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These terms (greater than 0) are not unique. The first duplicate appears at a(256) = a(2236) = 6711.

Least k such that A000005(A000010(k)) = n.

From Jon E. Schoenfield, Nov 13 2016: (Start)

For every n > 0, phi(2^n) = 2^(n-1) has exactly n divisors, so a(n) <= 2^n.

For every prime p, since phi(a(p)) has exactly p divisors, phi(a(p)) must be of the form q^(p-1), where q is a prime number. If q >= 3, we would have phi(a(p)) >= 3^(p-1), and since k > phi(k) for every k > 1, we would have a(p) >= 3^(p-1)+1, which would be contradicted by the upper bound a(p) <= 2^p (see above) unless 3^(p-1)+1 <= 2^p, which is true only for p = 2. Thus, for every prime p > 2, phi(a(p)) = 2^(p-1), so a(p) > 2^(p-1). In summary, we can state that, for every prime p > 2:

(1) a(p) is the least k such that phi(k) = 2^(p-1), and

(2) 2^(p-1) < a(p) <= 2^p.

After a(36)=1333, the next few known terms are a(38)=786433, a(39)=42653, a(40)=1763, and a(42)=2993; as shown above, known bounds on a(37) and a(41) are 2^36 < a(37) <= 2^37 and 2^40 < a(41) <= 2^41.

For prime p < 37, a(p) = A001317(p-1).

Observation: for prime p < 37, a(p) is the product of distinct Fermat primes 2^(2^j)+1 for j=0..4, i.e., 3, 5, 17, 257, and 65537 (see A019434), according to the locations of the 1-bits in p-1:

. p-1 in

p a(p) prime factorization of a(p) binary

== ========== =========================== ======

2 3 = 3 1

3 5 = 5 10

5 17 = 17 100

7 85 = 17 * 5 110

11 1285 = 257 * 5 1010

13 4369 = 257 * 17 1100

17 65537 = 65537 10000

19 327685 = 65537 * 5 10010

23 5570645 = 65537 * 17 * 5 10110

29 286331153 = 65537 * 257 * 17 11100

31 1431655765 = 65537 * 257 * 17 * 5 11110

.

This pattern does not continue to p=37, since 2^(2^5)+1 is not prime. (See also A038183 and the observation there from Arkadiusz Wesolowski.) (End)

As noted, for every prime p, phi(a(p))=2^(p-1), decompose a(p) = p_1^(e_1) *...* p_m^(e_m), then phi(a(p)) = p_1^(e_1-1)*(p_1 - 1) * ... * p_m^(e_m-1)*(p_m - 1). Thus a(p) is of the form 2^e * F_(a_1) *...* F_(a_l), where F_(a_i) = 2^(a_i) + 1 denote distinct Fermat primes. If e = 0, a_1 + ... + a_l = p - 1, while if e > 0, e + a_1 + ... + a_l = p. It can be deduced that a(p) = 2^p unless p-1 can be written as a_1 + ... a_l where 2^(a_i) + 1 are distinct Fermat primes. The only Fermat primes known have a_i in {1,2,4,8,16} and it is known that 2^a + 1 is composite for 16 < a < 2^33 (cf. A019434). It follows from the fact that 1 + 2 + 4 + 8 + 16 = 31 that a(p) = 2^p for primes p with 32 < p <= 2^33. - Pjotr Buys, Sep 18 2019

It appears that all terms greater than 1 are distinct. This is true for all n <= 10^6.

The larger endpoint is always twice the value of the smaller endpoint.

Conjecture 1: The number of solutions, excluding endpoints is always 0, or an odd number. (known to n = 2 * 10^5)

Conjecture 2: If both endpoints are divisible by 5, then the number of solutions (excluding terms of A007366) is of the form 4k + 1. (known to n = 2 * 10^5)

A007366 is contained in this sequence and the number of solutions, excluding endpoints is always 0.

Terms of this sequence are totients with a single odd totient inverse.

Such an even number always exists.

No terms are zero if Carmichael's conjecture is true.

Even terms are rare: e.g., all inverses of 257*2^16 are even [Foster], so the difference between the largest and smallest inverse is even.

Conjecture 1: If A062402(n) = A000203(A000010(n)) = sigma(phi(n)) is a prime p for some n, then p is Mersenne prime (A000668); a(n) > 0 for all n.

Conjecture 2: a(n) = the smallest number k such that phi(k) has exactly A000043(n)-1 divisors; see A276044.

Conjecture 3: a(n) = the smallest number k such that phi(k) has exactly A000043(n)-1 prime factors (counted with multiplicity); see A275969.

a(n) <= A000668(n) for n = 1-18; conjecture: a(n) <= A000668(n) for all n.

Equals A002181 (index in A002202 of (intersection of A023194 and A002202)). - Michel Marcus, Feb 12 2017