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%I #6 Mar 30 2012 18:35:53
%S 1,2,4,8,10,11,20,31,40,42,71,75,80,100,101,110,123,200,211,215,301,
%T 305,310,321,400,413,420,503,611,651,701,710,750,800,842,1000,1001,
%U 1010,1100,1111,1151,1230,2000,2011,2051,2101,2110,2150,2211,3001,3010,3050,3100,3131,3201,3210,3623,3694,4000,4031,4130,4200,4222,4433
%N Numbers n such that d(1)^1 + d(2)^2 +... + d(p)^p is a power of 2, where d(i), i=1..p, are the decimal digits of n.
%e 7496 is in the sequence because 7 + 4^2 + 9^3 + 6^4 = 2048 = 2^11.
%p with(numtheory):for n from 1 to 10000 do:l:=length(n):n0:=n:s:=0:for m from
%p 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v :s:=s+u^(l-m+1):od: x:=evalf(log[2](s)):if
%p x=floor(x) then printf(`%d, `,n):else fi:od:
%K nonn,base
%O 1,2
%A _Michel Lagneau_, Dec 21 2010
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