View Single Post 2022-05-14, 13:28 #16 Dr Sardonicus   Feb 2017 Nowhere 135568 Posts If I understand the problem correctly, you could (in theory) form your "permuted" integers as follows:Form a vector of length n whose i-th entry is i. Convert the entries to strings (of decimal digits). Permute the entries Concatenate the permuted entries Convert the resulting string to an integer in base ten. If the resulting integer is to be a cube, it has to be congruent to 0, 1 or 8 (mod 9). However, any concatenation of the integers 1 to n is congruent to the sum of the integers 1 to n, or n(n+1)/2 (mod 9). This gives the requirement that n be congruent to 0, 1, 4, 7, or 8 (mod 9). (No number n(n+1)/2 is congruent to 8 (mod 9).) This is of course only a "constant factor" thing. I find the argument persuasive that, when n is "large enough" (and the possibility of cubes is not ruled out by congruences etc) then there will very likely be cubes among the "concatenated permuted" integers 1 to n. I find the argument similarly persuasive for higher powers. I'd call the question an "open question" rather than a "Conjecture." It's a numerical curiosity, but AFAIK does not have any theoretical significance.  