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 A124915 a(n) = least integer k>=0 such that n=Floor[(2^j)/(3^k)] for some integer j>=0. 1

%I #2 Mar 30 2012 18:57:06

%S 0,0,2,0,1,4,2,0,3,1,6,4,9,2,12,0,10,3,8,25,1,6,11,16,4,9,26,2,7,36,

%T 12,0,5,34,10,27,3,32,8,37,25,1,30,6,47,23,11,40,16,4,33,21,9,38,26,2,

%U 43,31,7,48,36,24,12,0,29,17,5,46,34,22,10,39,27,15,3,44,32,20,8,49

%N a(n) = least integer k>=0 such that n=Floor[(2^j)/(3^k)] for some integer j>=0.

%C Every nonnegative integer occurs infinitely many times. The j-sequence is A124907.

%e 1=[2^0/3^0], 2=[2^1/3^0], 3=[2^5/3^2], 4=[2^2/3^0],...,

%e so j-sequence=(0,1,5,2,...); k-sequence=(0,0,2,0,...).

%Y Cf. A124907.

%K nonn

%O 1,3

%A _Clark Kimberling_, Nov 12 2006

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