A147875 -id:A147875 - cp's OEIS frontend

A062728 Second 11-gonal (or hendecagonal) numbers: a(n) = n(9n+7)/2.

0, 8, 25, 51, 86, 130, 183, 245, 316, 396, 485, 583, 690, 806, 931, 1065, 1208, 1360, 1521, 1691, 1870, 2058, 2255, 2461, 2676, 2900, 3133, 3375, 3626, 3886, 4155, 4433, 4720, 5016, 5321, 5635, 5958, 6290, 6631, 6981, 7340, 7708, 8085, 8471, 8866, 9270

Offset: 0

Floor van Lamoen, Jul 21 2001

Old name: 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 0,8,...

Sequence found by reading the line from 0, in the direction 0, 25, ... and the line from 8, in the direction 8, 51, ..., in the square spiral whose vertices are the generalized 11-gonal numbers A195160. - Omar E. Pol, Jul 24 2012

			The spiral begins:
          15
          / \
        16  14
        /     \
      17   3  13
      /   / \   \
    18   4   2  12
    /   /     \   \
  19   5   0---1  11
  /   /             \
20   6---7---8---9--10

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

Cf. A051682.

Second n-gonal numbers: A005449, A014105, A147875, A045944, A179986, A033954, this sequence, A135705.

List([0..50], n-> n*(9*n+7)/2); # G. C. Greubel, May 24 2019

[n*(9*n+7)/2: n in [0..50]]; // G. C. Greubel, May 24 2019

Table[n*(9*n+7)/2, {n,0,50}] (* G. C. Greubel, May 24 2019 *)
LinearRecurrence[{3,-3,1},{0,8,25},50] (* Harvey P. Dale, Sep 06 2019 *)

a(n)=n*(9*n+7)/2 \\ Charles R Greathouse IV, Jun 17 2017

[n*(9*n+7)/2 for n in (0..50)] # G. C. Greubel, May 24 2019

a(n) = n*(9*n+7)/2.
a(n) = 9*n + a(n-1) - 1 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010
From Bruno Berselli, Jan 13 2011: (Start)
G.f.: x*(8 + x)/(1 - x)^3.
a(n) = Sum_{i=0..n-1} A017257(i) for n > 0. (End)
a(n) = A218470(9n+7). - Philippe Deléham, Mar 27 2013
E.g.f.: x*(16 + 9*x)*exp(x)/2. - G. C. Greubel, May 24 2019

New name from Bruno Berselli (with the original formula), Jan 13 2011

A254963 a(n) = n(11n + 3)/2.

Original entry on oeis.org

0, 7, 25, 54, 94, 145, 207, 280, 364, 459, 565, 682, 810, 949, 1099, 1260, 1432, 1615, 1809, 2014, 2230, 2457, 2695, 2944, 3204, 3475, 3757, 4050, 4354, 4669, 4995, 5332, 5680, 6039, 6409, 6790, 7182, 7585, 7999, 8424, 8860, 9307, 9765, 10234, 10714, 11205, 11707

Offset: 0

Bruno Berselli, Feb 11 2015

This sequence provides the first differences of A254407 and the partial sums of A017473.
Also:
a(n) - n         = A022269(n);
a(n) + n         = n*(11*n+5)/2: 0, 8, 27, 57, 98, 150, 213, 287, ...;
a(n) - 2*n       = A022268(n);
a(n) + 2*n       = n*(11*n+7)/2: 0, 9, 29, 60, 102, 155, 219, 294, ...;
a(n) - 3*n       = n*(11*n-3)/2: 0, 4, 19, 45, 82, 130, 189, 259, ...;
a(n) + 3*n       = A211013(n);
a(n) - 4*n       = A226492(n);
a(n) + 4*n       = A152740(n);
a(n) - 5*n       = A180223(n);
a(n) + 5*n       = n*(11*n+13)/2: 0, 12, 35, 69, 114, 170, 237, 315, ...;
a(n) - 6*n       = A051865(n);
a(n) + 6*n       = n*(11*n+15)/2: 0, 13, 37, 72, 118, 175, 243, 322, ...;
a(n) - 7*n       = A152740(n-1) with A152740(-1) = 0;
a(n) + 7*n       = n*(11*n+17)/2: 0, 14, 39, 75, 122, 180, 249, 329, ...;
a(n) - n*(n-1)/2 = A168668(n);
a(n) + n*(n-1)/2 = A049453(n);
a(n) - n*(n+1)/2 = A202803(n);
a(n) + n*(n+1)/2 = A033580(n).

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

Cf. A008729 and A218530 (seventh column); A017473, A254407.
Cf. similar sequences of the type 4*n^2 + k*n*(n+1)/2: A055999 (k=-7, n>6), A028552 (k=-6, n>2), A095794 (k=-5, n>1), A046092 (k=-4, n>0), A000566 (k=-3), A049450 (k=-2), A022264 (k=-1), A016742 (k=0), A022267 (k=1), A202803 (k=2), this sequence (k=3), A033580 (k=4).
Cf. A069125: (2*n+1)^2 + 3*n*(n+1)/2; A147875: n^2 + 3*n*(n+1)/2.
Cf. A022268, A022269, A049453, A051865, A152740, A168668, A180223, A211013, A226492.

Magma
```
[n*(11*n+3)/2: n in [0..50]];
```

Table[n (11 n + 3)/2, {n, 0, 50}]
LinearRecurrence[{3,-3,1},{0,7,25},50] (* Harvey P. Dale, Mar 25 2018 *)

Maxima
```
makelist(n*(11*n+3)/2, n, 0, 50);
```
PARI
```
vector(50, n, n--; n*(11*n+3)/2)
```
Sage
```
[n*(11*n+3)/2 for n in (0..50)]
```

G.f.: x*(7 + 4*x)/(1 - x)^3.
From Elmo R. Oliveira, Dec 15 2024: (Start)
E.g.f.: exp(x)*x*(14 + 11*x)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A135705 a(n) = 10binomial(n,2) + 9n.

Original entry on oeis.org

0, 9, 28, 57, 96, 145, 204, 273, 352, 441, 540, 649, 768, 897, 1036, 1185, 1344, 1513, 1692, 1881, 2080, 2289, 2508, 2737, 2976, 3225, 3484, 3753, 4032, 4321, 4620, 4929, 5248, 5577, 5916, 6265, 6624, 6993, 7372, 7761, 8160, 8569, 8988, 9417, 9856, 10305, 10764

Offset: 0

N. J. A. Sloane, Mar 04 2008

Also, second 12-gonal (or dodecagonal) numbers. Identity for the numbers b(n)=n*(h*n+h-2)/2 (see Crossrefs): Sum_{i=0..n} (b(n)+i)^2 = (Sum_{i=n+1..2*n} (b(n)+i)^2) + h*(h-4)*A000217(n)^2 for n>0. - Bruno Berselli, Jan 15 2011
Sequence found by reading the line from 0, in the direction 0, 28, ..., and the line from 9, in the direction 9, 57, ..., in the square spiral whose vertices are the generalized 12-gonal numbers A195162. - Omar E. Pol, Jul 24 2012
Bisection of A195162. - Omar E. Pol, Aug 04 2012

Ivan Panchenko, Table of n, a(n) for n = 0..1000
L. Hogben, Choice and Chance by Cardpack and Chessboard, Vol. 1, Max Parrish and Co, London, 1950, p. 36.
Leo Tavares, Illustration: Diamond Clipped Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Second n-gonal numbers: A005449, A014105, A147875, A045944, A179986, A033954, A062728, this sequence.
Cf. A003154, A000290.
Cf. A195162.

List([0..50], n-> n*(5*n+4)); # G. C. Greubel, Jul 04 2019

[n*(5*n+4): n in [0..50]]; // G. C. Greubel, Jul 04 2019

LinearRecurrence[{3,-3,1}, {0,9,28}, 50] (* or *) Table[5*n^2 + 4*n, {n,0,50}] (* G. C. Greubel, Oct 29 2016 *)
Table[10 Binomial[n,2]+9n,{n,0,60}] (* Harvey P. Dale, Jun 14 2023 *)

a(n) = 10*binomial(n,2) + 9*n \\ Charles R Greathouse IV, Jun 11 2015

[n*(5*n+4) for n in (0..50)] # G. C. Greubel, Jul 04 2019

From R. J. Mathar, Mar 06 2008: (Start)
O.g.f.: x*(9+x)/(1-x)^3.
a(n) = n*(5*n+4). (End)
a(n) = a(n-1) + 10*n - 1 (with a(0)=0). - Vincenzo Librandi, Nov 24 2009
a(n) = Sum_{i=0..n-1} A017377(i) for n>0. - Bruno Berselli, Jan 15 2011
a(n) = A131242(10n+8). - Philippe Deléham, Mar 27 2013
Sum_{n>=1} 1/a(n) = 5/16 + sqrt(1 + 2/sqrt(5))*Pi/8 - 5*log(5)/16 - sqrt(5)*log((1 + sqrt(5))/2)/8 = 0.2155517745488486003038... . - Vaclav Kotesovec, Apr 27 2016
From G. C. Greubel, Oct 29 2016: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
E.g.f.: x*(9 + 5*x)*exp(x). (End)
a(n) = A003154(n+1) - A000290(n+1). - Leo Tavares, Mar 29 2022

A008732 Molien series for 3-dimensional group [2,n] = *22n.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 9, 11, 13, 15, 18, 21, 24, 27, 30, 34, 38, 42, 46, 50, 55, 60, 65, 70, 75, 81, 87, 93, 99, 105, 112, 119, 126, 133, 140, 148, 156, 164, 172, 180, 189, 198, 207, 216, 225, 235, 245, 255, 265

Offset: 0

N. J. A. Sloane

			From _Philippe Deléham_, Apr 05 2013: (Start)
Stored in five columns:
    1   2   3   4   5
    7   9  11  13  15
   18  21  24  27  30
   34  38  42  46  50
   55  60  65  70  75
   81  87  93  99 105
  112 119 126 133 140
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 188
Brian O'Sullivan and Thomas Busch, Spontaneous emission in ultra-cold spin-polarised anisotropic Fermi seas, arXiv 0810.0231v1 [quant-ph], 2008. [Eq 8a, lambda=5]
Index entries for Molien series
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,1,-2,1).

Cf. A130520.

List([0..50], n-> Int((n+3)*(n+4)/10)); # G. C. Greubel, Jul 30 2019

[Floor((n+3)*(n+4)/10): n in [0..50] ]; // Vincenzo Librandi, Aug 21 2011

A092202 := proc(n) op(1+(n mod 5),[0,1,0,-1,0]) ; end proc:
A010891 := proc(n) op(1+(n mod 5),[1,-1,0,0,0]) ; end proc:
A008732 := proc(n) (n+2)*(n+5)/10+(A010891(n-1)+2*A092202(n-1))/5 ; end proc:

LinearRecurrence[{2, -1, 0, 0, 1, -2, 1}, {1, 2, 3, 4, 5, 7, 9}, 50] (* Jean-François Alcover, Jan 18 2018 *)

a(n)=(n+3)*(n+4)\10 \\ Charles R Greathouse IV, Oct 07 2015

[floor((n+3)*(n+4)/10) for n in (0..50)] # G. C. Greubel, Jul 30 2019

a(n) = floor( (n+3)*(n+4)/10 ) = (n+2)*(n+5)/10 + b(n)/5 where b(n) = A010891(n-2) + 2*A092202(n-1) = 0, 1, 1, 0, -2, ... with period length 5.
G.f.: 1/((1-x)^2*(1-x^5)).
a(n) = a(n-5) + n + 1. - Paul Barry, Jul 14 2004
From Mitch Harris, Sep 08 2008: (Start)
a(n) = Sum_{j=0..n+5} floor(j/5).
a(n-5) = (1/2)floor(n/5)*(2*n - 3 - 5*floor(n/5)). (End)
a(n) = A130520(n+5). - Philippe Deléham, Apr 05 2013
a(5n) = A000566(n+1), a(5n+1) = A005476(n+1), a(5n+2) = A005475(n+1), a(5n+3) = A147875(n+2), a(5n+4) = A028895(n+1); these formulas correspond to the 5 columns of the array shown in example. - Philippe Deléham, Apr 05 2013

A211013 Second 13-gonal numbers: a(n) = n(11n+9)/2.

Original entry on oeis.org

0, 10, 31, 63, 106, 160, 225, 301, 388, 486, 595, 715, 846, 988, 1141, 1305, 1480, 1666, 1863, 2071, 2290, 2520, 2761, 3013, 3276, 3550, 3835, 4131, 4438, 4756, 5085, 5425, 5776, 6138, 6511, 6895, 7290, 7696, 8113, 8541, 8980, 9430, 9891, 10363

Offset: 0

Omar E. Pol, Aug 04 2012

Sequence found by reading the line from 0, in the direction 0, 31... and the line from 10, in the direction 10, 63,..., in the square spiral whose vertices are the generalized 13-gonal numbers A195313.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A195313.
Second k-gonal numbers (k=5..14): A005449, A014105, A147875, A045944, A179986, A033954, A062728, A135705, this sequence, A211014.
Cf. A051865.

List([0..50], n-> n*(11*n+9)/2); # G. C. Greubel, Jul 04 2019

[n*(11*n+9)/2: n in [0..50]]; // G. C. Greubel, Jul 04 2019

Table[n*(11*n+9)/2, {n,0,50}] (* G. C. Greubel, Jul 04 2019 *)

a(n)=n*(11*n+9)/2 \\ Charles R Greathouse IV, Jun 17 2017

[n*(11*n+9)/2 for n in (0..50)] # G. C. Greubel, Jul 04 2019

G.f.: x*(10+x)/(1-x)^3. - Philippe Deléham, Mar 27 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 10, a(2) = 31. - Philippe Deléham, Mar 27 2013
a(n) = A051865(n) + 9n = A180223(n) + 8n = A022268(n) + 5n = A022269(n) + 4n = A152740(n) - n. - Philippe Deléham, Mar 27 2013
a(n) = A218530(11n+9). - Philippe Deléham, Mar 27 2013
E.g.f.: x*(20 + 11*x)*exp(x)/2. - G. C. Greubel, Jul 04 2019

A162148 a(n) = n(n+1)(5*n+7)/6.

Original entry on oeis.org

0, 4, 17, 44, 90, 160, 259, 392, 564, 780, 1045, 1364, 1742, 2184, 2695, 3280, 3944, 4692, 5529, 6460, 7490, 8624, 9867, 11224, 12700, 14300, 16029, 17892, 19894, 22040, 24335, 26784, 29392, 32164, 35105, 38220, 41514, 44992, 48659, 52520, 56580

Offset: 0

Vladimir Joseph Stephan Orlovsky, Jun 25 2009

Partial sums of A147875.
Equals the fourth right hand column of A175136 for n>=1. - Johannes W. Meijer, May 06 2011
a(n) is the number of triples (w,x,y) havingt all terms in {0,...,n} and x+y>w. - Clark Kimberling, Jun 14 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A002412, A002413, A006331, A016061, A033994.
Cf. A162147, A175136, A190048, A190049, A254407.
Related to A000292 and A091894.

[n*(n+1)*(5*n+7)/6: n in [0..50]]; // Vincenzo Librandi, May 07 2011

A162148:= n-> n*(n+1)*(5*n+7)/6; seq(A162148(n), n=0..50); # G. C. Greubel, Mar 31 2021

Table[(n(n+1)(5n+7))/6,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,4,17,44}, 50] (* Harvey P. Dale, May 20 2014 *)

a(n)=n*(n+1)*(5*n+7)/6 \\ Charles R Greathouse IV, Oct 07 2015

[n*(n+1)*(5*n+7)/6 for n in (0..50)] # G. C. Greubel, Mar 31 2021

a(n) = A162147(n) + A000217(n).
From Johannes W. Meijer, May 06 2011: (Start)
G.f.: x*(4+x)/(1-x)^4.
a(n) = 4*binomial(n+2,3) + binomial(n+1,3).
a(n) = A091894(3,0)*binomial(n+2,3) + A091894(3,1)*binomial(n+1,3). (End)
a(n) = (n+1)*A000290(n+1) - Sum_{i=1..n+1} A000217(i).
a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4), a(0)=0, a(1)=4, a(2)=17, a(3)=44. - Harvey P. Dale, May 20 2014
E.g.f.: x*(24 +27*x +5*x^2)*exp(x)/6. - G. C. Greubel, Mar 31 2021

Definition rephrased by R. J. Mathar, Jun 27 2009

A308640 Number of ways to write n as (2^a3^b)^2 + c(2c+1) + d*(3d+1)/2, where a,b,c are nonnegative integers and d is an integer.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 1, 2, 4, 1, 4, 3, 3, 4, 1, 7, 2, 2, 7, 2, 5, 2, 4, 5, 1, 8, 5, 2, 3, 4, 6, 2, 3, 4, 2, 3, 7, 6, 5, 4, 7, 6, 1, 7, 5, 4, 6, 4, 4, 1, 6, 9, 2, 5, 3, 3, 5, 6, 7, 4, 7, 5, 4, 6, 6, 6, 4, 4, 5, 3, 9, 7, 4, 8, 2, 8, 5, 4, 10, 3, 9, 6, 5, 6, 4, 11, 7, 5, 8, 4, 7, 7, 8, 8, 2, 14, 6, 3, 8, 4

Offset: 1

Zhi-Wei Sun, Jun 12 2019

nonn

Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: Let k be 1 or 2. Then, any positive integer n can be written as (2^a*3^b)^2 + k*c^2 + d*(3d+1)/2, where a,b,c are nonnegative integers and d is an integer.
Conjecture 3: Let k be 1 or -1. Then, any positive integer n can be written as (2^a*3^b)^2 + c*(5c+3k)/2 + d*(3d+1)/2, where a,b,c are nonnegative integers and d is an integer.
We have verified Conjectures 1-3 for all n = 1..10^6.
See also A308641 for similar conjectures.

			a(230) = 1 with 230 =(2^3*3^0)^2 + 3*(2*3+1) + 10*(3*10+1)/2.
a(2058) = 1 with 2058 = (2^0*3^0)^2 + 25*(2*25+1) + (-23)*(3*(-23)+1)/2.
a(26550) = 1 with 26550 = (2^0*3^3)^2 + 14*(2*14+1) + 130*(3*130+1)/2.
a(39433) = 1 with 39433 = (2*3^3)^2 + 135*(2*135+1) + 17*(3*17+1)/2.
a(505330) = 1 with 505330 = (2*3^2)^2 + 198*(2*198+1) + 533*(3*533+1)/2.
a(537830) = 1 with 537830 = (2^5*3^2)^2 + 402*(2*402+1) + (-296)*(3*(-296)+1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (2015), No. 7, 1367-1396.

Cf. A000079, A000244, A000290, A000566, A001318, A014105, A147875, A308566, A308584, A308621, A308623, A308641.

PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]];
tab={};Do[r=0;Do[If[PenQ[n-4^a*9^b-c(2c+1)],r=r+1],{a,0,Log[4,n]},{b,0,Log[9,n/4^a]},{c,0,(Sqrt[8(n-4^a*9^b)+1]-1)/4}];tab=Append[tab,r],{n,1,100}];Print[tab]

A211014 Second 14-gonal numbers: n(6n+5).

Original entry on oeis.org

0, 11, 34, 69, 116, 175, 246, 329, 424, 531, 650, 781, 924, 1079, 1246, 1425, 1616, 1819, 2034, 2261, 2500, 2751, 3014, 3289, 3576, 3875, 4186, 4509, 4844, 5191, 5550, 5921, 6304, 6699, 7106, 7525, 7956, 8399, 8854, 9321, 9800, 10291, 10794, 11309, 11836, 12375

Offset: 0

Omar E. Pol, Aug 04 2012

Sequence found by reading the line from 0, in the direction 0, 34, ... and the line from 11 in the direction 11, 69, ..., in the square spiral whose vertices are the generalized 14-gonal numbers A195818.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Mark W. Coffey, Bernoulli identities, zeta relations, determinant expressions, Mellin transforms, and representation of the Hurwitz numbers, arXiv:1601.01673 [math.NT], 2016. See 3rd formula in Proposition 3 p. 36 giving a(n+1).
Leo Tavares, Star illustration
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A195818.
Second k-gonal numbers (k=5..14): A005449, A014105, A147875, A045944, A179986, A033954, A062728, A135705, A211013, this sequence.
Cf. A051866.
Cf. A003154.

List([0..50], n-> n*(6*n+5)); # G. C. Greubel, Jul 04 2019

[n*(6*n+5): n in [0..50]]; // G. C. Greubel, Jul 04 2019

Table[n*(6*n+5), {n,0,50}] (* G. C. Greubel, Jul 04 2019 *)
LinearRecurrence[{3,-3,1},{0,11,34},50] (* Harvey P. Dale, Dec 12 2022 *)

a(n)=n*(6*n+5) \\ Charles R Greathouse IV, Jun 17 2017

[n*(6*n+5) for n in (0..50)] # G. C. Greubel, Jul 04 2019

a(n) = -2*Sum_{k=0..n-1} binomial(6*n+5, 6*k+8)*Bernoulli(6*k+8). - Michel Marcus, Jan 11 2016
From G. C. Greubel, Jul 04 2019: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(11+x)/(1-x)^3.
E.g.f.: x*(11+6*x)*exp(x). (End)
From Amiram Eldar, Feb 28 2022: (Start)
Sum_{n>=1} 1/a(n) = sqrt(3)*Pi/10 + 6/25 - 3*log(3)/10 - 2*log(2)/5.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/5 + log(2)/5 - 6/25 - sqrt(3)*log(sqrt(3)+2)/5. (End)
a(n) = A003154(n+1) - n - 1. - Leo Tavares, Jan 29 2023

A350189 Triangle T(n,k) read by rows: the number of symmetric binary n X n matrices with k ones and no all-1 2 X 2 submatrix.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 1, 3, 6, 10, 9, 9, 4, 1, 4, 12, 28, 46, 72, 80, 80, 60, 16, 1, 5, 20, 60, 140, 296, 500, 780, 1005, 1085, 992, 560, 170, 1, 6, 30, 110, 330, 876, 1956, 4020, 7140, 11480, 16248, 19608, 20560, 16500, 9720, 3276, 360, 1, 7, 42, 182, 665, 2121, 5852, 14792, 33117, 68355, 126994, 214158

Offset: 0

R. J. Mathar, Mar 09 2022

There are 2^(n^2) binary n X n matrices (entries of {0,1}). There are 2^(n*(n+1)/2) symmetric binary matrices. There are A184948(n,k) symmetric binary n X n matrices with k ones.
This sequence is the triangle T(n,k) of symmetric binary n x n matrices with k ones but no 2 X 2 submatrix with all entries = 1. [So in the display of these matrices there is no rectangle with four 1's at the corners.]
The row lengths minus 1 are 0, 1, 3, 6, 9, 12, 17, 21, 24, 29, ... and indicate the maximum number of 1's than can be packed into a symmetric binary n X n matrix without creating an all-1 quadrangle/submatrix of order 2.

			The triangle starts
  1;
  1 1;
  1 2 2 2;
  1 3 6 10 9 9 4;
  1 4 12 28 46 72 80 80 60 16;
  1 5 20 60 140 296 500 780 1005 1085 992 560 170;
  ...
To place 4 ones, one can place 2 of them in C(n,2) ways on the diagonal and the other 2 in n*(n-1)/2 ways outside the diagonal, avoiding one matrix that builds an all-1 submatrix, which are C(n,2)*(n*(n-1)/2-1) matrices. One can place all 4 on the diagonal in C(n,4) ways. One can place 2 outside the diagonal (the other 2 mirror symmetrically) in C(n*(n-1)/2,2) ways. Sum of the 3 terms is T(n,4) = C(n,3)*(5*n+3)/2. - _R. J. Mathar_, Mar 10 2022

Max Alekseyev, Table of n, a(n) for n = 0..361 (rows 0..14; rows 0..9 from R. J. Mathar)
P. Erdos and D. Kleitman, On collections of subsets containing no 4-member boolean algebra, Proc. Am. Math. Soc. 28 (1) (1971) 87.
R. J. Mathar, Illustration of graphs with these incidence matrices
Index entries for sequences of binary matrices
Index entries for sequences of squarefree graphs

Cf. A001197 (conjectured row lengths), A352258 (row sums), A352801 (rightmost terms), A350296, A350304, A350237, A352472 (traceless symmetric).

Cf. A002378, A006331, A000217, A147875.

T(n,0) = 1.
T(n,1) = n.
T(n,2) = A002378(n-1).
T(n,3) = A006331(n-1).
T(n,4) = n*(n-1)*(n-2)*(5*n+3)/12 = A147875(n)*A000217(n-1)/3. - R. J. Mathar, Mar 10 2022
T(n,5) = n*(n-1)*(n-2)*(13*n^2-n-24)/60. T(n,6) = n*(n-1)*(n-2)*(19*n^3-18*n^2-97*n+60)/180. T(n,7) = n*(n-1)*(n-2)*(n-3)*(58*n^3+75*n^2-223*n+180)/1260. - Conjectured by R. J. Mathar, Mar 11 2022; proved by Max Alekseyev, Apr 02 2022
G.f.: F(x,y) = Sum_{n,k} T(n,k)*x^n/n!*y^k = exp( Sum_G x^n(G) * y^u(G) / |Aut(G)| ), where G runs over the connected squarefree graphs with loops, n(G) is the number of nodes in G, u(G) the number of ones in the adjacency matrix of G, and Aut(G) is the automorphism group of G. It follows that F(x,y) = exp(x) * (1 + x*y + x^2*y^2 + (2/3*x^3 + x^2)*y^3 + (5/12*x^4 + 3/2*x^3)*y^4 + (13/60*x^5 + 3/2*x^4 + 3/2*x^3)*y^5 + (19/180*x^6 + 7/6*x^5 + 8/3*x^4 + 2/3*x^3)*y^6 + (29/630*x^7 + 3/4*x^6 + 19/6*x^5 + 10/3*x^4)*y^7 + O(y^8)), implying the above formulas for T(n,k). - Max Alekseyev, Apr 02 2022
Conjecture: the largest k such that T(n,k) is nonzero is k = A072567(n) = A001197(n) - 1. - Max Alekseyev, Apr 03 2022

A224419 Numbers n such that triangular(n) + triangular(2*n) is a square.

Original entry on oeis.org

0, 1, 25, 216, 1849, 36481, 311904, 2666689, 52606009, 449765784, 3845364121, 75857828929, 648561949056, 5545012396225, 109386936710041, 935225880773400, 7995904029992761, 157735886878050625, 1348595071513294176, 11530088066237165569, 227455039491212291641, 1944673157896289428824, 16626378995609962758169, 327990009210441246496129

Offset: 1

Alex Ratushnyak, Apr 18 2013

nonn

8 of the first 10 terms are of the form x^y. The two exceptions are a(7) = 311904 = 2^5 * 3^3 * 19^2 and a(10) = 449765784 = 2^3 * 3^5 * 13^2 * 37^2.
The corresponding squares are given by A075873(2*n-1)^2. E.g., triangular(a(10)) + triangular(2*a(10)) = 711142146^2 = A075873(19)^2.
Locations of squares in A147875, equivalent to solving the Diophantine equation n*(5*n+3)=2*s^2. - R. J. Mathar, Apr 19 2013

Max Alekseyev, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,1442,-1442,0,-1,1).

Cf. A000217, A075873.

Cf. A220186 (numbers n such that triangular(2*n) - triangular(n) is a square).

LinearRecurrence[{1,0,1442,-1442,0,-1,1},{0,1,25,216,1849,36481,311904},30]  (* Harvey P. Dale, Jan 23 2015 *)

import math
for i in range(1<<30):
        s = i*(i+1)/2 + i*(2*i+1)
        t = int(math.sqrt(s))
        if s == t*t:  print(i)

a(n) = (A228209(2*n-1) - 3) / 10. - Max Alekseyev, Sep 04 2013

G.f.: x^2*(x+1)*(x^4 + 23*x^3 + 168*x^2 + 23*x + 1) / (x^6 - 1442*x^3 + 1) / (1-x). - Max Alekseyev, Sep 04 2013

Terms a(11) onward from Max Alekseyev, Sep 04 2013

cp's OEIS Frontend

A062728 Second 11-gonal (or hendecagonal) numbers: a(n) = n*(9*n+7)/2.

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A254963 a(n) = n*(11*n + 3)/2.

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A135705 a(n) = 10*binomial(n,2) + 9*n.

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A008732 Molien series for 3-dimensional group [2,n] = *22n.

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A211013 Second 13-gonal numbers: a(n) = n*(11*n+9)/2.

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A162148 a(n) = n*(n+1)*(5*n+7)/6.

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A062728 Second 11-gonal (or hendecagonal) numbers: a(n) = n(9n+7)/2.

A254963 a(n) = n(11n + 3)/2.

A135705 a(n) = 10binomial(n,2) + 9n.

A211013 Second 13-gonal numbers: a(n) = n(11n+9)/2.

A162148 a(n) = n(n+1)(5*n+7)/6.

A308640 Number of ways to write n as (2^a3^b)^2 + c(2c+1) + d*(3d+1)/2, where a,b,c are nonnegative integers and d is an integer.

A211014 Second 14-gonal numbers: n(6n+5).