A135453 -id:A135453 - cp's OEIS frontend

A317297 a(n) = (n - 1)(4n^2 - 8*n + 5).

0, 5, 34, 111, 260, 505, 870, 1379, 2056, 2925, 4010, 5335, 6924, 8801, 10990, 13515, 16400, 19669, 23346, 27455, 32020, 37065, 42614, 48691, 55320, 62525, 70330, 78759, 87836, 97585, 108030, 119195, 131104, 143781, 157250, 171535, 186660, 202649, 219526, 237315, 256040, 275725, 296394, 318071

Offset: 1

Omar E. Pol, Sep 01 2018

Conjecture: For n > 1, a(n) is the maximum eigenvalue of a 2*(n-1) X 2*(n-1) square matrix M defined as M[i,j,n] = j + n*(i-1) if i is odd and M[i,j,n] = n*i - j + 1 if i is even (see A317614). - Stefano Spezia, Dec 27 2018

Connections can be made to A022144 and A010014. Namely, a formula for A022144 is (2*n+1)^2 - (2*n-1)^2. A formula for A010014 is (2*n+1)^3 - (2*n-1)^3. The general form can be represented by (2*n+1)^d - (2*n-1)^d, where d designates the number of dimensions. When d is 4, a(n) = ((2*(n-1)+1)^4 - (2*(n-1)-1)^4)/16, namely the general form shifted by 1 and divided by 16 is a(n). - Yigit Oktar, Aug 16 2024

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

First bisection of A006003.
Nonzero terms give the row sums of A007607.
Conjecture: 0 together with a bisection of A246697.
Cf. A219086 (partial sums).
Cf. A008595, A033430, A135453, A317614.
Cf. A010014, A022144 (see comments)

Table[(n - 1) (4 n^2 - 8 n + 5), {n, 1, 50}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {0, 5, 34, 111}, 50] (* or *) CoefficientList[Series[x (5 + 14 x + 5 x^2)/(1 - x)^4, {x, 0, 50}], x] (* Stefano Spezia, Sep 01 2018 *)

PARI
```
a(n) = (n - 1)*(4*n^2 - 8*n + 5)
```

concat(0, Vec(x^2*(5 + 14*x + 5*x^2)/(1 - x)^4 + O(x^50))) \\ Colin Barker, Sep 01 2018

a(n) = 4*n^3 - 12*n^2 + 13*n - 5 = A033430(n) - A135453(n) + A008595(n) - 5.
G.f.: x^2*(5 + 14*x + 5*x^2)/(1 - x)^4. - Colin Barker, Sep 01 2018
a(n) = 4*a(n - 1) - 6*a(n - 2) + 4*a(n - 3) - a(n - 4) for n > 4. - Stefano Spezia, Sep 01 2018
E.g.f.: exp(x)*(5*x + 12*x^2 + 4*x^3). - Stefano Spezia, Jan 15 2019
a(n) = ((2*(n-1)+1)^4 - (2*(n-1)-1)^4)/16. - Yigit Oktar, Aug 16 2024

A158480 a(n) = 12*n^2 + 1.

Original entry on oeis.org

1, 13, 49, 109, 193, 301, 433, 589, 769, 973, 1201, 1453, 1729, 2029, 2353, 2701, 3073, 3469, 3889, 4333, 4801, 5293, 5809, 6349, 6913, 7501, 8113, 8749, 9409, 10093, 10801, 11533, 12289, 13069, 13873, 14701, 15553, 16429, 17329, 18253, 19201, 20173, 21169

Offset: 0

Vincenzo Librandi, Mar 20 2009

The identity (12*n^2 + 1)^2 - (36*n^2 + 6)*(2*n)^2 = 1 can be written as a(n)^2 - A158479(n)*A005843(n)^2 = 1.

Sequence found by reading the line from 13, in the direction 13, 49, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012

			a(1) = 12*1^2 + 1 = 13.
a(2) = 12*2^2 + 1 = 49.
a(3) = 12*3^2 + 1 = 109.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Vincenzo Librandi, X^2-AY^2=1 [broken link]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A158479.

List([1..40], n-> 12*n^2 + 1); # G. C. Greubel, Nov 06 2019

I:=[13,49,109];[n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2) +Self(n-3): n in [1..50]];

seq(12*n^2 +1, n=0..45); # G. C. Greubel, Nov 06 2019

LinearRecurrence[{3,-3,1}, {13,49,109}, 40]
12*Range[40]^2 +1 (* G. C. Greubel, Nov 06 2019 *)

PARI
```
a(n)=12*n^2+1
```

[12*n^2 +1 for n in (1..40)] # G. C. Greubel, Nov 06 2019

a(n) = A010014(n)/2. - Vladimir Joseph Stephan Orlovsky, May 18 2009
G.f: (13*x^2 + 10*x + 1)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = 1 + A135453(n). - Omar E. Pol, Jul 18 2012
a(n) = (A016969(n-1)*A016921(n) + 4)/3. - Hilko Koning, Oct 25 2019
E.g.f.: (1 + 12*x + 12*x^2)*exp(x). - G. C. Greubel, Nov 06 2019
From Amiram Eldar, Jul 15 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + (Pi/sqrt(12))*coth(Pi/sqrt(12)))/2.
Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/sqrt(12))*csch(Pi/sqrt(12)))/2. (End)
From Amiram Eldar, Feb 05 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/sqrt(12))*sinh(Pi/sqrt(6)).
Product_{n>=1} (1 - 1/a(n)) = (Pi/sqrt(12))*csch(Pi/sqrt(12)). (End)

a(0)=1 prepended by Alois P. Heinz, Jun 12 2021

A299965 Number of triangles in a Star of David of size n.

Original entry on oeis.org

0, 20, 118, 354, 788, 1480, 2490, 3878, 5704, 8028, 10910, 14410, 18588, 23504, 29218, 35790, 43280, 51748, 61254, 71858, 83620, 96600, 110858, 126454, 143448, 161900, 181870, 203418, 226604, 251488, 278130, 306590, 336928, 369204, 403478, 439810, 478260, 518888

Offset: 0

John King, Feb 22 2018

In a Star of David of size n, there are A135453(n) "size=1" triangles.

The number of matchstick units is A045946.

			For n=1, there are 12 (size=1) + 6 (size=4) + 2 (size=9) = 20 triangles.

Paolo Xausa, Table of n, a(n) for n = 0..10000 (corrected and extended original b-file from Colin Barker after data change).
John King, Star a=6, 84 matches, 118 triangles
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A045950, A045946, A135453,

For the total number of triangles in a different arrangement, see A002717 (for triangular matchstick), A045949 (for hexagonal matchstick).

A299965[n_] := n*(n*(10*n + 9) + 1); Array[A299965, 50, 0] (* or *)
LinearRecurrence[{4, -6, 4, -1}, {0, 20, 118, 354}, 50] (* Paolo Xausa, Sep 18 2024 *)

concat(0, Vec(2*x*(10 + 19*x + x^2) / (1 - x)^4 + O(x^40))) \\ Colin Barker, Apr 04 2019

a(n) = n*(10*n^2+9*n+1) = 2*A045950(n).
From Colin Barker, Apr 04 2019: (Start)
G.f.: 2*x*(10 + 19*x + x^2) / (1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. (End)
E.g.f.: exp(x)*x*(20 + 39*x + 10*x^2). -  Stefano Spezia, Sep 20 2024

Corrected by John King, Stefano Spezia, and Paolo Xausa, Sep 20 2024

A303302 a(n) = 34*n^2.

Original entry on oeis.org

0, 34, 136, 306, 544, 850, 1224, 1666, 2176, 2754, 3400, 4114, 4896, 5746, 6664, 7650, 8704, 9826, 11016, 12274, 13600, 14994, 16456, 17986, 19584, 21250, 22984, 24786, 26656, 28594, 30600, 32674, 34816, 37026, 39304, 41650, 44064, 46546, 49096, 51714, 54400, 57154, 59976, 62866, 65824, 68850, 71944

Offset: 0

Omar E. Pol, May 13 2018

Sequence found by reading the line from 0, in the direction 0, 34, ..., in the square spiral whose vertices are the generalized 19-gonal numbers A303813.

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

Cf. A005843, A008599, A303813.

Cf. similar sequences of the type k*n^2: A000290 (k=1), A001105 (k=2), A033428 (k=3), A016742 (k=4), A033429 (k=5), A033581 (k=6), A033582 (k=7), A139098 (k=8), A016766 (k=9), A033583 (k=10), A033584 (k=11), A135453 (k=12), A152742 (k=13), A144555 (k=14), A064761 (k=15), A016802 (k=16), A244630 (k=17), A195321 (k=18), A244631 (k=19), A195322 (k=20), A064762 (k=21), A195323 (k=22), A244632 (k=23), A195824 (k=24), A016850 (k=25), A244633 (k=26), A244634 (k=27), A064763 (k=28), A244635 (k=29), A244636 (k=30), A244082 (k=32), this sequence (k=34), A016910 (k=36), A016982 (k=49), A017066 (k=64), A017162 (k=81), A017270 (k=100), A017390 (k=121), A017522 (k=144).

[34*n^2: n in [0..50]]; // Vincenzo Librandi Jun 07 2018

Table[34 n^2, {n, 0, 40}]
LinearRecurrence[{3,-3,1},{0,34,136},50] (* Harvey P. Dale, Jul 23 2018 *)

PARI
```
a(n) = 34*n^2;
```

concat(0, Vec(34*x*(1 + x) / (1 - x)^3 + O(x^40))) \\ Colin Barker, Jun 12 2018

a(n) = 34*A000290(n) = 17*A001105(n) = 2*A244630(n).
G.f.: 34*x*(1 + x)/(1 - x)^3. - Vincenzo Librandi, Jun 07 2018
From Elmo R. Oliveira, Dec 02 2024: (Start)
E.g.f.: 34*x*(1 + x)*exp(x).
a(n) = A005843(n)*A008599(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A256695 Integer areas A of integer-sided triangles (a, b, c) such that the area of the triangle (a+b, a+c, b+c) is also an integer.

Original entry on oeis.org

12, 48, 108, 192, 300, 432, 588, 768, 972, 1008, 1200, 1452, 1728, 2028, 2352, 2448, 2520, 2700, 2772, 3060, 3072, 3468, 3888, 4032, 4332, 4800, 5292, 5808, 6348, 6912, 7500, 8112, 8748, 9072, 9408, 9792, 10080, 10092, 10800, 11088, 11532, 11628, 12240, 12288

Offset: 1

Michel Lagneau, Apr 08 2015

nonn

The areas of the primitive triangles are 12, 2520, 2772, 3060, 4032, 5808, 9072, 11088, 11628, 17136, 24948, 41580, ...
The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2.
The sequence A135453 (numbers of the form 12*n^2) is included in the sequence because a(1) = 12 is a primitive triangle of the subsequence k^2*a(1), k=1,2,3,...
The following table gives the first values (A, A', a, b, c) where A is the integer area of the triangle (a, b, c), A' is the integer area of the triangle (a+b, a+c, b+c).
+-------+--------+------+------+------+
|    A  |    A'  |   a  |   b  |   c  |
+-------+--------+------+------+------+
|   12  |    60  |   5  |   5  |   8  |
|   48  |   240  |  10  |  10  |  16  |
|  108  |   540  |  15  |  15  |  24  |
|  192  |   960  |  20  |  20  |  32  |
|  300  |  1500  |  25  |  25  |  40  |
|  432  |  2160  |  30  |  30  |  48  |
|  588  |  2940  |  35  |  35  |  56  |
|  768  |  3840  |  40  |  40  |  64  |
|  972  |  4860  |  45  |  45  |  72  |
+-------+--------+------+------+------+
We find a majority of isosceles triangles, but there is a subsequence of non-isosceles triangles with areas 2520, 3060, 10080, 11088, ...
+--------+----------+-------+-------+-------+
|    A   |     A'   |   a   |   b   |   c   |
+--------+----------+-------+-------+-------+
|   2520 |   18270  |   29  |  174  |  175  |
|   3060 |   33150  |   39  |  221  |  250  |
|  10080 |   73080  |   58  |  348  |  350  |
|  11088 |   64350  |  150  |  169  |  275  |
|  12240 |   12240  |   78  |  442  |  500  |
|  17136 |   92820  |  168  |  221  |  325  |
|  41580 |  183150  |  250  |  333  |  407  |
+--------+----------+-------+-------+-------+

			a(1) = 12 because, for (a,b,c) = (5, 5, 8) => s = (5+5+8)/2 = 9, and
A = sqrt(9(9-5)(9-5)(9-8)) = sqrt(144) = 12 and the triangle (5+5, 5+8, 5+8) = (10, 13, 13)=> s1 = (10+13+13)/2 = 18, and A1 = sqrt(18(18-10)(18-13)(18-13)) = sqrt(3600) = 60 is an integer.

Cf. A135453, A188158.

nn=1000;lst={};Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s(s-a)(s-b)(s-c);u=a+b;v=a+c;w=b+c;s1=(u+v+w)/2;area3=s1(s1-u)(s1-v)(s1-w);If[area2>0&&area3>0&&IntegerQ[Sqrt[area2]]&&IntegerQ[Sqrt[area3]], AppendTo[lst,Sqrt[area2]]]],{a,nn},{b,a},{c,b}];Union[lst]

A363436 Array read by ascending antidiagonals: A(n, k) = k*n^2, with k >= 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 4, 2, 0, 0, 9, 8, 3, 0, 0, 16, 18, 12, 4, 0, 0, 25, 32, 27, 16, 5, 0, 0, 36, 50, 48, 36, 20, 6, 0, 0, 49, 72, 75, 64, 45, 24, 7, 0, 0, 64, 98, 108, 100, 80, 54, 28, 8, 0, 0, 81, 128, 147, 144, 125, 96, 63, 32, 9, 0, 0, 100, 162, 192, 196, 180, 150, 112, 72, 36, 10, 0

Offset: 0

Stefano Spezia, Jul 08 2023

			The array begins:
  0,  0,  0,   0,   0,   0,   0, ...
  0,  1,  2,   3,   4,   5,   6, ...
  0,  4,  8,  12,  16,  20,  24, ...
  0,  9, 18,  27,  36,  45,  54, ...
  0, 16, 32,  48,  64,  80,  96, ...
  0, 25, 50,  75, 100, 125, 150, ...
  0, 36, 72, 108, 144, 180, 216, ...
  ...

Cf. A000290 (k = 1), A001105 (k = 2), A033428 (k = 3), A016742 (k = 4), A033429 (k = 5), A033581 (k = 6), A033582 (k = 7), A139098 (k = 8), A016766 (k = 9), A033583 (k = 10), A033584 (k = 11), A135453 (k = 12),  A152742 (k = 13), A144555 (k = 14), A064761 (k = 15), A016802 (k = 16), A244630 (k = 17), A195321 (k = 18), A244631 (k = 19), A195322 (k = 20), A064762 (k = 21), A195323 (k = 22), A244632 (k = 23), A195824 (k = 24), A016850 (k = 25), A244633 (k = 26), A244634 (k = 27), A064763 (k = 28), A244635 (k = 29), A244636 (k = 30).
Cf. A001477 (n = 1), A008586 (n = 2), A008591 (n = 3), A008598 (n = 4), A008607 (n = 5), A044102 (n = 6), A152691 (n = 8).
Cf. A000007 (n = 0 or k = 0), A000578 (main diagonal), A002415 (antidiagonal sums), A004247.

A[n_,k_]:=k n^2; Table[A[n-k,k],{n,0,11},{k,0,n}]//Flatten

O.g.f.: x*y*(1 + x)/((1 - x)^3*(1 - y)^2).
E.g.f.: x*y*(1 + x)*exp(x + y).
A(n, k) = n*A004247(n, k).

cp's OEIS Frontend

A317297 a(n) = (n - 1)*(4*n^2 - 8*n + 5).

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A158480 a(n) = 12*n^2 + 1.

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A299965 Number of triangles in a Star of David of size n.

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A303302 a(n) = 34*n^2.

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A256695 Integer areas A of integer-sided triangles (a, b, c) such that the area of the triangle (a+b, a+c, b+c) is also an integer.

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A363436 Array read by ascending antidiagonals: A(n, k) = k*n^2, with k >= 0.

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A317297 a(n) = (n - 1)(4n^2 - 8*n + 5).