Conjecture All triangular numbers T(i) in the recursive series i(0) = 0, i(1) = k*2+1 and following the recursive relationship i(j) = 6*i(j-1) - i(j-2) + 2 can be factured into the product n*(n+k) where n and k are integers. where k = 0 these are the square triangular numbers. I know that the series relationship for square triangular numbers is well known, but has anyone before proven or made the above conjecture?(adsbygoogle = window.adsbygoogle || []).push({});

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# Triangular numbers facturable into n*(n+k)

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