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(r,1)-greedy sequence, where r(k) = 2/e^k.
1

%I #8 Oct 03 2018 16:11:55

%S 1,2,1,2,2,2,2,1,2,1,15,107,4536,9767531,119608113171152,

%T 27706455600364135685442345886,

%U 683882728856994887705617234665700899371621018916716222985

%N (r,1)-greedy sequence, where r(k) = 2/e^k.

%C 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 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 r(1)/a(1) + ... + r(n)/a(n) + ... = 1

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

%e a(2) = ceiling(r(2)/(1 - r(1)/1) = 2;

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

%e The first 6 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are

%e 0.735..., 0.871..., 0.970..., 0.975..., 0.988..., 0.995...

%t $MaxExtraPrecision = Infinity; z = 19;

%t r[k_] := N[2/E^k, 1000]; f[x_, 0] = x;

%t n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]

%t f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]

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

%t N[Sum[r[k]/n[x, k], {k, 1, 19}], 200]

%Y Cf. A001620, A270744, A068985.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Apr 09 2016