in reply to Triangle Numbers Revisited
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:
so if we split the T_n sequence into A_n (== 0) and B_n (== 1) the possible decompositions are restricted such that:T_n == 1 (mod 3) when n == 1 (mod 3) T_n == 0 (mod 3) otherwise
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
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