User blog comment:Wythagoras/Coins in boxes/@comment-5150073-20130812080915

The definition you gave to f(n) is poor, as you wrote: "any possible expression", and meant just "when each box contains 1". So, we redefine the function that. As for the values, Goucher gave me a clue that the expression:

"0 0 0 ... 0 0 0 n 0 0 0 ... 0 0 0" (with m zeroes to the right and finitely many zeroes to the left) can be transformed into:

"0 0 0 ... 0 0 0 0 $$2 \uparrow^{m-1} n$$ 0 0 ... 0 0"

Using that, we can get the following process (and possible value for f(6)) of solving f(6): 1 1 1 1 1 1 0 3 1 1 1 1 0 2 3 1 1 1 0 2 2 3 1 1 0 2 2 2 3 1 0 2 2 2 0 7 0 2 2 1 7 0 0 2 2 1 0 14 0 2 2 0 14 0 0 2 1 14 0 0 0 2 1 0 2^14 0 0 2 0 2^14 0 0 0 1 2^14 0 0 0 0 1 0 2^^(2^14) 0 0 0 0 2^^(2^14) 0 0 0 0 0 0 2^^(2^^(2^14)) 0 0 0 0 0 0 2^(2^^(2^^(2^14))) 0 0 0 0 0 0 2^(2^^(2^^(2^14))+1)

So, the largest lower bound I found for f(6) is $$2^{2\uparrow\uparrow 2\uparrow\uparrow 2^{14}+1}$$.

Actually, I have good bound for f(n) with arbitrary n. Consider f(7) to spot the pattern:

1 1 1 1 1 1 1 0 3 1 1 1 1 1 0 2 3 1 1 1 1 0 2 2 3 1 1 1 0 2 2 2 3 1 1 0 2 2 2 2 3 1 0 2 2 2 2 0 7 0 2 2 2 1 7 0 0 2 2 2 1 0 14 0 2 2 2 0 14 0 0 2 2 1 14 0 0 0 2 2 1 0 2^14 0 0 2 2 0 2^14 0 0 0 2 1 2^14 0 0 0 0 2 1 0 2^^(2^14) 0 0 0 2 0 2^^(2^14) 0 0 0 0 1 2^^(2^14) 0 0 0 0 0 1 0 2^^^(2^^(2^14)) 0 0 0 0 0 2^^^(2^^(2^14)) 0 0 0 0 0 0 0 2^^^(2^^^(2^^(2^14))) 0 0 0 0 0 0 0 2^^(2^^^(2^^^(2^^(2^14)))) 0 0 0 0 0 0 0 2^(2^^(2^^^(2^^^(2^^(2^14))))) 0 0 0 0 0 0 0 2^(2^^(2^^^(2^^^(2^^(2^14))))+1)

We can rewrite our expression as 2^(2^^(2^^^n)) = 2^(2^^^(n+1)) < 2^^^(n+2). This shows that the leftmost tetration and exponentiation doesn't change the hyper-magnitude of entire number sufficiently (as n is itself the number in the pentational range).

Another thing to note is that we removed the leftmost 2^^ and appended two 2^^^ in order to change the lower bound from f(6) to f(7). This makes sense if we extrapolate it to arbitrary n and say that we remove the leftmost $$2\uparrow^{n-4}$$ and append two $$2\uparrow^{n-3}$$'s in order to change the lower bound from f(n) to f(n+1), but consider the process more carefully, as we have:

1 1 1 1 ... 1 1 1 1 0 3 1 1 ... 1 1 1 1 0 2 3 1 ... 1 1 1 1 0 2 2 3 ... 1 1 1 1 0 2 2 2 ... 1 1 1 1 0 2 2 2 ... 3 1 1 1 0 2 2 2 ... 2 3 1 1 0 2 2 2 ... 2 2 3 1 0 2 2 2 ... 2 2 0 7 0 2 2 2 ... 2 1 7 0 0 2 2 2 ... 2 1 0 14 0 2 2 2 ... 2 0 14 0 0 2 2 2 ... 1 14 0 0 0 2 2 2 ... 1 0 2^14 0 0 2 2 2 ... 0 2^14 0 0 0 2 2 2 ... 2^14 0 0 0 0 2 2 2 ... 0 2^^(2^14) 0 0 0 2 2 2 ... 2^^(2^14) 0 0 0 0 2 2 2 ... 0 0 0 0 0 2 2 1 ... 0 0 0 0 0 2 2 0 ... 0 0 0 0 0 2 1 $$2 \uparrow^m n$$ ... 0 0 0 0 0 2 1 0 ... 0 0 0 0 0 2 0 $$2 \uparrow^{m+1} (2 \uparrow^m n)$$ ... 0 0 0 0 0 1 $$2 \uparrow^{m+1} (2 \uparrow^m n)$$ 0 ... 0 0 0 0 0 1 0 $$2 \uparrow^{m+2} (2 \uparrow^{m+1} (2 \uparrow^m n))$$ ... 0 0 0 0 0 0 $$2 \uparrow^{m+2} (2 \uparrow^{m+1} (2 \uparrow^m n))$$ 0 ... 0 0 0 0 0 0 0 $$2 \uparrow^{m+2} (2 \uparrow^{m+2} (2 \uparrow^{m+1} (2 \uparrow^m n)))$$ ... 0 0 0 0

So, I can found that $$f(n) > 2 \uparrow^{n-4} 2 \uparrow^{n-4} 2 \uparrow^{n-3} 2 \uparrow^{n-2} \cdots 2 \uparrow\uparrow\uparrow 2 \uparrow\uparrow 2^{14}$$ It's strictly larger, since we get this expression only at 4-th box, and we know that the largest number becomes only in the last box (but in terms of hyper-magnitude, both numbers almost the same).