A270916 (r,1)-greedy sequence, where r(k) = 1/(k*sqrt(2)).
1, 2, 3, 5, 65, 6529, 136091233, 41625259047416909, 2189507051227161558033650829868135, 75931290362065676573711484986356332365619562746656079489987281066955
Offset: 1
Examples
a(1) = ceiling(r(1)) = ceiling(1/sqrt(2)) = ceiling(0.707...) = 1; a(2) = ceiling(r(2)/(1 - r(1)/1)) = 2; a(3) = ceiling(r(3)/(1 - r(1)/1 - r(2)/2)) = 3. The first 6 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are 0.707..., 0.883..., 0.962..., 0.997..., 0.99998..., 0.9999999992...
Programs
-
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]
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
Comments