Re: Triangle Numbers Revisited

[hv]

I'm not sure quite how you might take advantage of it for optimisation, but the possible decompositions are restricted by the mod 3 equivalences. That is:

  T_n == 1 (mod 3) when n == 1 (mod 3)
  T_n == 0 (mod 3) otherwise

so if we split the T_n sequence into A_n (== 0) and B_n (== 1) the possible decompositions are restricted such that:

  if n == 0 (mod 3), require A A A or B B B
  if n == 1 (mod 3), require A A B
  if n == 2 (mod 3), require A B B

You can get similar restrictions by considering other prime moduli, but 3 is likely to be the most beneficial because it has a shorter than possible cycle in T_n.

Also, a word of warning on your p_tri() routine - the final result relies on comparing a floating point number for equality, normally considered a bad idea. It would be preferable to do the test instead by calculating from $t back up to the last known integer value, something like:

  sub p_tri {
    my $num = shift;
    my $x = 8 * $num + 1;
    my $t = int((sqrt($x) + 1)/2);
    return +((2 * $t - 1) * (2 * $t - 1) == $x) ? 0 : --$t;
  }

Hugo