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A025610 is the range of A319100.

Let b = A319100. Note that:

- if k is an odd number, then b(2*k) = b(k), b(4*k) = 2*b(k), b(2^e*k) = 4*b(k) for e >= 3;

- if k is not divisible by 3, then b(3*k) = 2*b(k), b(3^e*k) = 6*b(k) for e >= 2;

- for all primes p > 3, if k is not divisible by p, then b(p^e*k) = b(p*k).

As a result, it is easy to see that for every n, a(n) is not congruent to 2 modulo 4 and is not divisible by 16 or 27 or p^2 for any prime p > 3.

Let b = A319100 and v(k, p) be the p-adic valuation of k. Note that:

- if k is an odd number, then b(2*k) = b(k), b(4*k) = 2*b(k), b(2^e*k) = 4*b(k) for e >= 3;

- if k is not divisible by 3, then b(3*k) = 2*b(k), b(3^e*k) = 6*b(k) for e >= 2;

- for all primes p > 3, if k is not divisible by p, then b(p^e*k) = b(p*k).

As a result, every term k of this sequence satisfies: v(k, 2) = 0, 2 or 3, v(k, 3) <= 2 and v(k, p) <= 1 for all primes p > 3.

All terms k such that v(k, 3) = 1 are k = 3, 21, 168 and 2184. Proof:

- if k has a prime factor q == 5 (mod 6) or v(k, 2) = 2 (let q = 4), then b(9/(3*q)*k) = 6*b(k/(3*q)) = (3/2)*b(k), but 9/(3*q) < 1;

- if k has a prime factor p == 1 (mod 6) and p >= 19, then b((9*5)/(3*p)*k) = b(k), but (9*5)/(3*p) < 1.

So all terms k such that v(k, 3) = 1 are of the form k = 3*8^e*7^f*13^g, where 0 <= e <= f <= 1, 0 <= g <= f <= 1, but note that 3*7*13 = 273 is not a term because b(252) = b(273).

It is easy to see that all the other terms are of the form F, where F(i,j) = Product_{s=1..i} (p_s)*Product_{t=1..j} (q_t), where p_1 = 7, p_2 = 9, p_s = A002476(s-1) for s >= 3; q_1 = 4, q_2 = 2, q_t = A007528(t-2) for t >= 3. This is because F(i,j) is the smallest number k such that v(k, 3) != 1 and that b(k) = 6^i*2^j. Other than 4, 28 and 2520, the number F(i,j) is a term if and only if for all i', j' such that F(i',j') < F(i,j) we have 6^i'*2^j' < 6^i*2^j. (Note that 4, 28 and 2520 are of the form F and they satisfy this but they are not terms.)

Equivalently, a number k other than 3, 21, 168, 2184 and 4, 28, 2520 is a term if and only if k is of the form F and: (a) for any u, v such that u <= i and 6^u < 2^v, Product_{s=i-u+1..i} (p_s) < Product_{t=j+1..j+v} (q_t); (b) for any u, v such that v <= j and 6^u > 2^v, Product_{s=i+1..i+u} (p_s) > Product_{t=j-v+1..j} (q_t).

Specially, if a number k other than 3, 21, 168, 2184 and 4, 28, 2520 is a term, then k is of the form F and: (a') p_i < (q_(j+1))*(q_(j+2))*(q_(j+3)); (b') p_(i+1) > (q_(j-1))*(q_j). From this we can see that for every prime r, there are only finitely many terms that are not divisible by r (the largest term not divisible by 5 is (7*9*13*19*...*919)*(4*2) = 1.1832*10^190, nevertheless). But note that these are not sufficient. For example, k = (7*9*13*19*...*55819*55831*55837*55849)*(4*2*5*11*17*23) is not a term because although 55849 < 29*41*47, 55831*55837*55849 > 29*41*...*89 so k' = (7*9*13*19*...*55819)*(4*2*5*11*...*89) have k' < k but b(k') = (256/216)*b(k). Similarly, k = (7*9*13*19*...*643*661)*(4*2*5*11*17*23*29) is not a term because although 23*29 < 673, 5*11*17*23*29 > 673*691 so k' = (7*9*13*19*...*643*661*673*691)*(4*2) have k' < k but b(k') = (36/32)*b(k).

All terms are of the form 6^u*2^j. Other than the term 48, k = 6^i*2^j is a term if and only if for all i', j' such that F(i',j') < F(i,j) we have 6^i'*2^j' < 6^i*2^j, where F(i,j) = Product_{s=1..i} (p_s)*Product_{t=1..j} (q_t), where p_1 = 7, p_2 = 9, p_s = A002476(s-1) for s >= 3; q_1 = 4, q_2 = 2, q_t = A007528(t-2) for t >= 3. Or equivalently: (a) for any u, v such that u <= i and 6^u < 2^v, Product_{s=i-u+1..i} (p_s) < Product_{t=j+1..j+v} (q_t); (b) for any u, v such that v <= j and 6^u > 2^v, Product_{s=i+1..i+u} (p_s) > Product_{t=j-v+1..j} (q_t). For example, 746496 = 6^6*2^4 is not a term because (q_3)*(q_4) = 5*11 > p_7 = 43.

Numbers k such that the number of solutions to x^6 == 1 (mod k) is a power of 6.

Also numbers k such that (Z/kZ)* has the same 2-rank and 3-rank, where (Z/kZ)* is the multiplicative group of integers modulo k, and the p-rank of a finite abelian group G is equal to log_p(#{x belongs to G : x^p = 1}) with p being a prime number.

k is a term in this sequence iff v(2, k) = 0 or 1, v(3, k) = 0 or >= 2 and k is not divisible by any prime p == 5 (mod 6). Here v(p, k) is the p-adic valuation of k.

Sequence contains all primes p == 1 (mod 6) and their powers and all powers of 3 except for 3 itself.

Decompose the multiplicative group of integers modulo k as a product of cyclic groups C_{s_1} x C_{s_2} x ... x C_{s_m}, where s_i divides s_j for i < j, then k is a term iff s_1 is divisible by 6. For k = 1 or 2, (Z/kZ)* is the trivial group, s_1 does not exist so 1 and 2 are also terms.

If gcd(k_1, k_2) = 1 and both k_1 and k_2 are in this sequence, so is k_1*k_2. For example, 7 and 9 are both here so 7*9 = 63 is also here. Indeed, the number of solutions to x^6 == 1 (mod 7), x^6 == 1 (mod 9) and x^6 == 1 (mod 36) are 6, 6 and 36, respectively.

This is an analog of A008784, since k is a term there iff s_1 (defined as above) is divisible by 4 instead of 6. But on the other hand, if k is in A008784, so are all its divisors, while this is not true for this sequence. However, if k is here and k is not divisible by 9, then all its divisors are also here.

This is a also an analog of A192453 (s_1 divisible by 8).

Sum_{k=1..n} a(k) appears to be asymptotic to C*n*log(n) with C = 0.4... - Benoit Cloitre, Aug 19 2002 [C = (11/(6*Pi*sqrt(3))) * Product_{p prime == 1 (mod 3)} (1 - 2/(p*(p+1))) = 0.3170565167... (Finch and Sebah, 2006). - Amiram Eldar, Mar 26 2021]

All terms are powers of 5. Those n such that a(n) > 1 are in A066500.

All terms are powers of 7. Those n such that a(n) > 1 are in A066502.

Number of solutions to x^8 == 1 (mod n). - Jianing Song, Nov 10 2019

a(n) is the number of primitive Dirichlet characters modulo n such that all entries are 0 or a six-power root of unity (1, (1 + sqrt(3)*i)/2, (-1 + sqrt(3)*i)/2, -1, (-1 - sqrt(3)*i)/2, (1 - sqrt(3)*i)/2).

Mobius transform of A319100.

Complement of A192452. Subsequence of A008784. A further reduction to 8th powers yields 1, 2, 17, 34, 97, 113, 193, 194, ...

From Jianing Song, Mar 31 2019: (Start)

k is a term if and only if k is not divisible by 4 and all odd prime factors are congruent to 1 modulo 8. If k is a term of this sequence, then so are all divisors of k.

Decompose the multiplicative group of integers modulo k as a product of cyclic groups C_{s_1} x C_{s_2} x ... x C_{s_m}, where s_i divides s_j for i < j, then k is a term iff s_1 is divisible by 8. For k = 1 or 2, (Z/kZ)* is the trivial group, s_1 does not exist so 1 and 2 are also terms. This is an analog of A008784 (where s_1 is divisible by 4) and A319100 (where s_1 is divisible by 6). (End)