A074313 - OEIS

%I #2 Feb 11 2014 19:05:30

%S 1,1,2,2,3,1,1,1,4,1,1,1,1,1,5,1,2,1,1,1,1,1,2,1,1,1,1,2,1,1,1,1,1,1,

%T 1,1,1,1,3,1,2,1,1,1,1,1,1,1,2,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,1,1,

%U 2,1,1,3,1,1,1,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,1,1,1,2,1,1,1,1

%N a(n) = the maximal length of a sequence of primes {s_1 = prime(n), s_2 = f(s1), s_3 = f(s_2), ....} formed by repeated application of f(m) = Floor(m/2) on prime(n).

%C The smallest value of n such that a(n) = 6 is n = 417.

%e To compute a(9): prime(9) = 23, f(23) = 11, f(11) = 5, f(5) = 2, f(2) = 1, where f(m) = Floor(m/2). Hence the sequence (of length 4) 23, 11, 5, 2 is the sequence of primes of maximal length formed by repeated application of f to prime(9) = 23. Therefore a(9) = 4.

%t f[n_] := Module[{i}, i = 0; m = n; While[PrimeQ[m], m = Floor[m/2]; i++ ]; i]; Table[f[Prime[i]], {i, 1, 100}]

%K easy,nonn

%O 1,3

%A _Joseph L. Pe_, Sep 22 2002