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A126783 Smallest 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.

%I A126783 #8 Aug 24 2012 10:49:58
%S A126783 80,80,70,3905,4004,700,19278,32761,5600,8100,24940,10600,56330,68040,
%T A126783 81760,149705,116180,126360,123580,0,65500,311003,205030,114400,
%U A126783 454951,317350,312170,296270,359380,332750,699785,723338,498150,499130,901368
%N A126783 Smallest 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.
%C A126783 For each n there is an upper bound (see A130179) for values of k such that (sum of digits of k^n)*(sum of digits of k^(n+1)) = k, hence the number of such k is finite, possibly zero, (see A130180) and if the number is not zero there is a largest one (see A130181).
%H A126783 Klaus Brockhaus, <a href="/A126783/b126783.txt">Table of n, a(n) for n=1..54</a>
%e A126783 For n = 2 the smallest such k is 80: 80^2 = 6400 and 6+4+0+0 = 10; 80^3 = 512000 and 5+1+2+0+0+0 = 8; 10*8 = 80. Hence a(2) = 80.
%e A126783 For n = 3 the smallest such k is 70: 70^3 = 343000 and 3+4+3+0+0+0 = 10; 70^4 = 24010000 and 2+4+0+1+0+0+0+0 = 7; 10*7 = 70. Hence a(3) = 70.
%p A126783 P:=proc(n) local a,i,j,k,w,x; for a from 1 by 1 to n do for i from 1 by 1 to n*n do w:=0;k:=i^a;j:=0;x:=i^(a+1); while k>0 do w:=w+k-(trunc(k/10)*10); k:=trunc(k/10); od; while x>0 do j:=j+x-(trunc(x/10)*10); x:=trunc(x/10); od; if (i=w*j and i>1) then print(i); break; fi; od; od; end: P(1000);
%Y A126783 Cf. A130179, A130180, A130181.
%K A126783 hard,nonn,base
%O A126783 1,1
%A A126783 _Paolo P. Lava_ and _Giorgio Balzarotti_, Apr 17 2007
%E A126783 Edited and a(17) to a(35) added by _Klaus Brockhaus_, May 14 2007