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Smallest integer k such that n divides floor((3/2)^k).
2

%I #8 Dec 15 2012 13:26:56

%S 1,2,3,16,4,37,5,16,38,49,6,44,40,29,50,16,7,51,9,49,30,40,15,44,8,40,

%T 52,56,36,50,23,43,41,26,20,81,43,9,41,49,16,73,11,40,51,29,63,44,34,

%U 49,225,40,224,196,27,56,10,36,45,50,126,23,74,193,279,41,76,26,30,56

%N Smallest integer k such that n divides floor((3/2)^k).

%C a(n)=n for some n = 1,2,3,16,56,283,....Conjectures : (i) Log(n) < a(n) < n*Log(n)^2 for n>6; (ii) the equation a(x)=n always has a solution.

%H Harvey P. Dale, <a href="/A070973/b070973.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = min( k : A002379(k) == 0 mod(n) )

%t sik[n_]:=Module[{k=1},While[!Divisible[Floor[(3/2)^k],n],k++];k]; Array[sik,70] (* _Harvey P. Dale_, Dec 15 2012 *)

%o (PARI) for(n=1,100,s=1; while(floor((3/2)^s)%n>0,s++); print1(s,","))

%Y Cf. A002379.

%K easy,nonn

%O 1,2

%A _Benoit Cloitre_, May 24 2002