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A270745
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(r,1)-greedy sequence, where r(k) = k/tau^k and tau = golden ratio.
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1
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OFFSET
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1,2
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COMMENTS
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Let x > 0, and let r = (r(k)) be a sequence of positive irrational numbers. Let a(1) be the least positive integer m such that r(1)/m < x, and inductively let a(n) be the least positive integer m such that r(1)/a(1) + ... + r(n-1)/a(n-1) + r(n)/m < x. The sequence (a(n)) is the (r,x)-greedy sequence. We are interested in choices of r and x for which the series r(1)/a(1) + ... + r(n)/a(n) + ... converges to x. See A270744 for a guide to related sequences.
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LINKS
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FORMULA
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a(n) = ceiling(r(n)/s(n)), where s(n) = 1 - r(1)/a(1) - r(2)/a(2) - ... - r(n-1)/a(n-1).
r(1)/a(1) + ... + r(n)/a(n) + ... = 1
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EXAMPLE
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a(1) = ceiling(r(1)) = ceiling(1/tau) = ceiling(0.618...) = 1;
a(2) = ceiling(r(2)/(1 - r(1)/1) = 3;
a(3) = ceiling(r(3)/(1 - r(1)/1 - r(2)/2) = 6.
The first 6 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are
0.618..., 0.872..., 0.990..., 0.975..., 0.999..., 0.99999999959...
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MATHEMATICA
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$MaxExtraPrecision = Infinity; z = 13;
r[k_] := N[k/GoldenRatio^k, 1000]; f[x_, 0] = x;
n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
x = 1; Table[n[x, k], {k, 1, z}]
N[Sum[r[k]/n[x, k], {k, 1, 13}], 200]
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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