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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 A270745 A270744 for a guide to related sequences.

x = 1; Table[n[x, k], {k, 1, z}]

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allocated for Clark Kimberling

(r,1)-greedy sequence, where r(k) = 2/log(k+1).

3, 48, 5215, 43930979, 8221176288381971, 237472642129791861355082716048930, 59916111345562665920456160598356741759066440491193682529746704653

1,1

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 A270745 for a guide to related sequences.

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

a(1) = ceiling(r(1)) = ceiling(1/tau) = ceiling(0.618...) = 3;

a(2) = ceiling(r(2)/(1 - r(1)/1) = 48;

a(3) = ceiling(r(3)/(1 - r(1)/1 - r(2)/2) = 5215.

The first 3 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are

0.961..., 0.997..., 0.99999997...

$MaxExtraPrecision = Infinity; z = 16;

r[k_] := N[2/Log[k + 1], 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, 18}], 200]

Cf. A001620, A270744.

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nonn,easy

Clark Kimberling, Apr 09 2016

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