OFFSET
0,2
COMMENTS
If m is an abelian order, then m = (p_1)^2 * (p_2)^2 * ... * (p_r)^2 * q_1 * q_2 * ... * q_s, where p_1, p_2, ... p_r, q_1, q_2, ..., q_s are distinct primes such that (p_i)^2 !== 1 (mod p_j) for i != j, (p_i)^2 !== 1 (mod q_j), q_i !== 1 (mod p_j), q_i !== 1 (mod q_j) for i != j. In this case there are 2^r groups of order m.
Note that the smallest abelian order with precisely 2^n groups must be the square of a squarefree number.
a(n) is the smallest k with n distinct prime factors such that k^2 is an abelian order.
a(n) is the smallest number of the form p_1*p_2*...*p_n where the p_i are distinct primes and no (p_j)^2-1 is divisible by any p_i.
a(n) exists for all n.
Except for a(1) = 2, no term can be divisible by 2 or 3. Conjecture: lpf(a(n+1)) >= lpf(a(n)) for all n, where lpf = least prime factor. - David A. Corneth and Jianing Song, Jan 03 2022
LINKS
David A. Corneth, Table of n, a(n) for n = 0..30
David A. Corneth, PARI program
FORMULA
A350341(n) = a(n)^2.
EXAMPLE
a(2) = 35 = 5*7 since the smallest k with 2 distinct prime factors such that k^2 is an abelian order is 35.
a(3) = 595 = 5*7*17 since the smallest k with 3 distinct prime factors such that k^2 is an abelian order is 595.
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
a(n) = for(k=1, oo, if(issquarefree(k) && omega(k)==n && isA051532(k^2), return(k)))
CROSSREFS
KEYWORD
nonn,hard
AUTHOR
Jianing Song, Dec 25 2021
EXTENSIONS
a(7)-a(18) from David A. Corneth, Jan 02 2022
STATUS
approved