A049450 -id:A049450 - cp's OEIS frontend

A049452 Pentagonal numbers with even index.

0, 5, 22, 51, 92, 145, 210, 287, 376, 477, 590, 715, 852, 1001, 1162, 1335, 1520, 1717, 1926, 2147, 2380, 2625, 2882, 3151, 3432, 3725, 4030, 4347, 4676, 5017, 5370, 5735, 6112, 6501, 6902, 7315, 7740, 8177, 8626, 9087, 9560, 10045, 10542

Offset: 0

Joe Keane (jgk(AT)jgk.org)

If Y is a 3-subset of an (2n+1)-set X then, for n>=4, a(n-1) is the number of 4-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 16 2007
Sequence found by reading the line (one of the diagonal axes) from 0, in the direction 0, 5,..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - Omar E. Pol, Sep 08 2011
a(n) is the sum of 2*n consecutive integers starting from 2*n. - Bruno Berselli, Jan 16 2018

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

Cf. A000326, A033570, A049453, A001318, A033568, A185019.

See index to sequences with numbers of the form n*(d*n+10-d)/2 in A140090.

seq(n*(6*n-1),n=0..42); # Zerinvary Lajos, Jun 12 2007

Table[n(6n-1),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,5,22},50] (* Harvey P. Dale, Mar 07 2012 *)

a(n)=n*(6*n-1) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = n*(6*n-1).
G.f.: x*(5+7*x)/(1-x)^3.
a(n) = C(6*n,2)/3. - Zerinvary Lajos, Jan 02 2007
a(n) = A001105(n) + A033991(n) = A033428(n) + A049450(n) = A022266(n) + A000326(n). - Zerinvary Lajos, Jun 12 2007
a(n) = 12*n + a(n-1) - 7 for n>0, a(0)=0. - Vincenzo Librandi, Aug 06 2010
a(n) = 4*A000217(n) + A001107(n). - Bruno Berselli, Feb 11 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2, a(0)=0, a(1)=5, a(2)=22. - Harvey P. Dale, Mar 07 2012
E.g.f.: (6*x^2 + 5*x)*exp(x). - G. C. Greubel, Jul 17 2017
From Amiram Eldar, Jul 03 2020: (Start)
Sum_{n>=1} 1/a(n) = 2*log(2) + 3*log(3)/2 - sqrt(3)*Pi/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi - log(2) - 2*sqrt(3)*arccoth(sqrt(3)). (End)

A001859 Triangular numbers plus quarter-squares: n*(n+1)/2 + floor((n+1)^2/4) (i.e., A000217(n) + A002620(n+1)).

Original entry on oeis.org

0, 2, 5, 10, 16, 24, 33, 44, 56, 70, 85, 102, 120, 140, 161, 184, 208, 234, 261, 290, 320, 352, 385, 420, 456, 494, 533, 574, 616, 660, 705, 752, 800, 850, 901, 954, 1008, 1064, 1121, 1180, 1240, 1302, 1365, 1430, 1496, 1564, 1633, 1704, 1776, 1850, 1925

Offset: 0

N. J. A. Sloane

Number of series-reduced planted trees with n+7 nodes and 3 internal nodes.
The trees enumerated with 3 internal nodes are of two types. Those with all internal nodes at different heights are enumerated by the triangular numbers. Those with two internal nodes at the same height are enumerated by the quarter squares. - Michael Somos, May 19 2000
Number of pairs (x,y) with x in {0,...,n}, y even in {0,...,2n}, and x < y. - Clark Kimberling, Jul 02 2012

			For n=1 we find 2 planted trees with 8 nodes, 3 of which are internal (i) and 5 are endpoints (e):
.e...e...e...e....e...e....
...i.......i........i...e..
.......i..............i...e
.......e................i..
........................e..
G.f. = 2*x + 5*x^2 + 10*x^3 + 16*x^4 + 24*x^5 + 33*x^6 + 44*x^7 + 56*x^8 + ...

John Riordan, personal communication.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
D. D. Frey and J. A. Sellers, Jacobsthal Numbers and Alternating Sign Matrices, J. Integer Seqs., Vol. 3 (2000), #00.2.3
S. V. Gervacio and H. Maehara, Partial order on a family of k-subsets of a linearly ordered set, Discr. Math., 306 (2006), 413-419.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
J. Riordan, Letter to N. J. A. Sloane, Oct. 1970
S. G. Wagner, An identity for the cycle indices of rooted tree automorphism groups, Elec. J. Combinat., 13 (2006), #R00.
Index entries for sequences related to rooted trees.
Index entries for sequences related to trees.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A000217, A002378, A002620, A004526, A006578, A045944, A049450.
First differences of A045947.
Antidiagonal sums of array A003984.
Cf. A107661, A077043.
Cf. A185212 (odd terms).

a001859 n = a000217 n + a002620 (n + 1)  -- Reinhard Zumkeller, Dec 20 2012

A001859:=(-1-z^2-2*z^3+z^4)/(z+1)/(z-1)^3; # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence with an additional leading 1
with (combinat):seq(count(Partition((3*n+2)), size=3), n=0..50); # Zerinvary Lajos, Mar 28 2008

With[{nn=60},Total/@Thread[{Accumulate[Range[0,nn]],Floor[Range[ nn+1]^2/4]}]] (* or *) LinearRecurrence[{2,0,-2,1},{0,2,5,10},60] (* Harvey P. Dale, Apr 01 2012 *)

PARI
```
{a(n) = n + (3*n^2 + 1) \ 4};
```

a(n) = A000217(n)+A002620(n+1).
a(n) = n + floor( (3n^2+1)/4 ).
G.f.: (2*x+x^2)/((1-x)^2*(1-x^2)).
a(n) = a(n-1) + a(n-2) - a(n-3) + 3 = A002378(n) - A002620(n) = A006578(n-1) + A004526(n+1) - Henry Bottomley, Mar 08 2000
a(n) = A006578(-1-n) for all n in Z. - Michael Somos, May 10 2006
From Mitch Harris, Aug 22 2006: (Start)
a(n) = (6n^2 + 8n + 1 - (-1)^n)/8;
a(n) = Sum_{k=0..n} max(k, n-k). (End)
Starting (2, 5, 10, 16, 24, ...), = binomial transform of [2, 3, 2, -1, 2, -4, 8, -16, 32, ...]. - Gary W. Adamson, Nov 30 2007
a(0)=0, a(1)=2, a(2)=5, a(3)=10, a(n) = 2*a(n-1) + 0*a(n-2) - 2*a(n-3) + a(n-4). - Harvey P. Dale, Apr 01 2012
a(n) = 3*n*(n+1)/2 - A006578(n). - Clark Kimberling, Jul 02 2012
a(2*n) = A045944(n), a(2*n - 1) = A049450(n) for all n in Z. - Michael Somos, Nov 03 2014
0 = -6 + a(n) - 2*a(n+2) + a(n+4) for all n in Z. - Michael Somos, Nov 03 2014
0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-3 - a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Nov 03 2014
a(n) = Sum_{k=1..n} floor((n+k+2)/2). - Wesley Ivan Hurt, Mar 31 2017
Sum_{n>=1} 1/a(n) = 3/4 - Pi/(4*sqrt(3)) + 3*log(3)/4. - Amiram Eldar, May 28 2022
E.g.f.: (x*(7 + 3*x)*cosh(x) + (1 + 7*x + 3*x^2)*sinh(x))/4. - Stefano Spezia, Aug 22 2023

Entry improved by Michael Somos

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

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

Subsequence of A062717: A010052(6*a(n)+1) = 1. - Reinhard Zumkeller, Feb 21 2011

Sequence found by reading the line from 0, in the direction 0, 4, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - Omar E. Pol, Sep 08 2011

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

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

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

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

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

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

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

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

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

More terms from Michel Marcus, Mar 04 2014

A033586 a(n) = 4n(2*n + 1).

Original entry on oeis.org

0, 12, 40, 84, 144, 220, 312, 420, 544, 684, 840, 1012, 1200, 1404, 1624, 1860, 2112, 2380, 2664, 2964, 3280, 3612, 3960, 4324, 4704, 5100, 5512, 5940, 6384, 6844, 7320, 7812, 8320, 8844, 9384, 9940, 10512, 11100, 11704, 12324, 12960, 13612, 14280

Offset: 0

N. J. A. Sloane

Number of possible king moves on an (n+1) X (n+1) chessboard. E.g., for a 3 X 3 board: king has 4*5 moves, 4*3 moves and 1*8 moves, so a(2)=40. - Ulrich Schimke (ulrschimke(AT)aol.com)
Sequence found by reading the line from 0, in the direction 0, 12, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A085250 in the same spiral. - Omar E. Pol, Sep 03 2011
Sum of the numbers from 3n to 5n. - Wesley Ivan Hurt, Dec 22 2015
From Emeric Deutsch, Nov 09 2016: (Start)
a(n) is the second Zagreb index of the friendship graph F[n]. The second Zagreb index of a simple connected graph is the sum of the degree products d(i)d(j) over all edges ij of the graph. The friendship graph (or Dutch windmill graph) F[n] can be constructed by joining n copies of the cycle graph C[3] with a common vertex.
For instance, a(2)=40. Indeed, the friendship graph F[2] has 2 edges with end-point degrees 2,2 and 4 edges with end-point degrees 2,4. Then the second Zagreb index is 2*4 + 4*8 = 40. (End)
a(n) is the number of vertices in conjoined n X n dodecagons which are arranged into a square array, a.k.a. 3-4-3-12 tiling. - Donghwi Park, Dec 20 2020

E. Bonsdorff, K. Fabel and O. Riihimaa, Schach und Zahl (Chess and numbers), Walter Rau Verlag, Dusseldorf, 1966.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Tim Krabbe, Open Chess Diary, see item 221
Wikipedia, Friendship graph
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A035005 (Queen), A035006 (Rook), A035008 (Knight), A002492 (Bishop) and A049450 (Pawn).

Cf. A000217, A005408, A008586, A014105, A085250.

[4*n*(2*n + 1) : n in [0..50]]; // Wesley Ivan Hurt, Dec 22 2015

A033586:=n->4*n*(2*n+1); seq(A033586(n), n=0..60); # Wesley Ivan Hurt, Feb 25 2014

Table[4n*(2n + 1), {n, 0, 60}] (* Stefan Steinerberger, Apr 08 2006 *)
LinearRecurrence[{3,-3,1},{0,12,40},60] (* Harvey P. Dale, May 19 2011 *)

a(n)=4*n*(2*n+1) \\ Charles R Greathouse IV, Jul 16 2011

Binomial transform of [12, 28, 16, 0, 0, 0, ...] = (12, 40, 84, 144, 220, ...). - Gary W. Adamson, Oct 24 2007
a(n) = 4 * A014105(n). - Johannes W. Meijer, Feb 04 2010
a(n) = 16*n + a(n-1) - 4 (with a(0)=0). - Vincenzo Librandi, Aug 05 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. - Harvey P. Dale, May 10 2011
G.f.: 4*x*(3+x)/(1-3*x+3*x^2-x^3). - Colin Barker, Jan 06 2012
From Wesley Ivan Hurt, Feb 25 2014, Dec 22 2015: (Start)
a(n) = A008586(n) * A005408(n).
a(n) = Sum_{i=3n..5n} i.
a(-n) = A085250(n). (End)
E.g.f.: (8*x^2 + 12*x)*exp(x). - G. C. Greubel, Jul 16 2017
From Vaclav Kotesovec, Dec 21 2020: (Start)
Sum_{n>=1} 1/a(n) = (1 - log(2))/2.
Sum_{n>=1} (-1)^n/a(n) = 1/2 - Pi/8 - log(2)/4. (End)

More terms from Erich Friedman

Crossref added, minor errors corrected and edited by Johannes W. Meijer, Feb 04 2010

A022264 a(n) = n(7n - 1)/2.

Original entry on oeis.org

0, 3, 13, 30, 54, 85, 123, 168, 220, 279, 345, 418, 498, 585, 679, 780, 888, 1003, 1125, 1254, 1390, 1533, 1683, 1840, 2004, 2175, 2353, 2538, 2730, 2929, 3135, 3348, 3568, 3795, 4029, 4270, 4518, 4773, 5035, 5304, 5580, 5863, 6153, 6450, 6754, 7065, 7383

Offset: 0

N. J. A. Sloane

Sequence found by reading the line from 0, in the direction 0, 13, ..., and the parallel line from 3, in the direction 3, 30, ..., in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020. - Omar E. Pol, Sep 09 2011

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

Cf. A000217, A022265.
Cf. sequences listed in A254963.
Cf. similar sequences listed in A022288.
Cf. A016742, A049450, A174738, A195019, A195020.
Cf. A005449, A000384.

[n*(7*n-1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Dec 04 2016

[seq(binomial(7*n,2)/7, n=0..37)]; # Zerinvary Lajos, Jan 02 2007

Table[n (7*n - 1)/2, {n, 0, 40}] (* Zerinvary Lajos, Jul 10 2009 *)

a(n)=n*(7*n-1)/2 \\ Charles R Greathouse IV, Mar 08 2013

a(n) = C(7*n,2)/7, n >= 0. - Zerinvary Lajos, Jan 02 2007
a(n) = A049450(n) + A000217(n). - Reinhard Zumkeller, Oct 09 2008
a(n) = 7*n + a(n-1) - 4 for n > 0, a(0)=0. - Vincenzo Librandi, Aug 04 2010
a(n) = (2*n)^2 - n*(n+1)/2 = A016742(n) - A000217(n). - Philippe Deléham, Mar 08 2013
a(n) = A174738(7*n+2). - Philippe Deléham, Mar 26 2013
G.f.: x*(3 + 4*x)/(1 - x)^3. - R. J. Mathar, Aug 04 2016
a(n) = A000217(4*n-1) - A000217(3*n-1). - Bruno Berselli, Oct 17 2016
a(n) = (1/5) * Sum_{i=n..(6*n-1)} i. - Wesley Ivan Hurt, Dec 04 2016
E.g.f.: (1/2)*x*(7*x + 6)*exp(x). - G. C. Greubel, Aug 19 2017
a(n) = A005449(n) + A000384(n). See Crysta-gons illustration. - Leo Tavares, Nov 21 2021

A226488 a(n) = n(13n - 9)/2.

Original entry on oeis.org

0, 2, 17, 45, 86, 140, 207, 287, 380, 486, 605, 737, 882, 1040, 1211, 1395, 1592, 1802, 2025, 2261, 2510, 2772, 3047, 3335, 3636, 3950, 4277, 4617, 4970, 5336, 5715, 6107, 6512, 6930, 7361, 7805, 8262, 8732, 9215, 9711, 10220, 10742, 11277, 11825, 12386, 12960

Offset: 0

Bruno Berselli, Jun 09 2013

Sum of n-th octagonal number and n-th 9-gonal (nonagonal) number.

Sum of reciprocals of a(n), for n>0: 0.629618994194109711163742089971688...

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

Cf. A000567, A001106, A153080 (first differences).

Cf. numbers of the form n*(n*k-k+4)/2 listed in A005843 (k=0), A000096 (k=1), A002378 (k=2), A005449 (k=3), A001105 (k=4), A005476 (k=5), A049450 (k=6), A218471 (k=7), A002939 (k=8), A062708 (k=9), A135706 (k=10), A180223 (k=11), A139267 (n=12), this sequence (k=13), A139268 (k=14), A226489 (k=15), A139271 (k=16), A180232 (k=17), A152995 (k=18), A226490 (k=19), A152965 (k=20), A226491 (k=21), A152997 (k=22).

List([0..50], n-> n*(13*n-9)/2); # G. C. Greubel, Aug 30 2019

Magma
```
[n*(13*n-9)/2: n in [0..50]];
```

I:=[0,2,17]; [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

A226488:=n->n*(13*n - 9)/2; seq(A226488(n), n=0..50); # Wesley Ivan Hurt, Feb 25 2014

Table[n(13n-9)/2, {n, 0, 50}]
LinearRecurrence[{3, -3, 1}, {0, 2, 17}, 50] (* Harvey P. Dale, Jun 19 2013 *)
CoefficientList[Series[x(2+11x)/(1-x)^3, {x, 0, 45}], x] (* Vincenzo Librandi, Aug 18 2013 *)

a(n)=n*(13*n-9)/2 \\ Charles R Greathouse IV, Sep 24 2015

[n*(13*n-9)/2 for n in (0..50)] # G. C. Greubel, Aug 30 2019

G.f.: x*(2+11*x)/(1-x)^3.
a(n) + a(-n) = A152742(n).
a(0)=0, a(1)=2, a(2)=17; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jun 19 2013
E.g.f.: x*(4 + 13*x)*exp(x)/2. - G. C. Greubel, Aug 30 2019
a(n) = A000567(n) + A001106(n). - Michel Marcus, Aug 31 2019

A007980 Expansion of (1+x^2)/((1-x)^2*(1-x^3)).

Original entry on oeis.org

1, 2, 4, 7, 10, 14, 19, 24, 30, 37, 44, 52, 61, 70, 80, 91, 102, 114, 127, 140, 154, 169, 184, 200, 217, 234, 252, 271, 290, 310, 331, 352, 374, 397, 420, 444, 469, 494, 520, 547, 574, 602, 631, 660, 690, 721, 752, 784, 817, 850, 884, 919, 954, 990, 1027, 1064

Offset: 0

N. J. A. Sloane

Molien series for ternary self-dual codes over GF(3) of length 12n containing 11...1.
(1+x)*(1+x^2) / ((1-x)*(1-x^2)*(1-x^3)) is the Poincaré series [or Poincare series] (or Molien series) for H^*(O_3(q); F_2).
a(n) is the position of the n-th triangular number in the running sum of the (pseudo-Orloj) sequence 1,2,1,2,1,2,1...., cf. A028355. - Wouter Meeussen, Mar 10 2002
a(n) = [a(n-1) + (number of even terms so far in the sequence)]. Example: 14 is [10 + 4 even terms so far in the sequence (they are 0,2,4,10)]. See A096777 for the same construction with odd integers. - Eric Angelini, Aug 05 2007
The number of partitions of 2*n into at most 3 parts. - Colin Barker, Mar 31 2015
Also a(n) equals the number of linearly-independent terms at 2n-th order in the power series expansion of a trigonal Rotational Energy Surface. An optimal basis for the expansion follows either decomposition: g1(x) = (1+x)(1+x^2)g2(x) or g1(x) = (1+x^2)x^(-1)g3(x), where g1(x), g2(x), g3(x) are the generating functions for sequences A007980, A001399, A001840. - Bradley Klee, Aug 06 2015
Also a(n) equals the number of linearly-independent terms at 4n-th order in the power series expansion of the symmetrized weight enumerator of a self-dual code of length n over Z4 that contains a vector (+/-)1^n and has all norms divisible by 8. An optimal basis for the expansion follows the decomposition: g1(x) = (1+x)(1+x^2)g2(x) where g1(x), g2(x) are the generating functions for sequences A007980, A001399. (Cf. Calderbank and Sloane, Corollary 5.) - Bradley Klee, Aug 06 2015
Also, a(n) is equal to the number of partitions of 2n+3 of length 3. Letting n=4, there are a(4)=10 partitions of 2n+3=11 of length 3: (9,1,1), (8,2,1), (7,3,1), (7,2,2), (6,4,1), (6,3,2), (5,5,1), (5,4,2), (5,3,3), (4,4,3). - John M. Campbell, Jan 30 2016
a(n) is the number of partitions of n into parts 1 (of two kinds), part 2 (occurring at most once), and parts 3. - Joerg Arndt, Oct 12 2020
Conjecture: a(n) is the maximum number of pieces a triangle can be cut into by n cevians. - Anton Zakharov, Apr 04 2017
Also, a(n) is the number of graphs which are double-triangle descendants of K_5 with n+6 triangles and 3 more vertices than triangles. See Laradji/Mishna/Yeats reference, proposition 3.6 for details. - Karen A. Yeats, Feb 21 2020

			G.f. = 1 + 2*x + 4*x^2 + 7*x^3 + 10*x^4 + 14*x^5 + 19*x^6 + 24*x^7 + ...

A. Adem and R. J. Milgram, Cohomology of Finite Groups, Springer-Verlag, 2nd. ed., 2004; p. 233.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Raghavendra N. Bhat, Cristian Cobeli, and Alexandru Zaharescu, A lozenge triangulation of the plane with integers, arXiv:2403.10500 [math.NT], 2024.
A. R. Calderbank and N. J. A. Sloane, Double circulant codes over Z_4, J. Algeb. Combin., 6 (1997) 119-131 (Abstract, pdf, ps).
Mohamed Laradji, Marni Mishna, and Karen Yeats, Some results on double triangle descendants of K_5, arXiv:1904.06923 [math.CO], 2019.
C. L. Mallows and N. J. A. Sloane, Weight enumerators of self-orthogonal codes over GF(3), SIAM J. Alg. Discrete Methods, 2 (1981), 452-460.
Paul Tabatabai and Dieter P. Gruber, Knights and Liars on Graphs, J. Int. Seq., Vol. 24 (2021), Article 21.5.8.
Anton Zakharov, cevians.
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).
Index entries for Molien series.
Index entries for two-way infinite sequences.

Cf. A000326, A001399, A001840, A002378, A003215, A004396, A005449, A007980.

Cf. A008796, A028355, A049450, A069905, A096777, A192736, A211540, A238705.

with (combinat):seq(count(Partition((2*n+1)), size=3), n=1..56); # Zerinvary Lajos, Mar 28 2008

Table[Ceiling[n (n+1)/3], {n, 56}]
CoefficientList[Series[(1+x^2)/((1-x)^2*(1-x^3)),{x,0,60}],x] (* Vincenzo Librandi, Feb 25 2012 *)
a[ n_] := Quotient[ n^2, 3] + n + 1; (* Michael Somos, Aug 23 2015 *)
LinearRecurrence[{2,-1,1,-2,1},{1,2,4,7,10},60] (* Harvey P. Dale, Aug 24 2016 *)

{a(n) = if( n<-1, a(-3-n), polcoeff( (1 + x^2) / ( (1 - x)^2 * (1 - x^3)) + x*O(x^n), n))}; /* Michael Somos, Jun 07 2003 */

{a(n) = n^2\3 + n+1}; /* Michael Somos, Aug 23 2015 */

a(n) = #partitions(2*n, ,[1,3]); \\ Michel Marcus, Feb 12 2016

a(n) = #partitions(2*n+3, ,[3,3]); \\ Michel Marcus, Feb 12 2016

G.f.: (1 + x^2) / ((1 - x)^2 * (1 - x^3)). - Michael Somos, Jun 07 2003
a(n) = a(n-1) + a(n-3) -a(n-4) + 2 = a(-3-n) for all n in Z. - Michael Somos, Jun 07 2003
a(n) = ceiling((n+1)*(n+2)/3). - Paul Boddington, Jan 26 2004
a(n) = A192736(n+1) / (n+1). - Reinhard Zumkeller, Jul 08 2011
From Bruno Berselli, Oct 22 2010: (Start)
a(n) = ((n+1)*(n+2)+(2*cos(2*Pi*n/3)+1)/3)/3 = Sum_{i=1..n+1} A004396(i).
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5) for n>4.
a(n) = A002378(n+1)/3 if 3 divides A002378(n+1), a(n) = (A002378(n)+1)/3 otherwise. (End)
a(n) = A001840(n+1) + A001840(n-1). - R. J. Mathar, Aug 23 2015
From Michael Somos, Aug 23 2015: (Start)
Euler transform of length 4 sequence [2, 1, 1, -1].
a(n) = A001399(2*n) = A008796(2*n) = A008796(2*n + 3) = A069905(2*n + 3) = A211540(2*n + 5).
a(2*n)     = A238705(n+1).
a(3*n - 1) = A049451(n).
a(3*n)     = A003215(n).
a(3*n + 1) = A049450(n+1).
2*a(3*n - 1)  = A005449(n).
2*a(3*n + 1)  = A000326(n+1).
a(n+1) - a(n) = A004396(n+2). (End)
a(n) = floor((n^2+3*n+3)/3). - Giacomo Guglieri, May 01 2019
a(n) = A000212(n) + n+1. - Yuchun Ji, Oct 12 2020
Sum_{n>=0} 1/a(n) = (tanh(Pi/(2*sqrt(3)))-1)*Pi/sqrt(3) + 3. - Amiram Eldar, May 20 2023

A199832 T(n,k)=Number of -k..k arrays x(0..n+1) of n+2 elements with zero sum and no two neighbors summing to zero.

Original entry on oeis.org

2, 10, 4, 24, 40, 4, 44, 140, 114, 10, 70, 336, 646, 426, 22, 102, 660, 2146, 3556, 1650, 34, 140, 1144, 5390, 15708, 20240, 6126, 66, 184, 1820, 11384, 49302, 118280, 113884, 23206, 138, 234, 2720, 21364, 124982, 462234, 888420, 645780, 88636, 250, 290, 3876

Offset: 1

R. H. Hardin Nov 11 2011

Table starts
...2......10........24.........44..........70..........102...........140
...4......40.......140........336.........660.........1144..........1820
...4.....114.......646.......2146........5390........11384.........21364
..10.....426......3556......15708.......49302.......124982........273728
..22....1650.....20240.....118280......462234......1402934.......3579520
..34....6126....113884.....888420.....4340094.....15805218......47040968
..66...23206....645780....6715618....41008804....179213048.....622300326
.138...88636...3685550...51077518...389832124...2044221894....8281149188
.250..337866..21117750..390278378..3723199342..23427591518..110718596524
.472.1295566.121503530.2993722414.35697026718.269528370904.1486040082748

			Some solutions for n=4 k=3
..3....2...-2...-3...-3....3...-1...-1....0...-1....3...-1....2...-3....1....3
..3....2....0....0....1...-2....0....2....2....2....3....0...-1...-1....3....1
.-2....0....1....1....2...-2....1....2....1....0....0....1...-3....2....2...-2
.-2...-1...-2....0....2....0...-3....2....1...-2...-2...-2...-1...-1...-3...-3
..1...-1....0....3....1....3....1...-3...-2...-1...-1....0....3....2...-1....2
.-3...-2....3...-1...-3...-2....2...-2...-2....2...-3....2....0....1...-2...-1

R. H. Hardin, Table of n, a(n) for n = 1..788

Row 1 is A049450

Empirical for rows:
T(1,k) = 3*k^2 - k
T(2,k) = (16/3)*k^3 - (4/3)*k
T(3,k) = (115/12)*k^4 - (29/6)*k^3 + (5/12)*k^2 - (7/6)*k
T(4,k) = (88/5)*k^5 - (28/3)*k^4 + (2/3)*k^3 + (7/3)*k^2 - (19/15)*k
T(5,k) = (5887/180)*k^6 - (1013/60)*k^5 + (245/36)*k^4 - (35/12)*k^3 + (157/45)*k^2 - (6/5)*k
T(6,k) = (19328/315)*k^7 - (1424/45)*k^6 + (704/45)*k^5 - (112/9)*k^4 - (124/45)*k^3 + (229/45)*k^2 - (131/105)*k
T(7,k) = (259723/2240)*k^8 - (299869/5040)*k^7 + (39757/1440)*k^6 - (8303/360)*k^5 + (31829/2880)*k^4 - (8083/720)*k^3 + (32213/5040)*k^2 - (509/420)*k
T(8,k) = (124952/567)*k^9 - (35524/315)*k^8 + (50588/945)*k^7 - (2494/45)*k^6 + (13739/270)*k^5 - (1927/180)*k^4 - (41254/2835)*k^3 + (3319/420)*k^2 - (781/630)*k

A152749 a(n) = (n+1)(3n+1)/4 for n odd, a(n) = n(3n+2)/4 for n even.

Original entry on oeis.org

0, 2, 4, 10, 14, 24, 30, 44, 52, 70, 80, 102, 114, 140, 154, 184, 200, 234, 252, 290, 310, 352, 374, 420, 444, 494, 520, 574, 602, 660, 690, 752, 784, 850, 884, 954, 990, 1064, 1102, 1180, 1220, 1302, 1344, 1430, 1474, 1564, 1610, 1704, 1752, 1850, 1900, 2002

Offset: 0

Vincenzo Librandi, Dec 31 2009

Interleaving of A049450 and A049451 (for n > 0).
Also, integer values of k*(k+1)/3. - Charles R Greathouse IV, Dec 11 2010
The nonzero coefficients of the expansion of f(a) = Product_{k>=1} (1-a^(2k)), see A194159, occur at the terms of the sequence given above, i.e., f(a) = 1 - a^2 - a^4 + a^10 + a^14 - a^24 - a^30 + a^44 + a^52 - a^70 - a^80 + ... = Sum_{n>=0} (-1)^binomial(n+1,2)*a^A152749(n). - Johannes W. Meijer, Aug 21 2011
Partial sums of A109043. - Reinhard Zumkeller, Mar 31 2012
Nonnegative k such that 12*k+1 is a square. - Vicente Izquierdo Gomez, Jul 22 2013
Equivalently, numbers of the form h*(3*h+1), where h = 0, -1, 1, -2, 2, -3, 3, -4, 4, ... (see also the fifth comment of A062717). - Bruno Berselli, Feb 02 2017
For n > 0, a(n-1) is the sum of the largest parts of the partitions of 2n into two even parts. - Wesley Ivan Hurt, Dec 19 2017
The sequence terms occur as exponents in the expansion of Sum_{n >= 0} q^(n*(n+1)/2) * Product_{k >= n+1} 1 - q^k = 1 - q^2 - q^4 + q^10 + q^14 - q^24 - q^30 + + - -  .... - Peter Bala, Dec 15 2024
Sequence terms occur as exponents in the expansions of Sum_{n >= 0} q^(n*(2*n+1)) * Product_{k >= 2*n+2} 1 - q^k  = Sum_{n >= 0} q^(n*(2*n-1)) * Product_{k >= 2*n+1} 1 - q^k = 1 - q^2 - q^4 + q^10 + q^14 - q^24 - q^30 + + - - .... - Peter Bala, Jun 23 2025

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

Cf. A049450 (n*(3*n-1)), A049451 (n*(3*n+1)), A153383 (12n+1 is not prime).

a152749 n = a152749_list !! (n-1)
a152749_list = scanl1 (+) a109043_list
-- Reinhard Zumkeller, Mar 31 2012

[IsOdd(n) select (n+1)*(3*n+1)/4 else n*(3*n+2)/4: n in [0..52]];

f:=func; [0] cat [f(n*m): m in [-1,1], n in [1..30]]; // Bruno Berselli, Nov 13 2012

A152749 := proc(n): if type(n,even) then n*(3*n+2)/4  else (n+1)*(3*n+1)/4 fi: end: seq(A152749(n), n=0..51); # Johannes W. Meijer, Aug 21 2011

Table[If[OddQ[n],(n+1)*(3*n+1)/4,n*(3*n+2)/4],{n,0,60}] (* Vladimir Joseph Stephan Orlovsky, Feb 03 2012 *)
LinearRecurrence[{1,2,-2,-1,1}, {0, 2, 4, 10, 14}, 50] (* Vincenzo Librandi, Feb 22 2012 *)
Select[Range[1,1000], IntegerQ[Sqrt[12#+1]]&] (* Vicente Izquierdo Gomez, Jul 22 2013 *)

From R. J. Mathar, Jan 03-06 2009: (Start)
G.f.: 2*x*(1+x+x^2)/((1+x)^2*(1-x)^3).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) = A003154(n+1)/8 - (-1)^n*A005408(n)/8.
a(n) = 2*A001318(n) = ((6*n^2+6*n+1) - (2*n+1)*(-1)^n)/8. (End)
From Amiram Eldar, Mar 15 2022: (Start)
Sum_{n>=1} 1/a(n) = 3 - Pi/sqrt(3).
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*(log(3)-1). (End)

Edited, typo corrected and extended by Klaus Brockhaus, Jan 02 2009

Leading term a(0)=0 added by Johannes W. Meijer, Aug 21 2011

A014642 Even octagonal numbers: a(n) = 4n(3*n-1).

Original entry on oeis.org

0, 8, 40, 96, 176, 280, 408, 560, 736, 936, 1160, 1408, 1680, 1976, 2296, 2640, 3008, 3400, 3816, 4256, 4720, 5208, 5720, 6256, 6816, 7400, 8008, 8640, 9296, 9976, 10680, 11408, 12160, 12936, 13736, 14560, 15408, 16280, 17176, 18096, 19040, 20008, 21000, 22016

Offset: 0

Mohammad K. Azarian

8 times pentagonal numbers. - Omar E. Pol, Dec 11 2008
Sequence found by reading the line from 0, in the direction 0, 8, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012
The sequence forms the even nesting cube-frames (see illustrations in A000567), which separate and appear according to formula along the axes on the zero-centered and one-centered hexagonal number spirals, as well as the axes of the zero-centered and one-centered square number spirals. See illustrations in links. - John Elias, Jul 20 2022

G. C. Greubel, Table of n, a(n) for n = 0..1000
John Elias, Octagonal Nesting Cubes on the Hexagonal Number Spiral Octagonal Nesting Cubes on the Square Number Spiral
Craig Knecht, Number of positions the remaining tiles can occupy in a 4*n length polyiamond bilayer when one tile is missing.
Yaohui Zhu, Kaiming Sun, Zhengdong Luo, and Lingfeng Wang, Progressive Self-Learning for Domain Adaptation on Symbolic Regression of Integer Sequences, Proc. 39th AAAI Conf. Artif. Intel. (2025) Vol. 39, No. 1, 1692-1699. See p. 1698.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567, A000326, A001082, A014641, A014793, A014794, A033579, A049450.

List([0..50], n-> 8*Binomial(3*n,2)/3); # G. C. Greubel, Oct 09 2019

[8*Binomial(3*n,2)/3: n in [0..50]]; // G. C. Greubel, Oct 09 2019

seq(8*binomial(3*n,2)/3, n=0..50); # G. C. Greubel, Oct 09 2019

LinearRecurrence[{3,-3,1},{0,8,40}, 50] (* G. C. Greubel, Jun 07 2017 *)
PolygonalNumber[8,Range[0,90,2]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Aug 19 2020 *)

vector(51, n, 8*binomial(3*(n-1),2)/3 ) \\ G. C. Greubel, Jun 07 2017

[8*binomial(3*n,2)/3 for n in (0..50)] # G. C. Greubel, Oct 09 2019

a(n) = A000326(n)*8. - Omar E. Pol, Dec 11 2008
a(n) = A049450(n)*4 = A033579(n)*2. - Omar E. Pol, Dec 13 2008
a(n) = a(n-1) + 24*n - 16 (with a(0)=0). - Vincenzo Librandi, Nov 20 2010
G.f.: x*(8+16*x)/(1-3*x+3*x^2-x^3). - Colin Barker, Jan 06 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - G. C. Greubel, Jun 07 2017
E.g.f.: 4*x*(2 + 3*x)*exp(x). - G. C. Greubel, Oct 09 2019
From Amiram Eldar, Mar 24 2021: (Start)
Sum_{n>=1} 1/a(n) = 3*log(3)/8 - Pi/(8*sqrt(3)).
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/2 - Pi/(4*sqrt(3)). (End)

More terms from Patrick De Geest

cp's OEIS Frontend

A049452 Pentagonal numbers with even index.

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A001859 Triangular numbers plus quarter-squares: n*(n+1)/2 + floor((n+1)^2/4) (i.e., A000217(n) + A002620(n+1)).

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

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A033586 a(n) = 4*n*(2*n + 1).

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A022264 a(n) = n*(7*n - 1)/2.

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A226488 a(n) = n*(13*n - 9)/2.

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

A033586 a(n) = 4n(2*n + 1).

A022264 a(n) = n(7n - 1)/2.

A226488 a(n) = n(13n - 9)/2.

A152749 a(n) = (n+1)(3n+1)/4 for n odd, a(n) = n(3n+2)/4 for n even.

A014642 Even octagonal numbers: a(n) = 4n(3*n-1).