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A208746
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Size of largest subset of [1..n] containing no three terms in geometric progression.
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8
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1, 2, 3, 3, 4, 5, 6, 7, 7, 8, 9, 10, 11, 12, 13, 13, 14, 14, 15, 15, 16, 17, 18, 19, 19, 20, 21, 21, 22, 23, 24, 24, 25, 26, 27, 27, 28, 29, 30, 31, 32, 33, 34, 34, 35, 36, 37, 38, 38, 38, 39, 39, 40, 41, 42, 43, 44, 45, 46, 46, 47, 48, 49, 49, 50, 51, 52, 52, 53, 54, 55, 55, 56, 57, 57, 57, 58, 59, 60, 61, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 71, 72, 73, 74, 74, 75, 75, 75, 75
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OFFSET
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1,2
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COMMENTS
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All three-term geometric progressions must be avoided, even those such as 4,6,9 whose ratio is not an integer.
David Applegate's computation used a floating-point IP solver for the packing subproblems, so although it's almost certainly correct there is no proof. First he enumerated geometric progressions using
for (i=1;i<=N;i++) {
for (j=2; j*j<=i; j++) {
if (i % (j*j) != 0) continue;
for (k=1; k<j; k++) {
print i*k*k/(j*j), i*k/j, i;
}
}
}
and then solved the integer program of maximizing the subset of {1..N} subject to not taking all 3 of any progression.
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LINKS
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MAPLE
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A:= proc(n)
local cons, x;
cons:=map(op, {seq(map(t -> x[t]+x[b]+x[b^2/t]<=2,
select(t -> (t<b) and (t>=b^2/n),
numtheory:-divisors(b^2))), b=2..n-1)});
Optimization:-Maximize(add(x[i], i=1..n), cons, assume=binary)[1]
end proc;
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MATHEMATICA
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a[n_] := a[n] = If[n <= 2, n, Module[{cons, x}, cons = And @@ Flatten@ Rest@ Union@ Table[x[#] + x[b] + x[b^2/#] <= 2& /@ Select[Divisors[b^2], # < b && # >= b^2/n&], {b, 2, n-1}]; Maximize[{Sum[x[i], {i, 1, n}], cons && AllTrue[Array[x, n], 0 <= # <= 1&]}, Array[x, n], Integers]][[1]]];
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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a(1)-a(82) confirmed by Robert Israel and extended to a(100), Mar 01 2012
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STATUS
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approved
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