TL;DR

Where we remove all programming joy from this nice puzzle.

Do you see any pattern?

$ for N in $(seq 10 20) ; do ./run.sh 04-luckier-sum "$N" ; done
solution => [6,2,1,0,0,0,1,0,0,0]
solution => [7,2,1,0,0,0,0,1,0,0,0]
solution => [8,2,1,0,0,0,0,0,1,0,0,0]
solution => [9,2,1,0,0,0,0,0,0,1,0,0,0]
solution => [10,2,1,0,0,0,0,0,0,0,1,0,0,0]
solution => [11,2,1,0,0,0,0,0,0,0,0,1,0,0,0]
solution => [12,2,1,0,0,0,0,0,0,0,0,0,1,0,0,0]
solution => [13,2,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0]
solution => [14,2,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0]
solution => [15,2,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0]
solution => [16,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0]

It seems that this would always be a solution, at least for $N$ sufficiently large:

  • 0 contains value $N - 4$
  • 1 contains value $2$
  • 2 contains value $1$
  • N-4 contains value $1$
  • everything else is $0$.

When $N > 6$, then $N - 4 > 2$ which is the condition in which slot N-4 does not overlap with any of the other three slots that have non-zero values.

Is this always a solution for $N > 6$ a.k.a. $N - 4 > 2$? Yes it is:

  • 0, 1, 2, and N-1 are 4 distinct slots, because $N-4>2$;
  • these are the only slots holding a value different from $0$;
  • all the other slots (i.e. $N - 4$ of all slots) hold value $0$, which is consistent with the value at slot 0;
  • value $1$ appears exactly 2 times (in slot 2 and N-4), and slot 1 contains value $2$;
  • value $2$ appears exactly once (in slot 1), and slot 2 contains value $1$;
  • value $N-4$ appears exactly once (in slot 0), and slot N-4 contains value $1$.

So there’s no need for complicated searches for $N > 6$: just provide the solution according to the pattern above.

sub autobiographical_numbers ($n) {
    my @solution;
    if ($n == 4) {
        @solution = (1, 2, 1, 0); # also good: (2, 0, 2, 0)
    }
    elsif ($n > 6) {
        @solution = (0) x $n;
        @solution[0, 1, 2, $n - 4] = ($n - 4, 2, 1, 1);
    }
    return {solution => [map {+{$_ => 1}} @solution]};
}

Find all of this at stage 5.

How boring. And yet… are these the only solutions?!? E.g. $N = 4$ allows two different solutions… is it possible elsewhere?!?

The end of it

Curious about the whole series? Here it is:

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