A350344 Composite k such that k^2 is an abelian order.

35, 65, 77, 85, 115, 119, 133, 143, 161, 185, 187, 209, 215, 217, 221, 235, 247, 259, 265, 299, 319, 323, 329, 335, 341, 365, 371, 377, 391, 403, 407, 413, 415, 427, 437, 451, 469, 481, 485, 493, 511, 515, 517, 527, 533, 535, 551, 553, 559, 565, 583, 589, 595

Offset: 1

Author

Jianing Song, Dec 25 2021

Keywords

• nonn

Comments

Numbers k such that k^2 is an abelian order with at least 4 groups.

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

The smallest number k such that k^2 is an abelian order with at least 8 groups is A350340(3) = 595.

No term can be divisible by 2 or 3.

Examples

			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}.
95 is not a term since 95^2 = 5^2 * 19^2 is not an abelian order. Note that 95 itself is a cyclic number.
		

Links

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

Crossrefs

Cf. A051532 (abelian orders), A050384, A350340.

Equals A350342 \ ({1} U A000040).

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
isA350344(n) = (n>1) && !isprime(n) && isA051532(n^2)

Formula

A350345(n) = a(n)^2.