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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.

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

Original entry on oeis.org

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

Views

Author

Gonzalo Martínez, Jun 04 2025

Keywords

Comments

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).
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
a(23) >= 1.83 * 10^9. Using Robert Israel's comment above a(23) <= 10^13 + 1. - David A. Corneth, Jun 06 2025

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.
		

Crossrefs

Programs

  • 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
  • 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 Martínez, Jun 06 2025
a(24)-a(29) from Jinyuan Wang, Jun 14 2025