I got to thinking about the conditions for a number to be divisible by its totient. It didn't take too long to see the pattern. Figure 1 shows the results for numbers up to 1024.
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Figure 1: permalink |
Clearly condition is that the numbers must by of the form 2p3q where p>0 and q≥0. There are only 35 numbers in this range and, if we extend the range to 10 million, there are still only 178 numbers that satisfy.
These numbers form OEIS A007694:
A007694 | Numbers k such that ϕ(k) divides k. |
The initial members of the sequence are:
1, 2, 4, 6, 8, 12, 16, 18, 24, 32, 36, 48, 54, 64, 72, 96, 108, 128, 144, 162, 192, 216, 256, 288, 324, 384, 432, 486, 512, 576, 648, 768, 864, 972, 1024, 1152, 1296, 1458, 1536, 1728, 1944, 2048, 2304, 2592, 2916, 3072, 3456, 3888, 4096, 4374, 4608, 5184, 5832, 6144, 6912, 7776, 8192, 8748, 9216
The numbers must be even, that is they must contain a power of 2. If the numbers are only powers of 3 then the dividend is 1.5. Figure 2 shows the results in the range up to one million. All numbers are of the form 3p where p>0.
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Figure 2: permalink |
What sort of numbers will produce a dividend of 2.5? Well, as it turns out, numbers of the form 2p5q where p>1 and q>1. See Figure 3 for the numbers in the range up to one thousand.
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Figure 3: permalink |
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Figure 4: permalink |
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Figure 5: permalink |
Numbers of the from 2p11q23q with p>0, q>0 and r>0 produce a dividend of 2.3. See Figure 6 where the range is up to 100,000.
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Figure 6: permalink |
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Figure 7: permalink |
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Figure 8: permalink |
- nϕ(n)=k where k>0 if numbers of form 2p3q with p>0 and q≥0
- nϕ(n)=1.5 if numbers of form 3p with p>0
- nϕ(n)=2.2 if numbers of form 2p11q where p>0 and q>0
- nϕ(n)=2.3 if numbers of form 2p11q23q with p>0, q>0 and r>0
- nϕ(n)=2.5 if numbers of form 2p5q with p>0 and q>0
- nϕ(n)=3.1 if numbers of form 2p3q31r where p>0, q>0 and r>0
- nϕ(n)=3.3 if numbers of form 2p3q11r where p>0, q>0 and r>0
- nϕ(n)=3.5 if numbers of form 2p3q7r with p>0, q>0 and r>0
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