cp's OEIS Frontend

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.

%I A346370 #48 Apr 21 2024 22:12:48
%S A346370 3,8,10,19,22,34,38,54,59,78,84,107,114,140,148,178,187,220,230,267,
%T A346370 278,318,330,374,387,434,448,499,514,568,584,642,659,720,738,803,822,
%U A346370 890,910,982,1003,1078,1100,1179,1202,1284,1308,1394,1419,1508,1534,1627
%N 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.
%C A346370 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.
%C A346370 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.
%C A346370 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).
%C A346370 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.
%H A346370 Mikhail Kurkov, <a href="https://youtu.be/UGLmI6LArTI">Demonstration of the TRINTUM cube puzzle</a>, YouTube video. [verification needed]
%H A346370 Mikhail Kurkov, <a href="https://i.ibb.co/vYmzt8z/parts.png">Parts of the TRINTUM cube puzzle</a>, image. [verification needed]
%H A346370 Mikhail Kurkov, <a href="https://i.ibb.co/K7sJ03z/cube.png">Solutions for the cube</a>, image. [verification needed]
%H A346370 Mikhail Kurkov, <a href="https://i.ibb.co/k42FpGB/cuboid2.png">Some solutions for the 2 X 1 X 1 cuboid</a>, image. [verification needed]
%H A346370 Mikhail Kurkov, <a href="https://i.ibb.co/q0q3hTy/cuboid3.png">Some solutions for the 3 X 1 X 1 cuboid</a>, image. [verification needed]
%H A346370 Jon Maiga, <a href="http://sequencedb.net/s/A346370">Computer-generated formulas for A346370</a>, Sequence Machine.
%H A346370 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1,1,-1,-1,1).
%F A346370 a(n) = (n + 2)*(floor(n/2) + 1) + floor(floor(n/2)^2/4).
%F A346370 a(n) = A102214(floor(n/2)) + (1 + n mod 2)*(1 + floor(n/2)).
%F A346370 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
%F A346370 Conjecture: a(n) = A008733(n-1) + A093907(n) for n > 0 (noticed by Sequence Machine). - _Mikhail Kurkov_, Oct 14 2021 [verification needed]
%e A346370 a(1) = 3 because there are only 3 possible nonnegative integer solutions:
%e A346370   19*0 + 21*2 + 27*4 + 35*0 = 150, 0 + 2 + 4 + 0 = 6;
%e A346370   19*1 + 21*2 + 27*2 + 35*1 = 150, 1 + 2 + 2 + 1 = 6;
%e A346370   19*2 + 21*2 + 27*0 + 35*2 = 150, 2 + 2 + 0 + 2 = 6.
%e A346370 a(2) = 8 because there are only 8 possible nonnegative integer solutions:
%e A346370   19*0 + 21*4 + 27*6 + 35*0 = 246, 0 + 4 + 6 + 0 = 10;
%e A346370   19*1 + 21*4 + 27*4 + 35*1 = 246, 1 + 4 + 4 + 1 = 10;
%e A346370   19*2 + 21*4 + 27*2 + 35*2 = 246, 2 + 4 + 2 + 2 = 10;
%e A346370   19*3 + 21*4 + 27*0 + 35*3 = 246, 3 + 4 + 0 + 3 = 10;
%e A346370   19*3 + 21*0 + 27*7 + 35*0 = 246, 3 + 0 + 7 + 0 = 10;
%e A346370   19*4 + 21*0 + 27*5 + 35*1 = 246, 4 + 0 + 5 + 1 = 10;
%e A346370   19*5 + 21*0 + 27*3 + 35*2 = 246, 5 + 0 + 3 + 2 = 10;
%e A346370   19*6 + 21*0 + 27*1 + 35*3 = 246, 6 + 0 + 1 + 3 = 10.
%t A346370 LinearRecurrence[{1, 1, -1, 1, -1, -1, 1}, {3, 8, 10, 19, 22, 34,
%t A346370 38}, 52] (* _Robert P. P. McKone_, Jul 16 2021 *)
%Y A346370 Cf. A102214.
%K A346370 nonn,easy
%O A346370 1,1
%A A346370 _Mikhail Kurkov_, Jul 14 2021 [verification needed]