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A280012 a(n) = least positive integer k such that sumdigits(k^2) = n*sumdigits(k). 1
1, 2, 3, 13, 113, 1113, 11113, 211113, 101011113, 1101111211, 110101111211 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

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

Table of n, a(n) for n=1..11.

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

PROG

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

CROSSREFS

Cf. A007953, A061912, A099628.

Sequence in context: A098462 A125283 A098406 * A224792 A062447 A153888

Adjacent sequences:  A280009 A280010 A280011 * A280013 A280014 A280015

KEYWORD

nonn,base,more

AUTHOR

M. F. Hasler, Apr 14 2017

EXTENSIONS

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

STATUS

approved

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Last modified September 20 10:51 EDT 2021. Contains 347584 sequences. (Running on oeis4.)