cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A330867 Decimal expansion of the continued fraction 1/(1 + 2/(2 + 3/(3 + 5/(5 + 7/(7 + ... + prime(k)/(prime(k) + ...)))))).

Original entry on oeis.org

5, 8, 1, 5, 2, 5, 0, 0, 4, 5, 9, 2, 2, 1, 4, 6, 5, 4, 3, 9, 9, 1, 5, 1, 7, 0, 4, 8, 1, 8, 0, 0, 4, 4, 6, 1, 9, 5, 5, 8, 6, 7, 5, 4, 0, 4, 9, 7, 2, 4, 6, 4, 4, 1, 1, 0, 0, 4, 7, 9, 4, 2, 3, 2, 6, 0, 9, 6, 7, 4, 6, 4, 5, 4, 1, 9, 6, 8, 6, 1, 4, 1, 2, 0, 2, 7, 6, 1, 4, 5, 2, 4, 3, 4, 0, 5, 4, 6, 9, 3
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 28 2020

Keywords

Examples

			0.58152500459221465439915170481800446195586754...
		

Crossrefs

A336112 a(n) is the least number k such that the Sum_{i=0..k} sqrt(k) equals or exceeds n.

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 19, 19, 20, 20, 20, 20, 21, 21, 21, 21, 21, 22, 22, 22, 22, 23, 23, 23, 23
Offset: 0

Views

Author

Robert G. Wilson v, Jul 08 2020

Keywords

Comments

Inspired by A045880.
Let c = (9/4)^(1/3) = (3/2)^(2/3) ~ 1.310370697..., then a(n) ~ c*n^(2/3).
a(10^k) for k>= 0: 1, 6, 28, 131, 608, 2823, 13104, 60822, 282311, 1310371, 6082202, 28231081, 131037070, 608220200, ..., .

Examples

			a(0) = 0 since the sqrt(0) = 0;
a(1) = 1 since the sqrt(0) + sqrt(1) = 1;
a(2) = 2 since the sqrt(0) + sqrt(1) + sqrt(2) ~ 2.41421... which exceeds 2;
a(3) = 3 since the sqrt(0) + sqrt(1) + sqrt(2) + sqrt(3) ~ 4.146264... which easily exceeds 3;
a(4) = 3 because the sqrt(0) + sqrt(1) + sqrt(2) + sqrt(3) ~ 4.146264... which barely exceeds 4; etc.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Block[{k = s = 0}, While[s < n, k++; s = s + Sqrt@k]; k]; Array[f, 75, 0]
  • PARI
    a(n) = my(s=0, k=0); while ((s+=sqrt(k)) < n, k++); k; \\ Michel Marcus, Jul 09 2020

Formula

a(k*n) ~ k^(2/3)*a(n).
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