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%I #19 Apr 06 2020 03:30:34
%S 2,4,6,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,
%T 92,96,104,112,120,128,136,144,152,160,168,176,184,192,200,208,216,
%U 224,232,240,248,256,264,272,280,288,296,304,312,320,328,336,344,352,360
%N Numbers m that are divisible by 2^k, where k is the digit length of m.
%C The number of terms of length k is equal to (9*5^(k-1) - 1)/2. - _Bernard Schott_, Apr 06 2020
%H Robert Israel, <a href="/A050622/b050622.txt">Table of n, a(n) for n = 1..10000</a>
%p seq(seq(j*2^k, j=(5^(k-1)+1)/2 .. 5^k-1),k=1..3); # _Robert Israel_, Apr 05 2020
%t Select[Range[360], IntegerQ[#/2^IntegerLength[#]] &] (* _Jayanta Basu_, May 25 2013 *)
%o (PARI) isok(n) = n % (2^#Str(n)) == 0; \\ _Michel Marcus_, Sep 17 2015
%Y Cf. A035014, A050621, A055642.
%K nonn,base,easy
%O 1,1
%A _Patrick De Geest_, Jun 15 1999
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