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EXAMPLE
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Let r = sqrt(2). Then
1 < r < 1 + 1/2, so a(1) = 2.
1 + 1/2 -1/11 < r, so a(2) = 11.
1 + 1/2 - 1/11 + 1/195 > r, so a(3) = 195.
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MATHEMATICA
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$MaxExtraPrecision = Infinity;
nn = 12; f[n_] := 1/n; r = Pi; s = 0; b[1] = NestWhile[# + 1 &, 1, ! (s += f[#]) > r &]; u[1] = Sum[f[n], {n, 1, b[1]}]; c[1] = Floor[1/(u[1] - r)]; v[1] = u[1] - 1/c[1]; n = 1; While[n < nn/2, n++; b[n] = Floor[1/(r - v[n - 1])]; u[n] = v[n - 1] + 1/b[n]; c[n] = Floor[1/(u[n] - r)]; v[n] = u[n] - 1/c[n]]; a = Riffle[Table[b[i], {i, 1, nn/2}], Table[c[i], {i, 1, nn/2}]]
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