Persistent Tau Numbers

I got to thinking about tau numbers, those numbers that are divisible by their number of divisors, because it's relevant to my divisors sequence which arises when these rules are applied recursively to a positive integer. The rules for a number \(n\) are:

Let \(d\) be the number of divisors of \(n\)

• if \(d \,| \,n \) then \( n \rightarrow d \, | \,n \)
• if \( d \, \nmid \, n\) then \(n \rightarrow d \times n\)
• if \(d\) is prime then:
• if \(d\) is a \(4k+1 \) prime then \( d \rightarrow 2 \times d +1\)
• if \(d\) is a \(4k+3 \) prime then \( d \rightarrow \dfrac {d -1}{2}\)

I'll get back to that sequence later but what interested me were those tau numbers that remain tau numbers after being divided by their number of digits. I got Gemini to investigate and the results revealed that there are 160 such numbers in the range from 1 to 40000 (permalink) and they are:

1, 2, 8, 12, 80, 96, 108, 240, 384, 640, 720, 864, 896, 1200, 1408, 1440, 1664, 1920, 2016, 2176, 2240, 2432, 2560, 2688, 2944, 3168, 3456, 3600, 3712, 3744, 3968, 4032, 4224, 4736, 4896, 4992, 5040, 5248, 5472, 5504, 6016, 6528, 6624, 6784, 7296, 7552, 7808, 7920, 8100, 8352, 8576, 8832, 8928, 9088, 9344, 9360, 10112, 10624, 10656, 10800, 11136, 11392, 11808, 11904, 12240, 12384, 12416, 12928, 13184, 13536, 13680, 13696, 13952, 14208, 14464, 15264, 15744, 16256, 16512, 16560, 16768, 16992, 17536, 17568, 17792, 17920, 18048, 19072, 19296, 19328, 20096, 20160, 20352, 20448, 20864, 20880, 21024, 21376, 22144, 22320, 22656, 22752, 22912, 23040, 23168, 23424, 23904, 24300, 24448, 24704, 25216, 25472, 25632, 25728, 26244, 26640, 27008, 27264, 27936, 28032, 28224, 28544, 29056, 29088, 29312, 29520, 29664, 29824, 30000, 30336, 30592, 30720, 30816, 30848, 30960, 31392, 31872, 32128, 32544, 32896, 33664, 33840, 34176, 34432, 34560, 34688, 35456, 35968, 36224, 36576, 37248, 37504, 37728, 38160, 38400, 38784, 39296, 39456, 39552, 39808

Most survive only one division but some go further and these are (in the given range):

• \(80 \rightarrow 8 \rightarrow 2 \rightarrow 1 \)
• \(96 \rightarrow 8 \rightarrow 2 \rightarrow 1 \)
• \(240 \rightarrow 12 \rightarrow 2 \rightarrow 1 \)
• \(2240 \rightarrow 80 \rightarrow 8 \rightarrow 2 \rightarrow 1 \)
• \(3456 \rightarrow 108 \rightarrow 9 \rightarrow 3 \)
• \(3600 \rightarrow 80 \rightarrow 8 \rightarrow 2 \rightarrow 1 \)
• \(4032 \rightarrow 96 \rightarrow 8 \rightarrow 2 \rightarrow 1 \)
• \(23040 \rightarrow 384 \rightarrow 24 \rightarrow 3 \)
• \(30720 \rightarrow 640 \rightarrow 40 \rightarrow 5 \)
• \(38400 \rightarrow 640 \rightarrow 40 \rightarrow 5 \)

I discussed the divisor sequence in my post titled Number's Divisors to Sequence Algorithm. Applied to 23040 (one of the numbers listed in the bullet points above), the sequence is very short indeed: 23040, 384, 24, 3, 1 Once the sequence reaches 3 then, because it is a \(4k+3\) prime, 3 becomes 1 and 1 has one divisor and so 1 times 1 or 1 divided by 1 is still 1.

Only 5.257% of the numbers up to 100,000 are tau numbers, so the divisors to sequence algorithm is strongly weighted toward multiplication. A composite number is about 19 times more likely to get multiplied than it is to get divided. The primes strike a balance because of the \(4k+1\) versus \(4k+3\) rule. The rule is actually not needed as primes always have two divisors. I introduced the prime rule because in the two factors to sequence algorithms (using prime factors with and without multiplicity), the primes have no prime factors.

What if instead of multiplying by the number of divisors, we simply added 1? It would be interesting to see what sort of sequence develops if we do this.