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A130181 Largest k > 1 such that (sum of digits of k^n)*(sum of digits of k^(n+1)) = k, or 0 if no such k exists. 5

%I #5 Mar 30 2012 17:27:54

%S 486,1215,4374,4672,12862,12649,23408,32761,47477,56852,59048,90746,

%T 116864,112346,139472,149705,190512,234247,254015,0,322322,331775,

%U 391238,446512,454951,546121,530145,316250,613927,763795,786664,809936

%N Largest k > 1 such that (sum of digits of k^n)*(sum of digits of k^(n+1)) = k, or 0 if no such k exists.

%H Klaus Brockhaus, <a href="/A130181/b130181.txt">Table of n, a(n) for n=1..54</a>

%e For n = 2 the largest such k is 1215: 1215^2 = 1476225 and 1+4+7+6+2+2+5 = 27; 1215^3 = 1793613375and 1+7+9+3+6+1+3+3+7+5 = 45; 27*45 = 1215. Hence a(2) = 1215.

%Y Cf. A126783 (smallest k), A130179 (upper bound), A130180 (number of such k).

%K nonn,base

%O 1,1

%A _Klaus Brockhaus_, May 14 2007

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