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.

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

1,1

Comments

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.

Links

Table of n, a(n) for n=1..52.

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).

Formula

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]

Example

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.

Mathematica

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

Crossrefs

Cf. A102214.

Sequence in context: A283757 A353078 A131725 * A373200 A332976 A032914

Adjacent sequences: A346367 A346368 A346369 * A346371 A346372 A346373

Keyword

nonn,easy

Author

Mikhail Kurkov, Jul 14 2021 [verification needed]

Status

approved