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### Re: Triangle Numbers Revisited

by hv (Parson)
 on Oct 14, 2004 at 12:47 UTC ( #399183=note: print w/replies, xml ) Need Help??

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:

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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