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

%I A270916 #14 Mar 06 2025 22:13:59
%S A270916 1,2,3,5,65,6529,136091233,41625259047416909,
%T A270916 2189507051227161558033650829868135,
%U A270916 75931290362065676573711484986356332365619562746656079489987281066955
%N A270916 (r,1)-greedy sequence, where r(k) = 1/(k*sqrt(2)).
%C A270916 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.
%F A270916 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).
%F A270916 r(1)/a(1) + ... + r(n)/a(n) + ... = 1.
%F A270916 Conjecture: a(n) = A270582(n-1). - _R. J. Mathar_, Jun 02 2016
%e A270916 a(1) = ceiling(r(1)) = ceiling(1/sqrt(2)) = ceiling(0.707...) = 1;
%e A270916 a(2) = ceiling(r(2)/(1 - r(1)/1)) = 2;
%e A270916 a(3) = ceiling(r(3)/(1 - r(1)/1 - r(2)/2)) = 3.
%e A270916 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...
%t A270916 $MaxExtraPrecision = Infinity; z = 16;
%t A270916 r[k_] := N[1/(k*Sqrt[2]), 1000]; f[x_, 0] = x;
%t A270916 n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
%t A270916 f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
%t A270916 x = 1; Table[n[x, k], {k, 1, z}]
%t A270916 N[Sum[r[k]/n[x, k], {k, 1, 18}], 200]
%Y A270916 Cf. A001620, A270744, A010503.
%K A270916 nonn,easy
%O A270916 1,2
%A A270916 _Clark Kimberling_, Apr 09 2016