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These are triplets of numbers into positive integers:

[0,0,0] => 0
[1,0,0] => 1
[1,1,0] => 2
[1,1,1] => 3
[2,0,0] => 4
[0,2,0] => 5
[1,2,0] => 6
[1,2,1] => 7
[1,1,2] => 8
[2,2,0] => 9
[3,1,0] => 10
...
[3,2,1] => 15
...
[3,3,3] => 36
[4,4,4] => 66
[5,5,5] => 105
[6,6,6] => 153

$[10,10,10]$ => $?$

Note :

  • I have checked that every integer from $0$ to $10^6$ can be written into a triplet (so, I believe it's true for every integer);
  • Usually, $[a,b,c] ≠ [b,c,a] ≠ [c,b,a]$;
  • An integer can have some triplet, but not vice versa. Example [2,2,0] = [3,0,0] => 9.

Addition :

I added more examples, so I hope the information is enough now.

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We plainly don't have enough information here, but here is a class of solutions.

[a,b,c] maps to f(a)+g(b)+h(c) where f,g,h are functions with the following properties:
f(0)=0, f(1)=1, f(2)=4, f(3)=9 (so there's an obvious guess for f)
g(0)=0, g(1)=1, g(2)=5
h(0)=0, h(1)=1, h(2)=6
every nonnegative integer has a representation as f+g+h

So it's clear enough that

we can find plenty of f,g,h having these properties (I can give several if requested)

but the challenge is

to find "nice" ones (it seems like g,h "want" to grow faster than the square numbers which introduces a little difficulty in making sure every nonnegative integer is represented).

The obvious guess is

f = square numbers, g = pentagonal numbers, h = hexagonal numbers

and my numerical expectation suggests that this probably works. I haven't attempted to find a proof (either someone else's or my own); it might be difficult to prove if true.

Anyway, the question asked for [10,10,10]. With the representation above that would yield

$10^2 + \frac{1}{2}10(3\cdot10-1) + 10(2\cdot10-1)=100+145+190=435$.

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  • $\begingroup$ I have add some more example, so I hope the information are enaugh now. And your guess is right, so just answer what is [10,10,10] $\endgroup$ – Jamal Senjaya Sep 19 '16 at 10:21
  • $\begingroup$ Done (at about the same time as you were writing that comment). $\endgroup$ – Gareth McCaughan Sep 19 '16 at 10:25

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