A102214 -id:A102214 - cp's OEIS frontend

A269064 At stage 1, start with a unit equilateral triangle. At each successive stage add 3*(n-1) new triangles around outside with vertex-to-vertex contacts. Sequence gives number of triangles at n-th stage.

0, 1, 4, 10, 26, 48, 87, 135, 208, 293, 410, 542, 714, 904, 1141, 1399, 1712, 2049, 2448, 2874, 3370, 3896, 4499, 5135, 5856, 6613, 7462, 8350, 9338, 10368, 11505, 12687, 13984, 15329, 16796, 18314, 19962, 21664, 23503, 25399, 27440, 29541, 31794, 34110, 36586, 39128, 41837, 44615, 47568

Offset: 0

Luce ETIENNE, Feb 18 2016

At stage n, we count (6*n^2-6*n+5-3*(2*n-1)*(-1)^n)/8 unit up-pointing triangles and 3*(2*n^2-2*n+1+(2*n-1)*(-1)^n)/8 unit down-pointing triangles.
At stage n, the total number of unit triangles is (3*n^2-3*n+2)/2 = A005448(n). It is the same total as for A064412. Note also that A064412 gives number of triangles in a geometrical structure according to expansion side-side (mode S-S).
The edges of several unit triangles can form larger size triangles, and these are also up- or down-pointing. The number of all such larger is given by :(14*n^3-9*n^2+11*n+18-(9*n^2+3*n+14)*(-1)^n-4*((-1)^((2*n+1-(-1)^n)/4)))/64 up-pointing triangles and (14*n^3-15*n^2+35*n-6+(9*n^2+21*n+2)*(-1)^n+4*((-1)^((2*n-1+(-1)^n)/4)))/64 down-pointing triangles.
As for A265282 we observe that starting with n = 4 we can see and count hexagonal and dodecagonal forms for example in a reticular system (incomplete with hexagonal holes) by opposition to a compact shape A064412.

			a(0)= 0, a(1) = 1, a(2) = 4, a(3) = 7+3 = 10, a(4) = 19 + 6 + 1 = 26, a(5) = 31 + 12 + 4 + 1 = 48.

Colin Barker, Table of n, a(n) for n = 0..1000
Luce ETIENNE, Illustration of initial terms
Kival Ngaokrajang, Illustration of triangles expansion
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,2,-2,0,2,-1).

Cf. A003215, A005448, A005993, A033428, A064412, A061600, A006003, A039623, A102214, A265282.

[(14*n^3-12*n^2+23*n+6+3*(3*n-2)*(-1)^n+2*((-1)^((2*n-1+(-1)^n) div 4)-(-1)^((6*n-1+(-1)^n) div 4)))/32: n in [0..50]]; // Vincenzo Librandi, Feb 19 2016

Table[(14 n^3 - 12 n^2 + 23 n + 6 + 3 (3 n - 2) (-1)^n + 2 ((-1)^((2*n - 1 + (-1)^n) / 4) - (-1)^((6 n - 1 + (-1)^n) / 4))) / 32, {n, 0, 45}] (* Vincenzo Librandi, Feb 19 2016 *)

concat(0, Vec(x*(1+2*x+2*x^2+8*x^3+2*x^4+5*x^5+x^6)/((1-x)^4*(1+x)^2*(1+x^2)) + O(x^50))) \\ Colin Barker, Feb 24 2016

a(n) = (7*n^3-3*n^2+4*n)/2 for n even.
a(n) = (28*n^3+30*n^2+16*n+7+(-1)^n)/8 for n odd.
a(n) = (14*n^3-12*n^2+23*n+6+3*(3*n-2)*(-1)^n+2*((-1)^((2*n-1+(-1)^n)/4)-(-1)^((6*n-1+(-1)^n)/4)))/32.
G.f.: x*(1+2*x+2*x^2+8*x^3+2*x^4+5*x^5+x^6) / ((1-x)^4*(1+x)^2*(1+x^2)). - Colin Barker, Feb 24 2016

A198392 a(n) = (6n(3n+7)+(2n+13)*(-1)^n+3)/16 + 1.

Original entry on oeis.org

2, 4, 12, 18, 31, 41, 59, 73, 96, 114, 142, 164, 197, 223, 261, 291, 334, 368, 416, 454, 507, 549, 607, 653, 716, 766, 834, 888, 961, 1019, 1097, 1159, 1242, 1308, 1396, 1466, 1559, 1633, 1731, 1809, 1912, 1994, 2102, 2188, 2301, 2391, 2509, 2603, 2726, 2824, 2952

Offset: 0

Bruno Berselli, Oct 25 2011

For an origin of this sequence, see the triangular spiral illustrated in the Links section.

First bisection gives A117625 (without the initial term).

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A152832 (by Superseeker).

Cf. sequences related to the triangular spiral: A022266, A022267, A027468, A038764, A045946, A051682, A062708, A062725, A062728, A062741, A064225, A064226, A081266-A081268, A081270-A081272, A081275 [incomplete list].

[(6*n*(3*n+7)+(2*n+13)*(-1)^n+3)/16+1: n in [0..50]];

LinearRecurrence[{1,2,-2,-1,1},{2,4,12,18,31},60] (* Harvey P. Dale, Jun 15 2022 *)

for(n=0, 50, print1((6*n*(3*n+7)+(2*n+13)*(-1)^n+3)/16+1", "));

G.f.: (2+2*x+4*x^2+2*x^3-x^4)/((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).
a(n)-a(-n-1) = A168329(n+1).
a(n)+a(n-1) = A102214(n).
a(2n)-a(2n-1) = A016885(n).
a(2n+1)-a(2n) = A016825(n).

A115282 Correlation triangle for the sequence 3-2*0^n.

Original entry on oeis.org

1, 3, 3, 3, 10, 3, 3, 12, 12, 3, 3, 12, 19, 12, 3, 3, 12, 21, 21, 12, 3, 3, 12, 21, 28, 21, 12, 3, 3, 12, 21, 30, 30, 21, 12, 3, 3, 12, 21, 30, 37, 30, 21, 12, 3, 3, 12, 21, 30, 39, 39, 30, 21, 12, 3, 3, 12, 21, 30, 39, 46, 39, 30, 21, 12, 3

Offset: 0

Paul Barry, Jan 19 2006

Row sums are A102214. Diagonal sums are A115283. T(2n,n) is 9n+1 (A017173), the partial sums of (3-2*0^n)^2. T(2n,n)-T(2n,n+1) is 7-6*0^n.

			Triangle begins
1;
3, 3;
3,10, 3;
3,12,12, 3;
3,12,19,12, 3;
3,12,21,21,12, 3;
3,12,21,28,21,12,3;

G.f.: (1+2x)(1+2x*y)/((1-x)(1-x*y)(1-x^2*y)); Number triangle T(n, k)=sum{j=0..n, [j<=k]*(3-2*0^(k-j))*[j<=n-k]*(3-2*0^(n-k-j))}.

A346370 Upper bound for the number of solutions of the TRINTUM cube puzzle n X 1 X 1 (cuboid formed by 4n + 2 parts) different by the set of parts, which are distinguished by the amount of surface area they contribute to the assembled cuboid.

Original entry on oeis.org

3, 8, 10, 19, 22, 34, 38, 54, 59, 78, 84, 107, 114, 140, 148, 178, 187, 220, 230, 267, 278, 318, 330, 374, 387, 434, 448, 499, 514, 568, 584, 642, 659, 720, 738, 803, 822, 890, 910, 982, 1003, 1078, 1100, 1179, 1202, 1284, 1308, 1394, 1419, 1508, 1534, 1627

Offset: 1

Mikhail Kurkov, Jul 14 2021 [verification needed]

There are 6 parts in a basic set with a given surface area (with addition of an empty cell in the center): A-part (19), H-part (21), fish (21), rooster (27) x2 and tower (35). For purposes of this sequence, the H-part and the fish are equivalent, since they contribute the same surface area.
Equivalently, number of nonnegative integer solutions of the system of equations 19x + 21y + 27z + 35w = 96n + 54, x + y + z + w = 4n + 2. Here x = 3m + w, y = 2n - 4m, z = 2(n - w + 1) + m, so for nonnegative integers we have 0 <= m <= floor(n/2), 0 <= w <= n + floor(n/4) + 1 and number of solutions is Sum_{k=0..floor(n/2)} n + floor(k/2) + 2 = (n + 2)*(floor(n/2) + 1) + floor(floor(n/2)^2/4) for n > 0.
Conjecture: upper bound can be reduced to n+2 (based on attempts to construct cuboid using nonnegative integer solutions with m > 0, where it looks like impossible to place at least all roosters and towers somewhere).
The simplest way to construct cuboid of any size is to use cubes formed by 2 A-parts, 2 H-parts and 2 towers. We just remove an A-part from one cube and a tower from another to easily connect them together.

			a(1) = 3 because there are only 3 possible nonnegative integer solutions:
  19*0 + 21*2 + 27*4 + 35*0 = 150, 0 + 2 + 4 + 0 = 6;
  19*1 + 21*2 + 27*2 + 35*1 = 150, 1 + 2 + 2 + 1 = 6;
  19*2 + 21*2 + 27*0 + 35*2 = 150, 2 + 2 + 0 + 2 = 6.
a(2) = 8 because there are only 8 possible nonnegative integer solutions:
  19*0 + 21*4 + 27*6 + 35*0 = 246, 0 + 4 + 6 + 0 = 10;
  19*1 + 21*4 + 27*4 + 35*1 = 246, 1 + 4 + 4 + 1 = 10;
  19*2 + 21*4 + 27*2 + 35*2 = 246, 2 + 4 + 2 + 2 = 10;
  19*3 + 21*4 + 27*0 + 35*3 = 246, 3 + 4 + 0 + 3 = 10;
  19*3 + 21*0 + 27*7 + 35*0 = 246, 3 + 0 + 7 + 0 = 10;
  19*4 + 21*0 + 27*5 + 35*1 = 246, 4 + 0 + 5 + 1 = 10;
  19*5 + 21*0 + 27*3 + 35*2 = 246, 5 + 0 + 3 + 2 = 10;
  19*6 + 21*0 + 27*1 + 35*3 = 246, 6 + 0 + 1 + 3 = 10.

Mikhail Kurkov, Demonstration of the TRINTUM cube puzzle, YouTube video. [verification needed]
Mikhail Kurkov, Parts of the TRINTUM cube puzzle, image. [verification needed]
Mikhail Kurkov, Solutions for the cube, image. [verification needed]
Mikhail Kurkov, Some solutions for the 2 X 1 X 1 cuboid, image. [verification needed]
Mikhail Kurkov, Some solutions for the 3 X 1 X 1 cuboid, image. [verification needed]
Jon Maiga, Computer-generated formulas for A346370, Sequence Machine.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,-1,-1,1).

Cf. A102214.

LinearRecurrence[{1, 1, -1, 1, -1, -1, 1}, {3, 8, 10, 19, 22, 34,
38}, 52] (* Robert P. P. McKone, Jul 16 2021 *)

a(n) = (n + 2)*(floor(n/2) + 1) + floor(floor(n/2)^2/4).
a(n) = A102214(floor(n/2)) + (1 + n mod 2)*(1 + floor(n/2)).
G.f.: x*(3 + 5*x - x^2 + 4*x^3 - 2*x^4 - 2*x^5 + 2*x^6)/((1 - x)^3*(1 + x)^2*(1 + x^2)). - Stefano Spezia, Jul 14 2021
Conjecture: a(n) = A008733(n-1) + A093907(n) for n > 0 (noticed by Sequence Machine). - Mikhail Kurkov, Oct 14 2021 [verification needed]

cp's OEIS Frontend

A269064 At stage 1, start with a unit equilateral triangle. At each successive stage add 3*(n-1) new triangles around outside with vertex-to-vertex contacts. Sequence gives number of triangles at n-th stage.

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A198392 a(n) = (6*n*(3*n+7)+(2*n+13)*(-1)^n+3)/16 + 1.

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A115282 Correlation triangle for the sequence 3-2*0^n.

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A346370 Upper bound for the number of solutions of the TRINTUM cube puzzle n X 1 X 1 (cuboid formed by 4n + 2 parts) different by the set of parts, which are distinguished by the amount of surface area they contribute to the assembled cuboid.

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Author

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Examples

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Crossrefs

Programs

Mathematica

Formula

A198392 a(n) = (6n(3n+7)+(2n+13)*(-1)^n+3)/16 + 1.