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A138768
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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.
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0
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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; internal format)
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OFFSET
| 1,2
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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.
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REFERENCES
| Gary Gordon, Problem 11218, Amer. Math. Monthly, 115 (No. 4, 2008), pp. 367-368.
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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.
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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);
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CROSSREFS
| Cf. A052126.
Sequence in context: A061984 A063208 A092988 * A111939 A003962 A102443
Adjacent sequences: A138765 A138766 A138767 * A138769 A138770 A138771
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KEYWORD
| nonn
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu) and Gary Gordon (gordomg(AT)lafayette.edu), Apr 01 2008
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