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> Of course, this really only works well for numbers with prime factors drawn from the set {2, 3, 5, 7}. So express the number as a sum of numbers with factors
by Bakkot 14y ago
> Of course, this really only works well for numbers with prime factors drawn from the set {2, 3, 5, 7}.
So express the number as a sum of numbers with factors drawn from that set, and draw a several diagrams, one for each summand. A quick bit of Python confirms that every number under 14925 can be expressed as the sum of no more than three such numbers in the most obvious way possible (ie using a greedy algorithm).
For example, you might write 9999 as 9800 + 196 + 3, all of which have nice factorization diagrams.
- ggchappell 14y agoCould it be that every positive integer is the sum of at most k such numbers, for some fixed k? EDIT: No. Proof. Fix a positive integer k. How many k-tuples of such numbers are there, whose sum is at most x? (For this to work, there would always need to be at least x of them.) If (2^a_1 + 3^b_1 + 5^c_1 + 7^d_1) + ... + (2^a_k + 3^b_k + 5^c_k + 7^d_k) <= x, then all the exponents are at most log_2(x). So the number of all such sums is at most [log_2(x)]^(4k). And that is asymptotically less than x. EDIT 2: But I'm sure the required k grows pretty slowly. This would be a workable technique for very large numbers indeed.
- grimboy 14y agoI feel like expanding that set a little is really only a matter of picking a more distinctive arrangement than a circle. Like 11 would be more readable as 3 vertical lines of 4, then 3, then 4 again. Just varying the shape a little would go a long way towards making them more readable, and you could still fall back to a circle for larger primes.
- tzs 14y agoIt is not clear to me that writing 9999 as 9800 + 196 + 3 is useful, because I don't see any way to readily infer anything about the factorization of 9999 from the factorizations of 9800, 196, and 3.