A081724 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.

162, 172, 204, 328, 403, 414, 809, 835, 840, 854, 1111, 1117, 1160, 1188, 1192, 1270, 1294, 1311, 1351, 1409, 1469, 1478, 1508, 1605, 1614, 1769, 1842, 1961, 2065, 2226, 2425, 2456, 2460, 2486, 2581, 2597, 2635, 2638, 2642, 2650, 2679, 2720, 2880, 2932

Offset: 1

Author

Benoit Cloitre, Apr 06 2003

Keywords

• nonn

Crossrefs

Cf. A065554, A065564.

Programs

Mathematica

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 *)

Formula

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