

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 than 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.


0



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, 27, 27, 27, 27, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 48, 48, 48, 48, 48, 48, 54, 54, 54, 54, 54, 54, 54, 54, 54, 54, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

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.


LINKS

Table of n, a(n) for n=1..73.
Gary Gordon, The Number between 1 and n That Is Least Prime: Problem 11218, Amer. Math. Monthly, 115 (No. 4, 2008), pp. 367368.


EXAMPLE

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.


MAPLE

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);


CROSSREFS

Cf. A052126.
Sequence in context: A092988 A304575 A296421 * A111939 A253248 A003962
Adjacent sequences: A138765 A138766 A138767 * A138769 A138770 A138771


KEYWORD

nonn


AUTHOR

Emeric Deutsch and Gary Gordon, Apr 01 2008


STATUS

approved



