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Smallest k>n such that floor((11/10)^k)/floor((11/10)^n) is an integer.
1

%I #18 Aug 29 2012 17:32:12

%S 2,3,4,5,6,7,8,9,10,11,15,13,14,19,16,22,18,25,20,36,28,23,45,40,36,

%T 44,50,35,36,37,54,55,56,139,71,43,74,123,75,63,65,113,139,140,139,

%U 132,133,85,100,178,148,376,98,234,139,1277,234,223,95,217,128,479,139,454

%N Smallest k>n such that floor((11/10)^k)/floor((11/10)^n) is an integer.

%t kln[n_]:=Module[{k=n+1},While[!IntegerQ[Floor[(11/10)^k]/Floor[(11/10)^n]],k++];k]; Array[kln,70] (* _Harvey P. Dale_, Aug 29 2012 *)

%o (PARI) q=1.1; for(s=1,80,n=s+1; while(frac(floor(q^n)/floor(q^s))>0,n++); print1(n,","); )

%Y See A215975 for the integers that arise.

%K easy,nonn

%O 1,1

%A _Benoit Cloitre_, May 01 2002

%E Definition clarified by _Harvey P. Dale_, Aug 29 2012