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Table of n, a(n) for n=1..35.
6552 = 14*18*26 = 13*21*24. Each triples' factors sum to 58 (14+18+26 = 58 and 13+21+24 = 58). 58 works out to be the minimal sum for all 3-factorizations of 6552 and there are two of them.
Sequence in context: A056492 A158308 A205738 * A140801 A236059 A206670
Adjacent sequences: A112533 A112534 A112535 * A112537 A112538 A112539
Lucas Finn (lucas.finn(AT)tufts.edu) and Bruce Boghosian, Dec 13 2005