OFFSET
1,1
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.
LINKS
Jianing Song, Table of n, a(n) for n = 1..10000
FORMULA
A350345(n) = a(n)^2.
EXAMPLE
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.
PROG
(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)
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Dec 25 2021
STATUS
approved