A280012 a(n) = least positive integer k such that sumdigits(k^2) = n*sumdigits(k).

1, 2, 3, 13, 113, 1113, 11113, 211113, 101011113, 1101111211, 110101111211

Offset: 1

Author

M. F. Hasler, Apr 14 2017

Keywords

• nonn
• base
• more

Comments

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).

a(12) <= 21201101101122, a(13) <= 10101010101101122. - Giovanni Resta, Apr 15 2017

Links

• Les Reid, Problem #12, Challenge Problem Archive, Missouri State University Math Department, Academic year 2013-2014.

Crossrefs

Cf. A007953, A061912, A099628.

Programs

PARI

a(n)=for(k=1,9e9,sumdigits(k^2)==n*sumdigits(k)&&return(k))

Extensions

a(10)-a(11) from Giovanni Resta, Apr 15 2017