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
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