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A384612 a(n) is the smallest integer k such that k^n is an abelian square; or -1 if no such k exists.

%I A384612 #52 Jul 12 2025 08:36:08
%S A384612 11,836,11,207,624,818222,1001,2776,100001,32323107,100001,85692627,
%T A384612 10000001,501249084,10000001,27962757,41695607,70983559,72768046,
%U A384612 977688137,219873071,112562383,2338280974,2435385853,1231380445,4557057314,361499019,8096434047,5278552513
%N A384612 a(n) is the smallest integer k such that k^n is an abelian square; or -1 if no such k exists.
%C A384612 Terms are the base of the smallest n-th power whose string of decimal digits is an abelian square; i.e., of the form m concatenated with a permutation of m (A272655).
%C A384612 If n is odd and A001700((n-1)/2) has d digits, then 0 < k <= 10^(2*d-1) + 1. - _Robert Israel_, Jun 05 2025
%C A384612 a(23) >= 1.83 * 10^9. Using _Robert Israel_'s comment above a(23) <= 10^13 + 1. - _David A. Corneth_, Jun 06 2025
%H A384612 Chai Wah Wu, <a href="/A384612/b384612.txt">Table of n, a(n) for n = 1..34</a>
%e A384612 a(1) = 11, since 11^1  = 1|1
%e A384612 a(2) = 836, since 836^2 = 698|896
%e A384612 a(3) = 11, since 11^3 = 13|31
%e A384612 a(4) = 207, since 207^4 = 18360|36801
%e A384612 a(5) = 624, since 624^5 = 9460692|9690624
%e A384612 a(6) = 818222, since 818222^6 = 300072996174564185|100579862765194304
%e A384612 a(7) = 1001, since 1001^7 = 10070210350|35021007001
%e A384612 a(8) = 2776, since 2776^8 = 35265958674713|24535718936576
%e A384612 a(9) = 100001, since 100001^9 = 10000900036000840012600|12600084000360000900001.
%p A384612 f:= proc(n) local k,d,x,L;
%p A384612       k:= 2;
%p A384612       do
%p A384612         x:= k^n;
%p A384612         d:= ilog10(x)+1;
%p A384612         if d::odd then k:= ceil(10^(d/n)); next fi;
%p A384612         L:= convert(x,base,10);
%p A384612         if sort(L[1..d/2]) = sort(L[d/2+1..d]) then return k fi;
%p A384612         k:= k+1
%p A384612       od;
%p A384612 end proc:
%p A384612 map(f, [$1..30]); # _Robert Israel_, Jun 05 2025
%o A384612 (Python)
%o A384612 from itertools import count
%o A384612 def ok(k, n):
%o A384612     s = str(k**n)
%o A384612     if len(s) % 2 != 0:
%o A384612         return False
%o A384612     mid = len(s) // 2
%o A384612     return sorted(s[:mid]) == sorted(s[mid:])
%o A384612 def a(n):
%o A384612     return next(k for k in count(2) if ok(k, n))
%o A384612 print([a(n) for n in range(1, 10)])
%Y A384612 Cf. A001597, A001700, A075786, A272655, A342942.
%K A384612 nonn,base
%O A384612 1,1
%A A384612 _Gonzalo Martínez_, Jun 04 2025
%E A384612 a(20)-a(22) from _David A. Corneth_, Jun 06 2025
%E A384612 a(23) from _Gonzalo Martínez_, Jun 06 2025
%E A384612 a(24)-a(29) from _Jinyuan Wang_, Jun 14 2025