%I A138768
%S A138768 1,2,3,4,4,6,6,8,8,8,8,12,12,12,12,16,16,16,16,16,16,16,16,24,24,24,27,
%T A138768 27,27,27,27,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,48,48,48,
%U A138768 48,48,48,54,54,54,54,54,54,54,54,54,54,64,64,64,64,64,64,64,64,64,64
%N A138768 For a positive integer n, write the integers 1,2,...,n in the following
order: first write 1 (round 0), then all primes less or equal to
n in increasing order (round 1), then 2p for all primes p with 2p<=n,
also in increasing order (round 2), then 3p, then 4p and so on. Each
number is written down only the first time it is encountered. Let
a(n) denote the last number written down.
%C A138768 a(1)=1. For a given n>=2, let M be the largest of the numbers in the
finite sequence [m/(largest prime dividing m), m=2,3,...,n]. a(n)
is defined to be the largest m in (2,3,...,n) for which M is attained.
Example: a(14)=12 because the values of m/(largest prime dividing
m) for m = 2,3,...,14 are 1,1,2,1,2,1,4,3,2,1,4,1,2. The largest
of these is 4 and it is attained for m=8 and m=12; the largest of
these is 12.
%D A138768 Gary Gordon, Problem 11218, Amer. Math. Monthly, 115 (No. 4, 2008), pp.
367-368.
%e A138768 For n=10 we get the ordering 1/ 2, 3, 5, 7/ 4, 6, 10/ 9/ 8 (the rounds
are separated by /); so a(10)=8.
%p A138768 with(numtheory): b:=proc(m) local u: if m=1 then 1 else u:=factorset(m):
m/max(seq(u[j],j=1..nops(u))) end if end proc: a:=proc(n) local M,
i,a: M:=max(seq(b(j),j=1..n)): for i to n do if b(i)=M then a[i]:=i
else a[i]:=0 end if end do: max(seq(a[i],i=1..n)) end proc: seq(a(n),
n=1..80);
%Y A138768 Cf. A052126.
%Y A138768 Sequence in context: A061984 A063208 A092988 this_sequence A111939 A003962
A102443
%Y A138768 Adjacent sequences: A138765 A138766 A138767 this_sequence A138769 A138770
A138771
%K A138768 nonn
%O A138768 1,2
%A A138768 Emeric Deutsch (deutsch(AT)duke.poly.edu) and Gary Gordon (gordomg(AT)lafayette.edu),
Apr 01 2008
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