A384612 a(n) is the smallest integer k such that k^n is an abelian square; or -1 if no such k exists.
11, 836, 11, 207, 624, 818222, 1001, 2776, 100001, 32323107, 100001, 85692627, 10000001, 501249084, 10000001, 27962757, 41695607, 70983559, 72768046, 977688137, 219873071, 112562383, 2338280974, 2435385853, 1231380445, 4557057314, 361499019, 8096434047, 5278552513
Offset: 1
Examples
a(1) = 11, since 11^1 = 1|1 a(2) = 836, since 836^2 = 698|896 a(3) = 11, since 11^3 = 13|31 a(4) = 207, since 207^4 = 18360|36801 a(5) = 624, since 624^5 = 9460692|9690624 a(6) = 818222, since 818222^6 = 300072996174564185|100579862765194304 a(7) = 1001, since 1001^7 = 10070210350|35021007001 a(8) = 2776, since 2776^8 = 35265958674713|24535718936576 a(9) = 100001, since 100001^9 = 10000900036000840012600|12600084000360000900001.
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..34
Programs
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Maple
f:= proc(n) local k,d,x,L; k:= 2; do x:= k^n; d:= ilog10(x)+1; if d::odd then k:= ceil(10^(d/n)); next fi; L:= convert(x,base,10); if sort(L[1..d/2]) = sort(L[d/2+1..d]) then return k fi; k:= k+1 od; end proc: map(f, [$1..30]); # Robert Israel, Jun 05 2025
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Python
from itertools import count def ok(k, n): s = str(k**n) if len(s) % 2 != 0: return False mid = len(s) // 2 return sorted(s[:mid]) == sorted(s[mid:]) def a(n): return next(k for k in count(2) if ok(k, n)) print([a(n) for n in range(1, 10)])
Extensions
a(20)-a(22) from David A. Corneth, Jun 06 2025
a(23) from Gonzalo MartÃnez, Jun 06 2025
a(24)-a(29) from Jinyuan Wang, Jun 14 2025
Comments