This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A350341 #25 Jan 11 2022 22:14:10 %S A350341 1,4,1225,354025,187279225,256385259025,720186192601225, %T A350341 2898439345541083201,6402652514300252791009, %U A350341 32275771324587574319476369,171997585388727183548489570401,1618325280922534070007738367903009,18528206141282092567518596574121550041,298841436852738871021507444144006480611289 %N A350341 a(n) is the smallest abelian order with precisely 2^n groups. %C A350341 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 A350341 Note that the smallest abelian order with precisely 2^n groups must be the square of a squarefree number. %C A350341 a(n) is the smallest square number with n distinct prime factors that is an abelian order. %C A350341 a(n) is the smallest number of the form (p_1*p_2*...*p_n)^2 where the p_i are distinct primes and no (p_j)^2-1 is divisible by any p_i. %C A350341 a(n) exists for all n. %C A350341 Except for a(1) = 4, 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. %H A350341 David A. Corneth, <a href="/A350341/b350341.txt">Table of n, a(n) for n = 0..18</a> %F A350341 a(n) = A350340(n)^2. %e A350341 a(2) = 1225 = 35^2 since the smallest square number with 2 distinct prime factors that is an abelian order is 1225. %e A350341 a(3) = 354025 = 595^2 since the smallest square number with 3 distinct prime factors that is an abelian order is 354025. %o A350341 (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 A350341 a(n) = for(k=1, oo, if(issquarefree(k) && omega(k)==n && isA051532(k^2), return(k^2))) %Y A350341 Cf. A051532 (abelian orders), A264907, A350340. %K A350341 nonn,hard %O A350341 0,2 %A A350341 _Jianing Song_, Dec 25 2021 %E A350341 a(7)-a(11) from _David A. Corneth_, Jan 02 2022