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A350340 a(n) is the smallest k such that k^2 is an abelian order with precisely 2^n groups.

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%I A350340 #28 Jan 11 2022 22:13:53
%S A350340 1,2,35,595,13685,506345,26836285,1702480351,80016576497,
%T A350340 5681176931287,414725915983951,40228413850443247,4304440281997427429,
%U A350340 546663915813673283483,75986284298100586404137,10144780646398552482233711,1511572316313384319852822939,252432576824335181415421430813,49729217634394030738838021870161
%N A350340 a(n) is the smallest k such that k^2 is an abelian order with precisely 2^n groups.
%C A350340 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.
%C A350340 Note that the smallest abelian order with precisely 2^n groups must be the square of a squarefree number.
%C A350340 a(n) is the smallest k with n distinct prime factors such that k^2 is an abelian order.
%C A350340 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.
%C A350340 a(n) exists for all n.
%C A350340 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
%H A350340 David A. Corneth, <a href="/A350340/b350340.txt">Table of n, a(n) for n = 0..30</a>
%H A350340 David A. Corneth, <a href="/A350340/a350340.gp.txt">PARI program</a>
%F A350340 A350341(n) = a(n)^2.
%e A350340 a(2) = 35 = 5*7 since the smallest k with 2 distinct prime factors such that k^2 is an abelian order is 35.
%e A350340 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.
%o A350340 (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
%o A350340 a(n) = for(k=1, oo, if(issquarefree(k) && omega(k)==n && isA051532(k^2), return(k)))
%Y A350340 Cf. A051532 (abelian orders), A264907, A350341.
%K A350340 nonn,hard
%O A350340 0,2
%A A350340 _Jianing Song_, Dec 25 2021
%E A350340 a(7)-a(18) from _David A. Corneth_, Jan 02 2022