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%I #8 Mar 30 2012 18:35:52
%S 99,1999,19999,199999,2999999,29999999,299999999,2999999999,
%T 29999999999,299999999999,2999999999999,29999999999999
%N Greatest number k such that sum of the n-th powers of the digits of k is greater than k.
%D J.M. De Koninck, Ces nombres qui nous fascinent. Entry 19999 p.212. Ellipses, 2008.
%e a(4) = 19999 is in the sequence because, if k > 19999, sum of 4th powers of digits < k.
%p with(numtheory):for k from 1 to 10 do:maxi:=0:for n from 1 to 2000000 do:l:=length(n):n0:=n:s:=0:for m from 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v :s:=s+u^k:od: if s>n and s>maxi then maxi:=s:n1:=n:else fi:od:print(n1):od:
%K nonn,base
%O 2,1
%A _Michel Lagneau_, Dec 02 2010
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