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a(n) = least integer j >= 0 such that n = floor((2^j)/(5^k)) for some integer k >= 0.
3

%I #8 Jul 06 2018 17:11:12

%S 0,1,4,2,7,5,33,3,38,8,36,6,13,34,55,4,25,39,60,9,23,37,44,58,7,14,28,

%T 35,49,56,70,5,19,26,33,40,54,61,68,10,17,24,31,103,38,45,52,59,66,73,

%U 8,15,22,29,101,36,43,115,50,57,64,136,71,6,13,85,20,27,99,34,106,41,48

%N a(n) = least integer j >= 0 such that n = floor((2^j)/(5^k)) for some integer k >= 0.

%C The k-sequence is A124916.

%H Robert Israel, <a href="/A124908/b124908.txt">Table of n, a(n) for n = 1..10000</a>

%F a(2^j) = j. - _Robert Israel_, Jul 05 2018

%e 1 = floor(2^0/5^0), 2 = floor(2^1/5^0), 3 = floor(2^4/5^1), 4 = floor(2^2/5^0), ...,

%e so j-sequence = (0,1,4,2,...); k-sequence = (0,0,1,0,...).

%p f:= proc(n) local k;

%p if n = 2^ilog2(n) then return ilog2(n) fi;

%p for k from 1 do if ilog2(n*5^k) <> ilog2((n+1)*5^k) then return ilog2((n+1)*5^k) fi od

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Jul 05 2018

%Y Cf. A124916.

%K nonn,look

%O 1,3

%A _Clark Kimberling_, Nov 12 2006