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Least integer k such that floor(k*f(n+1))>k*f(n), where f(n)=log_2 of n.
2

%I #10 Feb 27 2023 07:45:28

%S 1,2,1,4,2,3,1,6,4,3,2,3,4,6,1,12,6,5,4,3,5,2,7,5,3,4,5,6,8,11,1,23,

%T 12,8,6,5,9,4,7,3,8,5,7,13,2,11,7,5,8,3,10,7,4,9,5,11,6,8,9,11,15,22,

%U 1,45,23,16,12,10,8,7,6,11,5,9,13,4,11,7,13,3,14,8,13,5,12,7,9,13,21,2,23,13,11,9,7,17,5,8,11,17

%N Least integer k such that floor(k*f(n+1))>k*f(n), where f(n)=log_2 of n.

%H Rémy Sigrist, <a href="/A183200/b183200.txt">Table of n, a(n) for n = 1..10000</a>

%t Table[k=1; While[Floor[k*Log[2,n+1]]<=k*Log[2,n], k++]; k, {n,100}]

%Y Cf. A183163.

%K nonn

%O 1,2

%A _Clark Kimberling_, Dec 29 2010