A350342 Numbers k such that k^2 is an abelian order.

1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 35, 37, 41, 43, 47, 53, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 97, 101, 103, 107, 109, 113, 115, 119, 127, 131, 133, 137, 139, 143, 149, 151, 157, 161, 163, 167, 173, 179, 181, 185, 187, 191, 193, 197, 199, 209

Offset: 1

Author

Jianing Song, Dec 25 2021

Keywords

• nonn

Comments

Such k must be squarefree. Actually, such k must be a cyclic number (A003277).

Number of the form p_1*p_2*...*p_r where the p_i are distinct primes and no (p_j)^2-1 is divisible by any p_i.

The smallest term with exactly n distinct prime factors is given by A350340.

From the term 5 on, no term can be divisible by 2 or 3.

Examples

			For primes p, p is a term since every group of order p^2 is abelian. Such group is isomorphic to either C_{p^2} or C_p X C_p.
For primes p, q, if p^2 !== 1 (mod q), q^2 !== 1 (mod p), then p*q is a term since every group of that order is abelian. Such group is isomorphic to C_{p^2*q^2}, C_p X C_{p*q^2}, C_q X C_{p^2*q} or C_{p*q} X C_{p*q}.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A051532 (abelian orders), A003277 (cyclic numbers), A350343, A350340.

A350344 is the subsequence of composite numbers.

Programs

PARI

isA051532(n) = my(f=factor(n), v=vector(#f[, 1])); for(i=1, #v, if(f[i, 2]>2, return(0), v[i]=f[i, 1]^f[i, 2])); for(i=1, #v, for(j=i+1, #v, if(v[i]%f[j, 1]==1 || v[j]%f[i, 1]==1, return(0)))); 1 \\ Charles R Greathouse IV's program for A051532
isA350342(n) = isA051532(n^2)

Formula

A350343(n) = a(n)^2.