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A066326
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a(1) = 5; for n > 1, a(n) is the least k > 0 not already included such that a(m)^2 + k^2 is a square for some m < n.
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1
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5, 12, 9, 16, 30, 35, 40, 42, 56, 33, 44, 63, 60, 11, 25, 32, 24, 7, 10, 18, 45, 28, 21, 20, 15, 8, 6, 36, 27, 48, 14, 55, 64, 70, 72, 54, 65, 75, 77, 80, 39, 52, 84, 13, 90, 91, 96, 99, 100, 105, 88, 66, 108, 81, 110, 112, 117, 120, 22, 50, 119, 126, 128, 132, 85, 135, 140
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OFFSET
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1,1
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COMMENTS
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For a(1) = 0,1,2 no possible value of a(2) exists; for a(1) = 3,4 we get 4, 3 for a(2) but no further possible values. For a(1) >= 5 do we always get an infinite sequence?
Yes: if t >= 5 is the largest of a(1),...,a(n), then (if no smaller k works) it is always possible to take a(n+1) = (t^2-1)/2 if t is odd, t^2/4 - 1 if t is even.
Does the sequence include every number >= 5? (End)
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LINKS
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EXAMPLE
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a(5) = 30 because a(5)^2 + a(4)^2 = 30^2 + 16^2 = 34^2; a(6) = 35 because a(6)^2 + a(2)^2 = 35^2 + 12^2 = 37^2.
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MAPLE
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Cands:= {5}: S:= {}:
for n from 1 to 100 do
A[n]:= min(Cands);
Cands:= Cands minus {A[n]};
if A[n]::odd then divs:= select(`<`, numtheory:-divisors(A[n]^2), A[n])
else divs:= select(t -> t < A[n] and t::even and (A[n]^2/t)::even, numtheory:-divisors(A[n]^2))
fi;
Cands := Cands union (map(t -> (A[n]^2/t - t)/2, divs) minus S);
S:= S union {A[n]};
od:
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CROSSREFS
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KEYWORD
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AUTHOR
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Jonathan Ayres (jonathan.ayres(AT)ntlworld.com), Dec 15 2001
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EXTENSIONS
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STATUS
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approved
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