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A350344
Composite k such that k^2 is an abelian order.
3
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
1,1
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
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
Cf. A051532 (abelian orders), A050384, A350340.
Equals A350342 \ ({1} U A000040).
Sequence in context: A250764 A362198 A357188 * A245274 A092256 A108172
KEYWORD
nonn
AUTHOR
Jianing Song, Dec 25 2021
STATUS
approved