 Rhombic matchstick numbers: a(n) = n(3n+2).

A014106 a(n) = n(2n + 3).

A140090 a(n) = n(3n + 7)/2.

A022267 a(n) = n(9n + 1)/2.

A185019 a(n) = n(14n-3).

A139273 a(n) = n(8n - 3).

A195023 a(n) = 14n^2 - 4n.

A271937 a(n) = (7/4)n^2 + (5/2)n + (7 + (-1)^n)/8.