A270751 (r,1)-greedy sequence, where r(k) = 1/(k*tau) and tau = golden ratio.
1, 1, 3, 37, 1204, 21029921, 425355555167420, 439183524292095499600664584581, 240317442633783387248198509182959563857071128274317237128901
Offset: 1
Examples
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)/1)) = 3. The first 3 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are 0.618..., 0.927..., 0.995...
Programs
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Mathematica
$MaxExtraPrecision = Infinity; z = 16; r[k_] := N[1/(k*GoldenRatio), 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) = A270584(n-1) for n>1. - R. J. Mathar, Jun 21 2025
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