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Let b(n)=floor((3/2)^n), c(n)=floor((4/3)^n), d(n)=floor((5/4)^n); sequence gives values of n such that b(n+1)/b(n)=3/2, c(n+1)/c(n)=4/3 and d(n+1)/d(n)=5/4.
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%I #7 Jun 08 2013 15:53:05

%S 162,172,204,328,403,414,809,835,840,854,1111,1117,1160,1188,1192,

%T 1270,1294,1311,1351,1409,1469,1478,1508,1605,1614,1769,1842,1961,

%U 2065,2226,2425,2456,2460,2486,2581,2597,2635,2638,2642,2650,2679,2720,2880,2932

%N Let b(n)=floor((3/2)^n), c(n)=floor((4/3)^n), d(n)=floor((5/4)^n); sequence gives values of n such that b(n+1)/b(n)=3/2, c(n+1)/c(n)=4/3 and d(n+1)/d(n)=5/4.

%F It seems that a(n) is asymptotic to C*n where C is around 60.

%t bcdQ[n_]:=Module[{b=Floor[(3/2)^n],b1=Floor[(3/2)^(n+1)],c=Floor[ (4/3)^n], c1=Floor[(4/3)^(n+1)],d=Floor[(5/4)^n],d1=Floor[(5/4)^(n+1)]}, b1/b==3/2&&c1/c==4/3&&d1/d==5/4]; Select[Range[3000],bcdQ] (* _Harvey P. Dale_, Jun 08 2013 *)

%Y Cf. A065554, A065564.

%K nonn

%O 1,1

%A _Benoit Cloitre_, Apr 06 2003