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%I #49 Apr 18 2017 15:46:59
%S 1,2,3,13,113,1113,11113,211113,101011113,1101111211,110101111211
%N a(n) = least positive integer k such that sumdigits(k^2) = n*sumdigits(k).
%C a(n) exists for any n, since sum_{i=0..n-1} 10^(2^i-1) is an integer with the required property, having n digits 1, with its square having n digits 1 at positions 2^i-1 (n>=i>=1), and n(n-1)/2 digits 2 at positions 2^i+2^j-1 (n>=i>j>=0 i.e. at positions 1<=k<2^(n+1) for k in A099628).
%C a(12) <= 21201101101122, a(13) <= 10101010101101122. - _Giovanni Resta_, Apr 15 2017
%H Les Reid, <a href="http://people.missouristate.edu/lesreid/POW12_1314.html">Problem #12</a>, Challenge Problem Archive, Missouri State University Math Department, Academic year 2013-2014.
%o (PARI) a(n)=for(k=1,9e9,sumdigits(k^2)==n*sumdigits(k)&&return(k))
%Y Cf. A007953, A061912, A099628.
%K nonn,base,more
%O 1,2
%A _M. F. Hasler_, Apr 14 2017
%E a(10)-a(11) from _Giovanni Resta_, Apr 15 2017
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