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For even number n, if n is not a totient (i.e. not in the range of Euler totient function), then n is not in OEIS sequence A252502, but if n is a totient, must be n in A252502?

For odd number n, if n1 is not a totient (i.e. not in the range of Euler totient function), then n is not in OEIS sequence A252502 (except n=1), but if n1 is a totient, must be n in A252502?

If this conjecture is true, then the famous conjecture that there is no n such that there is only one x whose Euler totient function is n is also true.

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For numbers that are twice a prime, consider 2×3=6 as an example. Since this is twice an odd prime and also one less than the odd prime 7, it is the Euler totient of 14. Then we have a cyclotomic polynomial x6x5+x4x3+x2x+1 of order 14, so constructed that for x=10 the value lies strictly between 9×105 and 106 hence having six digits. Thus 6 enters the sequence, and similarly for any number that is both twice a prime and one less than a larger prime.

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  • Then A252502(14)=6 and A252502(7)=7, so 6 and 7 are not counterexamples. Commented 2 hours ago
  • I did not lanel them counterexampkes. To the contrary, I am claiming that all twice-primes occur if they have inverse totients. Commented 1 hour ago
  • I just noted that A252502(n)=20 for n=33 and n=44, but A252502(n)=21 only for n=66, I doubt that there is a number k such that k or k-1 is a totient but there is no n such that A252502(n)=k Commented 44 secs ago

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