A177151 a(n) = least k such that 1 + 1/4 + 1/9 + ... + 1/k^2 exceeds (Pi^2)*(n-1)/(6*n).

1, 1, 2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 8, 9, 9, 10, 10, 11, 12, 12, 13, 13, 14, 15, 15, 16, 16, 17, 18, 18, 19, 19, 20, 21, 21, 22, 22, 23, 24, 24, 25, 26, 26, 27, 27, 28, 29, 29, 30, 30, 31, 32, 32, 33, 33, 34, 35, 35, 36, 36, 37, 38, 38, 39, 40, 40, 41, 41, 42, 43, 43, 44, 44, 45

Offset

1,3

Comments

The series 1 + 1/4 + 1/9 + ... converges to (Pi^2)/6, so that a(n) is the least k for which the k-th partial sum exceeds (n-1)/n of the total sum.

Does A177151 have only one run of length 3?

Links

Table of n, a(n) for n=1..74.

Example

a(5)=3 because 1 + 1/4 < 4*T/5 < 1 + 1/4 + 1/9, where T=(Pi^2)/6.

Mathematica

Table[Ceiling[FindArgMin[{k, HarmonicNumber[k, 2] > Pi^2 (n - 1)/(6 n) && k > 0}, k][[1]]], {n, 74}] (* Eric W. Weisstein, Apr 17 2018 *)

Crossrefs

Cf. A177152 (runlengths of A177151).

Sequence in context: A064542 A210434 A256502 * A076935 A019446 A097369

Adjacent sequences: A177148 A177149 A177150 * A177152 A177153 A177154

Keyword

nonn

Author

Clark Kimberling, May 03 2010

Extensions

a(2) corrected by Eric W. Weisstein, Apr 17 2018

Status

approved