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We have a 6x6x6 room and 216 cubes each of dimensions 1x1x1, 1st we place a block at the corner, each time we must place cubes such that the previous layer is completely covered, for example to cover the first cube we place 3 more cubes such that each of it’s 3 dimensions are covered, thereon we place 6 cubes to cover the previous layer and so on. The room has walls. If we place "n" cubes in each successive time, what is series we get in the value of "n". Uptil 6 I got a clear pattern that it will just be

$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

But I am unsure about the pattern thereon, I want to verify whether it will just continue the pattern but descend.

This is a question I made to solve a probability problem in a rather unique method to save time in the future and want to verify my solution.

By hit and trial I got the pattern, 1,3,6,10,15,21,25,27,27,25,21,15,10,6,3,1, but to further up the same process in 4 dimensions I would have to understand this pattern furst

I am trying to solve the probability of getting the sum "x" when we roll 3 die together and get n1, n2, n3 on respective dice, such that their sum is x. I want a singular expression for P(Sum=x)

Kushman
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  • I do not understand everything you did but your pattern is the number of equally probable ways of getting $3,4,5,\ldots,18$ if you roll three $6$-sided dice and that is the coefficient of $x^k$ in the the expansion of $\left(\frac{x(1-x^6)}{1-x}\right)^3$. It appears in three OEIS sequences – Henry Feb 03 '25 at 18:20
  • You are correct about the sequence being ways of getting 3,4,5…18. That is exactly what I was trying to find. However this problem lead me to another problem, of covering cubes. However, can you please explain me about how you got it to be equal to the expansion of the function in x you gave? – Kushman Feb 03 '25 at 18:31
  • My expression is just a generating function. It could also be written as $(x+x^2+x^3+x^4+x^5+x^6)^3$. Divide it by $6^3$ and it becomes a probability generating function. – Henry Feb 03 '25 at 18:40
  • How did you get to this generating function? I am new to mathematics, sorry for bothering you – Kushman Feb 03 '25 at 19:23
  • Experience - it is a particularly simple example. – Henry Feb 03 '25 at 19:28
  • That’s great! You’ve been studying math for as long as I have been alive. But, could you share the reference, from where you got it. Or maybe it is out of pure intuition that comes from experience? – Kushman Feb 04 '25 at 15:57
  • Perhaps longer. The question and answer at https://math.stackexchange.com/questions/510183/about-the-generating-function-of-the-sum-of-roll-dice-values looks at generating functions and dice, which you may find of interest. – Henry Feb 04 '25 at 16:46
  • To see that the sequence does not immediately reverse itself and start back down the triangular numbers, one only has to look at a cube. The layers of new cubes laid down are parallel to each other and perpendicular to a full diagonal of the cube. Hold the cube so that full diagonal is vertical (so a pair of opposing vertices are vertically aligned). The 8 vertices of the cube are in 4 horizontal layers. The top and bottom layers just have 1 vertex each, while the middle two layers have 3 vertices each. – Paul Sinclair Feb 05 '25 at 19:01
  • Your analysis covers from the bottom to the lower middle layer of vertices. From the higher middle layer to the top is where it will do the reverse of what you covered. But in between the two middle vertex layers, the count of added cubes is going to follow a different pattern. – Paul Sinclair Feb 05 '25 at 19:03

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Sorry for the rough answer but please feel free to ask any doubts related to the answer if it is not clear So basically your problem can be properly analyzed in this way: Consider we have a single cube, it has a single face diagonal If I follow your algorithm and only notice the number of upper face diagonal, somehow it turns out for step two, for the on the top layer (h = 2 units here) now there is one upper face diagonal and in the lower layer 2 additional upper diagonals have been added, hence the new number of upper body diagonals added is 1 + 2 = 3 and the total number of upper body diagonals i.e. cubes are 4.

Repeating the step for the third time, in the uppermost layer there is a single upper face diagonal, in the second layer, there are 1+2 upper face diagonals and in the last layer there are now 1 + 2 + 3 upper face diagonals and hence the new number of cubes added would be 1 + (1+2) + (1+2+3) - 4 = 6 cubes( also 1+2+3 that is n(n+1)/2 where n is the nth step)

Now, here's what the pattern will look like when there are you apply constraints on say a 3*3 space

first step =1

second step =1 + (1+2)

third step = 1 + (1+2) + (1+2+3)

now for the fourth step, one can clearly observe that no cube can be added to the existing top layer anymore vertically and that (1+2+3) in the lowest layer implies the face diagonal of the cube(that is the 3 upper face diagonals of the unit cubes here) has been counted in the third step and hence we have to now treat the problem differently

Now clearly in the third layer one has to add 2 cubes i.e 2 upper face diagonals that is two cubes only to make sure the sides for the third step are covered across length and width

In the 2nd layer one again adds according to the algorithm and this will take care of the remaining uncovered faces of the third layer as well

In the first layer one adds 2 cubes in accordance with the algorithm and since our height is constrained we don't add an extra layer to cover the remaining uncovered faces

therefore the 4th step is : (1+2) + (1+2+3) + (1+2+3+2) and the no of new cubes added is 7

Similarly for the 5th step it is (1+2+3) + (1+2+3+2) + (1+2+3+2+1) Note that the third layer has been completely filled in the fifth step ( 9 upper face diagonals) and the number of new cubes added is 6

For the 6th step it is (1+2+3+2) + (1+2+3+2+1) + (1+2+3+2+1) Hence the 2nd layer has also been completely filled in the 6th step and the number of new cubes added is 3

And for the 7th step which is the last step the number of new cubes added is 1.

If you understand this method, you can fairly easily find out the number of new cubes added for each "step" for a 6*6 situation as well

For example after the 6th step the number of cubes added would be (5 for the lowest layer) + (6 for the second lowest layer) + (5 for the third lowest layer)....+ (2 for the topmost layer) and 5+ 6+ 5 + 4 +...2 + (1 - 1) = 11 + 5(5+1)/2 - 1 = 10 + 15 = 25 which is exactly what you've found out aswell.

I am extremely sorry for not being able to send an obviously much more required clearer solution to your problem as it is pretty late here in my part of the world and i have a hectic college schedule too but i will try to send a more elaborately explained solution sometime later with images if required for understanding if it is needed.

I imagined this problem as a problem of putting together squares in such a way no side was not a common side for two squares and that for each step a layer or like a plane with a square also got added which also had to follow the same constraints as the initial layer or plane of squares and somehow it just works out.

Hope this helps and please ask if you aren't able to understand something.

I am currently taking a course on probability and am somewhat interested in the subject so i am interested in how connected these two problems together so it would be my pleasure if you explained your rough idea:)

hellofrndz
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