A005449 -id:A005449 - cp's OEIS frontend

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

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

A126890 Triangle read by rows: T(n,k) = n(n+2k+1)/2, 0 <= k <= n.

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 6, 9, 12, 15, 10, 14, 18, 22, 26, 15, 20, 25, 30, 35, 40, 21, 27, 33, 39, 45, 51, 57, 28, 35, 42, 49, 56, 63, 70, 77, 36, 44, 52, 60, 68, 76, 84, 92, 100, 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, 55, 65, 75, 85, 95, 105, 115, 125, 135, 145, 155, 66, 77, 88

Offset: 0

Reinhard Zumkeller, Dec 30 2006

T(n,k) + T(n,n-k) = A014105(n);
row sums give A059270; Sum_{k=0..n-1} T(n,k) = A000578(n);
central terms give A007742; T(2*n+1,n) = A016754(n);
T(n,0)   = A000217(n);
T(n,1)   = A000096(n)   for n > 0;
T(n,2)   = A055998(n)   for n > 1;
T(n,3)   = A055999(n)   for n > 2;
T(n,4)   = A056000(n)   for n > 3;
T(n,5)   = A056115(n)   for n > 4;
T(n,6)   = A056119(n)   for n > 5;
T(n,7)   = A056121(n)   for n > 6;
T(n,8)   = A056126(n)   for n > 7;
T(n,10)  = A101859(n-1) for n > 9;
T(n,n-3) = A095794(n-1) for n > 2;
T(n,n-2) = A045943(n-1) for n > 1;
T(n,n-1) = A000326(n)   for n > 0;
T(n,n)   = A005449(n).

			From _Philippe Deléham_, Oct 03 2011: (Start)
Triangle begins:
   0;
   1,  2;
   3,  5,  7;
   6,  9, 12, 15;
  10, 14, 18, 22, 26;
  15, 20, 25, 30, 35, 40;
  21, 27, 33, 39, 45, 51, 57;
  28, 35, 42, 49, 56, 63, 70, 77; (End)

Léonard Euler, Introduction à l'analyse infinitésimale, tome premier, ACL-Editions, Paris, 1987, p. 353-354.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Émile Fourrey, Les nombres abstraits, Récreations arithmétiques, 1899 and later, Vuibert, Paris, page 86-87. Triangle without right diagonal.
Adrien-Marie Legendre, Théorie des nombres, tome 2, quatrième partie, p.131, troisième édition, Paris, 1830.

Cf. A110449.

a126890 n k = a126890_tabl !! n !! k
a126890_row n = a126890_tabl !! n
a126890_tabl = map fst $ iterate
   (\(xs@(x:_), i) -> (zipWith (+) ((x-i):xs) [2*i+1 ..], i+1)) ([0], 0)
-- Reinhard Zumkeller, Nov 10 2013

Flatten[Table[(n(n+2k+1))/2,{n,0,20},{k,0,n}]] (* Harvey P. Dale, Jun 21 2013 *)

T(n,k) = T(n,k-1) + n, for k <= n. - Philippe Deléham, Oct 03 2011

A130518 a(n) = Sum_{k=0..n} floor(k/3). (Partial sums of A002264.)

Original entry on oeis.org

0, 0, 0, 1, 2, 3, 5, 7, 9, 12, 15, 18, 22, 26, 30, 35, 40, 45, 51, 57, 63, 70, 77, 84, 92, 100, 108, 117, 126, 135, 145, 155, 165, 176, 187, 198, 210, 222, 234, 247, 260, 273, 287, 301, 315, 330, 345, 360, 376, 392, 408, 425, 442, 459, 477, 495, 513, 532, 551, 570

Offset: 0

Hieronymus Fischer, Jun 01 2007

Complementary with A130481 regarding triangular numbers, in that A130481(n) + 3*a(n) = n(n+1)/2 = A000217(n).
Apart from offset, the same as A062781. - R. J. Mathar, Jun 13 2008
Apart from offset, the same as A001840. - Michael Somos, Sep 18 2010
The sum of any three consecutive terms is a triangular number. - J. M. Bergot, Nov 27 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Cf. A002265, A002266, A004526, A010872, A010873, A010874, A062781, A130482, A130483.

List([0..60], n-> Int(n*(n-1)/6)); # G. C. Greubel, Aug 31 2019

[Round(n*(n-1)/6): n in [0..60]]; // Vincenzo Librandi, Jun 25 2011

seq(floor(n*(n-1)/6),n=0..60); # Robert Israel, Nov 27 2014

Table[n, {n, 0, 19}, {3}] // Flatten // Accumulate (* Jean-François Alcover, Jun 05 2013 *)

a(n)=n*(n-1)\/6 \\ Charles R Greathouse IV, Jun 05 2013

[floor(binomial(n,2)/3) for n in range(0,60)] # Zerinvary Lajos, Dec 01 2009

G.f.: x^3 / ((1-x^3)*(1-x)^2).
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).
a(n) = (1/2)*floor(n/3)*(2*n - 1 - 3*floor(n/3)) = A002264(n)*(2n - 1 - 3*A002264(n))/2.
a(n) = (1/2)*A002264(n)*(n - 1 + A010872(n)).
a(n) = round(n*(n-1)/6) = round((n^2-n-1)/6) = floor(n*(n-1)/6) = ceiling((n+1)*(n-2)/6). - Mircea Merca, Nov 28 2010
a(n) = a(n-3) + n - 2, n > 2. - Mircea Merca, Nov 28 2010
a(n) = A214734(n, 1, 3). - Renzo Benedetti, Aug 27 2012
a(3n) = A000326(n), a(3n+1) = A005449(n), a(3n+2) = 3*A000217(n) = A045943(n). - Philippe Deléham, Mar 26 2013
a(n) = (3*n*(n-1) - (-1)^n*((1+i*sqrt(3))^(n-2) + (1-i*sqrt(3))^(n-2))/2^(n-3) - 2)/18, where i=sqrt(-1). - Bruno Berselli, Nov 30 2014
Sum_{n>=3} 1/a(n) = 20/3 - 2*Pi/sqrt(3). - Amiram Eldar, Sep 17 2022

A059845 a(n) = n(3n + 11)/2.

Original entry on oeis.org

0, 7, 17, 30, 46, 65, 87, 112, 140, 171, 205, 242, 282, 325, 371, 420, 472, 527, 585, 646, 710, 777, 847, 920, 996, 1075, 1157, 1242, 1330, 1421, 1515, 1612, 1712, 1815, 1921, 2030, 2142, 2257, 2375, 2496, 2620, 2747, 2877, 3010, 3146, 3285, 3427, 3572, 3720

Offset: 0

Jason Earls, Mar 10 2001

Maximum dimension of Euclidean spaces which suffice for every smooth compact Riemannian n-manifold to be realizable as a sub-manifold. - comment edited by Gene Ward Smith, Jan 15 2017

Harry J. Smith, Table of n, a(n) for n = 0..2000
Sela Fried, Counting r X s rectangles in nondecreasing and Smirnov words, arXiv:2406.18923 [math.CO], 2024. See p. 5.
John Nash, The Imbedding Problem For Riemannian Manifolds, Annals of Mathematics, Vol. 63, No. 1, 1956, pp. 20-63.
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, A059845, A115067, A140090, A140091, A140672, A140673, A140674, A140675, A151542.

A059845:=n->n*(3*n + 11)/2: seq(A059845(n), n=0..100); # Wesley Ivan Hurt, Jan 15 2017

Table[n (3n+11)/2,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,7,17},50] (* Harvey P. Dale, Mar 19 2017 *)

a(n) = n*(3*n + 11)/2 \\ Harry J. Smith, Jun 29 2009

a(n) = 3*n + a(n-1) + 4 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010
G.f.: x*(7 - 4*x)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 24 2011
E.g.f.: (1/2)*(3*x^2 + 14*x)*exp(x). - G. C. Greubel, Jul 17 2017

A229079 Number A(n,k) of ascending runs in {1,...,k}^n; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 7, 3, 0, 0, 4, 15, 20, 4, 0, 0, 5, 26, 63, 52, 5, 0, 0, 6, 40, 144, 243, 128, 6, 0, 0, 7, 57, 275, 736, 891, 304, 7, 0, 0, 8, 77, 468, 1750, 3584, 3159, 704, 8, 0, 0, 9, 100, 735, 3564, 10625, 16896, 10935, 1600, 9, 0

Offset: 0

Alois P. Heinz, Sep 14 2013

			A(4,1) = 4: [1,1,1,1].
A(3,2) = 20 = 3+3+2+3+2+2+2+3: [1,1,1], [2,1,1], [1,2,1], [2,2,1], [1,1,2], [2,1,2], [1,2,2], [2,2,2].
A(2,3) = 15 = 2+2+2+1+2+2+1+1+2: [1,1], [2,1], [3,1], [1,2], [2,2], [3,2], [1,3], [2,3], [3,3].
A(1,4) = 4 = 1+1+1+1: [1], [2], [3], [4].
Square array A(n,k) begins:
  0, 0,   0,     0,     0,      0,       0,       0, ...
  0, 1,   2,     3,     4,      5,       6,       7, ...
  0, 2,   7,    15,    26,     40,      57,      77, ...
  0, 3,  20,    63,   144,    275,     468,     735, ...
  0, 4,  52,   243,   736,   1750,    3564,    6517, ...
  0, 5, 128,   891,  3584,  10625,   25920,   55223, ...
  0, 6, 304,  3159, 16896,  62500,  182736,  453789, ...
  0, 7, 704, 10935, 77824, 359375, 1259712, 3647119, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Columns k=0-10 give: A000004, A001477, A066373(n+1) for n>0, A229277, A229278, A229279, A229280, A229281, A229282, A229283, A229284.
Rows n=0-10 give: A000004, A001477, A005449, A099721, A229146, A229147, A229148, A229149, A229150, A229151, A229152.
Main diagonal gives A229078.

A:= (n, k)-> `if`(n=0, 0, k^(n-1)*((n+1)*k+n-1)/2):
seq(seq(A(n,d-n), n=0..d), d=0..12);

a[, 0] = a[0, ] = 0; a[n_, k_] := k^(n-1)*((n+1)*k+n-1)/2; Table[a[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 09 2013 *)

A(n,k) = k^(n-1)*((n+1)*k+n-1)/2 for n>0, A(0,k) = 0.

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

A080663 a(n) = 3*n^2 - 1.

Original entry on oeis.org

2, 11, 26, 47, 74, 107, 146, 191, 242, 299, 362, 431, 506, 587, 674, 767, 866, 971, 1082, 1199, 1322, 1451, 1586, 1727, 1874, 2027, 2186, 2351, 2522, 2699, 2882, 3071, 3266, 3467, 3674, 3887, 4106, 4331, 4562, 4799, 5042, 5291, 5546, 5807, 6074, 6347, 6626

Offset: 1

Cino Hilliard, Mar 01 2003

These numbers cannot be perfect squares. See the Hilliard link for a proof.
2nd elementary symmetric polynomial of n, n + 1 and n + 2: n(n+1) + n(n+2) + (n+1)(n+2). - Zak Seidov, Mar 23 2005
This sequence equals for n >= 2 the third right hand column of triangle A165674. Its recurrence relation leads to Pascal's triangle A007318. Crowley's formula for A080663(n-1) leads to Wiggen's triangle A028421 and the o.g.f. of this sequence, without the first term, leads to Wood's polynomials A126671. See also A165676, A165677, A165678 and A165679. - Johannes W. Meijer, Oct 16 2009
The Diophantine equation x(x+1) + (x+2)(x+3) = (x+y)^2 + (x-y)^2 has solutions x = a(n), y = 3n. - Bruno Berselli, Mar 29 2013
A simpler proof that these numbers can't be perfect squares can easily be constructed using congruences: If the equation x^2 = 3y^2 - 1 has a solution in positive integers, then x^2 = 2 mod 3. Obviously we can't have x = 0 mod 3, and x = 1 mod 3 doesn't work either because then x^2 = 1 mod 3 also. That leaves x = 2 mod 3, but then x^2 = 1 mod 3. - Alonso del Arte, Oct 19 2013
2*a(n+1) is surface area of a rectangular prism with consecutive integer sides: n, n+1, and n+2, (n>0). - Wesley Ivan Hurt, Sep 06 2014
Numbers m such that 3*m+3 is a square. So, these are the numbers m such that the system of equations x=sqrt(m-2yz), y=sqrt(m+1-2xz), z=sqrt(m+2-2xy) admits 3 real positive solutions whose sum is an integer. See the Rechtman link. - Michel Marcus, Jun 06 2020

Ethan D. Bolker, Elementary Number Theory: An Algebraic Approach. Mineola, New York: Dover Publications (1969, reprinted 2007): p. 7, Problem 6.6.
E. Grosswald, Topics from the Theory of Numbers, 1966 p 64 problem 11

Nathaniel Johnston, Table of n, a(n) for n = 1..10000
Cino Hilliard, 3n^2 - 1 not square. [Archived copy as of Apr 11 2008 from web.archive.org]
Ana Rechtman, Juin 2020, 1er défi, Images des Mathématiques, CNRS, 2020 (in French).
Leo Tavares, Illustration: Conjoined Trapezoids
Eric Weisstein's World of Mathematics, Symmetric Polynomial.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A007318, A028421, A080663, A121670, A126671, A165674, A165676, A165677, A165678, A165679.

Cf. A005449, A115067.

[3*n^2-1 : n in [1..50]]; // Wesley Ivan Hurt, Sep 04 2014

A080663 := proc(n) return 3*n^2-1: end proc: seq(A080663(n), n=1..50); # Nathaniel Johnston, Oct 16 2013

3*Range[47]^2 - 1 (* Alonso del Arte, Oct 19 2013 *)

list(n) = { for(x=1,n, y = 3*x*x-1; print1(y, ", ") ) } \\ edited by Michel Marcus, Feb 01 2020

Vec(x*(2+5*x-x^2)/(1-x)^3+O(x^66)) \\ Joerg Arndt, Sep 06 2014

a(n) = -Re((1 + n*i)^3) where i=sqrt(-1). - Gary W. Adamson, Aug 14 2006
a(n) = 3*n^2 - 1. - Stephen Crowley, Jul 06 2009
a(n) = a(n-1) - 3*a(n-2) + 3*a(n-3). - Johannes W. Meijer, Oct 16 2009
G.f.: x*(2 + 5*x - x^2)/(1-x)^3. - Joerg Arndt, Sep 06 2014
a(n) = a(n-1) + 6*n - 3 for n > 1. - Vincenzo Librandi, Aug 08 2010
E.g.f.: 1 + exp(x)*(3*x^2 + 3*x - 1). - Stefano Spezia, Feb 01 2020
From Amiram Eldar, Feb 04 2021: (Start)
Sum_{n>=1} 1/a(n) = (1 - (Pi/sqrt(3))*cot(Pi/sqrt(3)))/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = ((Pi/sqrt(3))*csc(Pi/sqrt(3)) - 1)/2.
Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(3))*csc(Pi/sqrt(3)).
Product_{n>=1} (1 - 1/a(n)) = csc(Pi/sqrt(3))*sin(sqrt(2/3)*Pi)/sqrt(2). (End)
a(n) = A005449(n) + A115067(n). - Leo Tavares, May 25 2022
a(n) = (n-1)*n + (n-1)*(n+1) + n*(n+1), for n >= 1. See the Zak Seidov comment above. - Wolfdieter Lang, Aug 15 2024

A179986 Second 9-gonal (or nonagonal) numbers: a(n) = n(7n+5)/2.

Original entry on oeis.org

0, 6, 19, 39, 66, 100, 141, 189, 244, 306, 375, 451, 534, 624, 721, 825, 936, 1054, 1179, 1311, 1450, 1596, 1749, 1909, 2076, 2250, 2431, 2619, 2814, 3016, 3225, 3441, 3664, 3894, 4131, 4375, 4626, 4884, 5149, 5421, 5700, 5986, 6279, 6579, 6886

Offset: 0

Bruno Berselli, Jan 13 2011

This sequence is a bisection of A118277 (even part).
Sequence found by reading the line from 0, in the direction 0, 19... and the line from 6, in the direction 6, 39,..., in the square spiral whose vertices are the generalized 9-gonal numbers A118277. - Omar E. Pol, Jul 24 2012
The early part of this sequence is a strikingly close approximation to the early part of A100752. - Peter Munn, Nov 14 2019

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

Cf. second k-gonal numbers: A005449 (k=5), A014105 (k=6), A147875 (k=7), A045944 (k=8), this sequence (k=9), A033954 (k=10), A062728 (k=11), A135705 (k=12).

Cf. A001106, A022264, A022265, A024966, A055998, A100752, A174738, A186029, A218471.

[n*(7*n+5)/2: n in [0..50]]; // Bruno Berselli, Sep 23 2016

I:=[0, 6, 19]; [n le 3 select I[n] else 3*Self(n-1) -3*Self(n-2) +Self(n-3): n in [1..60]]; // Vincenzo Librandi, Oct 15 2012

f[n_] := n (7 n + 5)/2; f[Range[0, 60]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011*)
LinearRecurrence[{3, -3, 1}, {0, 6, 19}, 60] (* or *) Array[(#(7# + 5))/2&, 60, 0] (* Harvey P. Dale, Aug 19 2011 *)
CoefficientList[Series[x (6 + x)/(1 - x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, Oct 15 2012 *)

a(n)=n*(7*n+5)/2 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: x*(6 + x)/(1 - x)^3.
a(n) = Sum_{i=0..(n-1)} A017053(i) for n>0.
a(-n) = A001106(n).
Sum_{i=0..n} (a(n)+i)^2 = ( Sum_{i=(n+1)..2*n} (a(n)+i)^2 ) + 21*A000217(n)^2 for n>0.
a(n) = a(n-1)+7*n-1 for n>0, with a(0)=0. - Vincenzo Librandi, Feb 05 2011
a(0)=0, a(1)=6, a(2)=19; for n>2, a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). - Harvey P. Dale, Aug 19 2011
a(n) = A174738(7n+5). - Philippe Deléham, Mar 26 2013
a(n) = A001477(n) + 2*A000290(n) + 3*A000217(n). - J. M. Bergot, Apr 25 2014
a(n) = A055998(4*n) - A055998(3*n). - Bruno Berselli, Sep 23 2016
E.g.f.: (x/2)*(12 + 7*x)*exp(x). - G. C. Greubel, Aug 19 2017

A242249 Number A(n,k) of rooted trees with n nodes and k-colored non-root nodes; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 2, 2, 0, 0, 1, 3, 7, 4, 0, 0, 1, 4, 15, 26, 9, 0, 0, 1, 5, 26, 82, 107, 20, 0, 0, 1, 6, 40, 188, 495, 458, 48, 0, 0, 1, 7, 57, 360, 1499, 3144, 2058, 115, 0, 0, 1, 8, 77, 614, 3570, 12628, 20875, 9498, 286, 0, 0, 1, 9, 100, 966, 7284, 37476, 111064, 142773, 44947, 719, 0

Offset: 0

Alois P. Heinz, May 09 2014

From Vaclav Kotesovec, Aug 26 2014: (Start)
Column k > 0 is asymptotic to c(k) * d(k)^n / n^(3/2), where constants c(k) and d(k) are dependent only on k. Conjecture: d(k) ~ k * exp(1). Numerically:
d(1)   =   2.9557652856519949747148175... (A051491)
d(2)   =   5.6465426162329497128927135... (A245870)
d(3)   =   8.3560268792959953682760695...
d(4)   =  11.0699628777593263124193026...
d(5)   =  13.7856511100846851989303249...
d(6)   =  16.5022088446930015657112211...
d(7)   =  19.2192613290638657575973462...
d(8)   =  21.9366222112987115910888213...
d(9)   =  24.6541883249893084812976812...
d(10)  =  27.3718979186642404090999595...
d(100) = 272.0126359583480733207362718...
d(101) = 274.7309127032967881125015217...
d(200) = 543.8405620978790523837823296...
d(201) = 546.5588426492458787468860222...
d(101)-d(100) = 2.718276744...
d(201)-d(200) = 2.718280551...
(End)

			Square array A(n,k) begins:
  0,  0,    0,     0,      0,      0,       0,       0, ...
  1,  1,    1,     1,      1,      1,       1,       1, ...
  0,  1,    2,     3,      4,      5,       6,       7, ...
  0,  2,    7,    15,     26,     40,      57,      77, ...
  0,  4,   26,    82,    188,    360,     614,     966, ...
  0,  9,  107,   495,   1499,   3570,    7284,   13342, ...
  0, 20,  458,  3144,  12628,  37476,   91566,  195384, ...
  0, 48, 2058, 20875, 111064, 410490, 1200705, 2984142, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
L. Foissy, Algebraic structures on typed decorated rooted trees, arXiv:1811.07572 (2018), doi SIGMA 17 (2021)

Columns k=0-10 give: A063524, A000081, A000151, A006964, A052763, A052788, A246235, A246236, A246237, A246238, A246239.
Rows n=0-3 give: A000004, A000012, A001477, A005449.
Lower diagonal gives A242375.
Cf. A255517, A256064.

with(numtheory):
A:= proc(n, k) option remember; `if`(n<2, n, (add(add(d*
      A(d, k), d=divisors(j))*A(n-j, k)*k, j=1..n-1))/(n-1))
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);

nn = 10; t[x_] := Sum[a[n] x^n, {n, 1, nn}]; Transpose[ Table[Flatten[ sol = SolveAlways[ 0 == Series[ t[x] - x Product[1/(1 - x^i)^(n a[i]), {i, 1, nn}], {x, 0, nn}], x]; Flatten[{0, Table[a[n], {n, 1, nn}]}] /. sol], {n, 0, nn}]] // Grid (* Geoffrey Critzer, Nov 11 2014 *)
A[n_, k_] := A[n, k] = If[n<2, n, Sum[Sum[d*A[d, k], {d, Divisors[j]}] *A[n-j, k]*k, {j, 1, n-1}]/(n-1)]; Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Dec 04 2014, translated from Maple *)

\\ ColGf gives column generating function
ColGf(N,k) = {my(A=vector(N, j, 1)); for (n=1, N-1, A[n+1] = k/n * sum(i=1, n, sumdiv(i, d, d*A[d]) * A[n-i+1] ) ); x*Ser(A)}
Mat(vector(8, k, concat(0, Col(ColGf(7, k-1))))) \\ Andrew Howroyd, May 12 2018

G.f. for column k: x*Product_{n>=1} 1/(1 - x^n)^(k*A(n,k)). - Geoffrey Critzer, Nov 13 2014

A141419 Triangle read by rows: T(n, k) = A000217(n) - A000217(n - k) with 1 <= k <= n.

Original entry on oeis.org

1, 2, 3, 3, 5, 6, 4, 7, 9, 10, 5, 9, 12, 14, 15, 6, 11, 15, 18, 20, 21, 7, 13, 18, 22, 25, 27, 28, 8, 15, 21, 26, 30, 33, 35, 36, 9, 17, 24, 30, 35, 39, 42, 44, 45, 10, 19, 27, 34, 40, 45, 49, 52, 54, 55

Offset: 1

Roger L. Bagula, Aug 05 2008

As a rectangle, the accumulation array of A051340.
From Clark Kimberling, Feb 05 2011: (Start)
Here all the weights are divided by two where they aren't in Cahn.
As a rectangle, A141419 is in the accumulation chain
... < A051340 < A141419 < A185874 < A185875 < A185876 < ...
(See A144112 for the definition of accumulation array.)
row 1: A000027
col 1: A000217
diag (1,5,...): A000326 (pentagonal numbers)
diag (2,7,...): A005449 (second pentagonal numbers)
diag (3,9,...): A045943 (triangular matchstick numbers)
diag (4,11,...): A115067
diag (5,13,...): A140090
diag (6,15,...): A140091
diag (7,17,...): A059845
diag (8,19,...): A140672
(End)
Let N=2*n+1 and k=1,2,...,n. Let A_{N,n-1} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], an n X n unit-primitive matrix (see [Jeffery]). Let M_n=[A_{N,n-1}]^4. Then t(n,k)=[M_n](1,k), that is, the n-th row of the triangle is given by the first row of M_n. - _L. Edson Jeffery, Nov 20 2011
Conjecture. Let N=2*n+1 and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (again see [Jeffery]) associated with N, and define the Chebyshev polynomials of the second kind by the recurrence U_0(x) = 1, U_1(x) = 2*x and U_r(x) = 2*x*U_(r-1)(x) - U_(r-2)(x)  (r>1). Define the column vectors V_(k-1) = (U_(k-1)(cos(Pi/N)), U_(k-1)(cos(3*Pi/N)), ..., U_(k-1)(cos((2*n-1)*Pi/N)))^T, where T denotes matrix transpose. Let S_N = [V_0, V_1, ..., V_(n-1)] be the n X n matrix formed by taking V_(k-1) as column k-1. Let X_N = [S_N]^T*S_N, and let [X_N](i,j) denote the entry in row i and column j of X_N, i,j in {0,...,n-1}. Then t(n,k) = [X_N](k-1,k-1), and row n of the triangle is given by the main diagonal entries of X_N. Remarks: Hence t(n,k) is the sum of squares t(n,k) = sum[m=1,...,n (U_(k-1)(cos((2*m-1)*Pi/N)))^2]. Finally, this sequence is related to A057059, since X_N = [sum_{m=1,...,n} A057059(n,m)*A_{N,m-1}] is also an integral linear combination of unit-primitive matrices from the N-th set. - L. Edson Jeffery, Jan 20 2012
Row sums: n*(n+1)*(2*n+1)/6. - L. Edson Jeffery, Jan 25 2013
n-th row = partial sums of n-th row of A004736. - Reinhard Zumkeller, Aug 04 2014
T(n,k) is the number of distinct sums made by at most k elements in {1, 2, ... n}, for 1 <= k <= n, e.g., T(6,2) = the number of distinct sums made by at most 2 elements in {1,2,3,4,5,6}. The sums range from 1, to 5+6=11. So there are 11 distinct sums. - Derek Orr, Nov 26 2014
A number n occurs in this sequence A001227(n) times, the number of odd divisors of n, see A209260. - Hartmut F. W. Hoft, Apr 14 2016
Conjecture: 2*n + 1 is composite if and only if gcd(t(n,m),m) != 1, for some m. - L. Edson Jeffery, Jan 30 2018
From Peter Munn, Aug 21 2019 in respect of the sequence read as a triangle: (Start)
A number m can be found in column k if and only if A286013(m, k) is nonzero, in which case m occurs in column k on row A286013(m, k).
The first occurrence of m is in row A212652(m) column A109814(m), which is the rightmost column in which m occurs. This occurrence determines where m appears in A209260. The last occurrence of m is in row m column 1.
Viewed as a sequence of rows, consider the subsequences (of rows) that contain every positive integer. The lexicographically latest of these subsequences consists of the rows with row numbers in A270877; this is the only one that contains its own row numbers only once.
(End)

			As a triangle:
   1,
   2,  3,
   3,  5,  6,
   4,  7,  9, 10,
   5,  9, 12, 14, 15,
   6, 11, 15, 18, 20, 21,
   7, 13, 18, 22, 25, 27, 28,
   8, 15, 21, 26, 30, 33, 35, 36,
   9, 17, 24, 30, 35, 39, 42, 44, 45,
  10, 19, 27, 34, 40, 45, 49, 52, 54, 55;
As a rectangle:
   1   2   3   4   5   6   7   8   9  10
   3   5   7   9  11  13  15  17  19  21
   6   9  12  15  18  21  24  27  30  33
  10  14  18  22  26  30  34  38  42  46
  15  20  25  30  35  40  45  50  55  60
  21  27  33  39  45  51  57  63  69  75
  28  35  42  49  56  63  70  77  84  91
  36  44  52  60  68  76  84  92 100 108
  45  54  63  72  81  90  99 108 117 126
  55  65  75  85  95 105 115 125 135 145
Since the odd divisors of 15 are 1, 3, 5 and 15, number 15 appears four times in the triangle at t(3+(5-1)/2, 5) in column 5 since 5+1 <= 2*3, t(5+(3-1)/2, 3), t(1+(15-1)/2, 2*1) in column 2 since 15+1 > 2*1, and t(15+(1-1)/2, 1). - _Hartmut F. W. Hoft_, Apr 14 2016

R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 139.

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened
Johann Cigler, Some elementary observations on Narayana polynomials and related topics, arXiv:1611.05252 [math.CO], 2016. See p. 24.
Carlton Gamer, David W. Roeder, and John J. Watkins, Trapezoidal Numbers, Mathematics Magazine 58:2 (1985), pp. 108-110.
L. E. Jeffery, Unit-primitive matrices
M. A. Nyblom, On the representation of the integers as a difference of nonconsecutive triangular numbers, Fibonacci Quarterly 39:3 (2001), pp. 256-263.

Cf. A000330 (row sums), A004736, A057059, A070543.
A144112, A051340, A141419, A185874, A185875, A185876 are accumulation chain related.
A141418 is a variant.
Cf. A001227, A209260. - Hartmut F. W. Hoft, Apr 14 2016
A109814, A212652, A270877, A286013 relate to where each natural number appears in this sequence.
A000027, A000217, A000326, A005449, A045943, A059845, A115067, A140090, A140091, A140672 are rows, columns or diagonals - refer to comments.

a141419 n k =  k * (2 * n - k + 1) `div` 2
a141419_row n = a141419_tabl !! (n-1)
a141419_tabl = map (scanl1 (+)) a004736_tabl
-- Reinhard Zumkeller, Aug 04 2014

a:=(n,k)->k*n-binomial(k,2): seq(seq(a(n,k),k=1..n),n=1..12); # Muniru A Asiru, Oct 14 2018

T[n_, m_] = m*(2*n - m + 1)/2; a = Table[Table[T[n, m], {m, 1, n}], {n, 1, 10}]; Flatten[a]

t(n,m) = m*(2*n - m + 1)/2.
t(n,m) = A000217(n) - A000217(n-m). - L. Edson Jeffery, Jan 16 2013
Let v = d*h with h odd be an integer factorization, then v = t(d+(h-1)/2, h) if h+1 <= 2*d, and v = t(d+(h-1)/2, 2*d) if h+1 > 2*d; see A209260. - Hartmut F. W. Hoft, Apr 14 2016
G.f.: y*(-x + y)/((-1 + x)^2*(-1 + y)^3). - Stefano Spezia, Oct 14 2018
T(n, 2) = A060747(n) for n > 1. T(n, 3) = A008585(n - 1) for n > 2. T(n, 4) = A016825(n - 2) for n > 3. T(n, 5) = A008587(n - 2) for n > 4. T(n, 6) = A016945(n - 3) for n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7.r n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7. T(n, 9) = A008591(n - 4) for n > 8. T(n, 10) = A017329(n - 5) for n > 9. T(n, 11) = A008593(n - 5) for n > 10.  T(n, 12) = A017593(n - 6) for n > 11. T(n, 13) = A008595(n - 6) for n > 12. T(n, 14) = A147587(n - 7) for n > 13. T(n, 15) = A008597(n - 7) for n > 14. T(n, 16) = A051062(n - 8) for n > 15. T(n, 17) = A008599(n - 8) for n > 16. - Stefano Spezia, Oct 14 2018
T(2*n-k, k) = A070543(n, k). - Peter Munn, Aug 21 2019

Simpler name by Stefano Spezia, Oct 14 2018

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Magma

Maple

Mathematica

PARI

Sage

Formula

A126890 Triangle read by rows: T(n,k) = n*(n+2*k+1)/2, 0 <= k <= n.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Mathematica

Formula

A130518 a(n) = Sum_{k=0..n} floor(k/3). (Partial sums of A002264.)

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

A059845 a(n) = n*(3*n + 11)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A229079 Number A(n,k) of ascending runs in {1,...,k}^n; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

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

A126890 Triangle read by rows: T(n,k) = n(n+2k+1)/2, 0 <= k <= n.

A059845 a(n) = n(3n + 11)/2.

A179986 Second 9-gonal (or nonagonal) numbers: a(n) = n(7n+5)/2.