A319838 Numbers k such that A319100(k) is a power of 6.

1, 2, 7, 9, 13, 14, 18, 19, 26, 27, 31, 37, 38, 43, 49, 54, 61, 62, 63, 67, 73, 74, 79, 81, 86, 91, 97, 98, 103, 109, 117, 122, 126, 127, 133, 134, 139, 146, 151, 157, 158, 162, 163, 169, 171, 181, 182, 189, 193, 194, 199, 206, 211, 217, 218, 223, 229, 234, 241

Offset: 1

Author

Jianing Song, Sep 28 2018

Keywords

• nonn

Comments

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).

Examples

			91 = 7*13 is a term since the number of solutions to x^6 == 1 (mod 91) is 36 = 6^2.
1197 = 7*9*19 is a term since the number of solutions to x^6 == 1 (mod 1197) is 216 = 6^3.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..15917 (all terms below 100000)

Crossrefs

Cf. A008784, A192453, A319100.

Programs

PARI

isA319838(n) = if(abs(n)==1||abs(n)==2, 1, my(i=znstar(n)[2]); !(i[#i]%6)==1, 1)