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A270744
(r,1)-greedy sequence, where r(k) = 1/tau^k, where tau = golden ratio.
12
1, 2, 2, 3, 4, 32, 1065, 2038968, 5977146319204, 36314862033946243071181679, 1028280647188781709727717632740627249617427013751977, 958046899855070460620234639622630375078362220775180051610386376308132568342498992099474472596860400289
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. (The same algorithm is used to generate sequences listed at A269993.)
Guide to related sequences:
x r(k)
1 1/tau^k A270744
1 k/tau^k A270745
1 2/e^k A270746
1 4/Pi^k A270747
1 2/log(k+1) A270748
1 1/(k*log(k+1)) A270749
1 (1/k)*log(k+1) A270750
1 2/(k*tau^k) A270751
1 1/(k*e) A270752
1 1/(k*sqrt(2)) A270916
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
EXAMPLE
a(1) = ceiling(r(1)) = ceiling(1/tau) = ceiling(0.618...) = 1;
a(2) = ceiling(r(2)/(1 - r(1)/1) = 2;
a(3) = ceiling(r(3)/(1 - r(1)/1 - r(2)/2) = 2.
The first 6 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are
0.618..., 0.809..., 0.927..., 0.975..., 0.998..., 0.999967... .
MATHEMATICA
$MaxExtraPrecision = Infinity; z = 13;
r[k_] := N[1/GoldenRatio^k, 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, 13}], 200]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 07 2016
STATUS
approved