A270916 (r,1)-greedy sequence, where r(k) = 1/(k*sqrt(2)).

1, 2, 3, 5, 65, 6529, 136091233, 41625259047416909, 2189507051227161558033650829868135, 75931290362065676573711484986356332365619562746656079489987281066955

Offset

1,2

Comments

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.

Links

Table of n, a(n) for n=1..10.

Formula

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

Conjecture: a(n) = A270582(n-1). - R. J. Mathar, Jun 02 2016

Example

a(1) = ceiling(r(1)) = ceiling(1/tau) = ceiling(0.618...) = 1;
a(2) = ceiling(r(2)/(1 - r(1)/1) = 1;
a(3) = ceiling(r(3)/(1 - r(1)/1 - r(2)/2) = 1.
The first 6 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ...  are
0.707..., 0.883..., 0.962..., 0.997..., 0,999..., 0.9999999992...

Mathematica

$MaxExtraPrecision = Infinity; z = 16;
r[k_] := N[1/(k*Sqrt[2]), 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]

Crossrefs

Cf. A001620, A270744, A010503.

Sequence in context: A103110 A042239 A270582 * A029975 A042787 A270355

Adjacent sequences: A270913 A270914 A270915 * A270917 A270918 A270919

Keyword

nonn,easy

Author

Clark Kimberling, Apr 09 2016

Status

approved