A320065 a(n) is the smallest integer i such that binomial(2i,i) > n.

1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5

Offset

1,2

Comments

The formula can be proved by using Stirling's formula to estimate the central binomial coefficient binomial(2i,i).

Links

Table of n, a(n) for n=1..103.

Formula

a(n) = (log n)/(log 4) + (log log n)/(log 16) + O(1).

For n>1, a(n) = ceiling((-LambertW(-1, -4*log(2)/(Pi*n^2)) / (4*log(2))) * (1 + 2*log(2)/(3*log(2) + (2 + log(2))*LambertW(-1, -4*log(2)/(Pi*n^2)) + 2*LambertW(-1, -4*log(2)/(Pi*n^2))^2))). - Ryan Jean and Vaclav Kotesovec, Jan 04 2025

Mathematica

Array[Block[{i = 1}, While[Binomial[2 i, i] <= #, i++]; i] &, 105] (* Michael De Vlieger, Oct 22 2018 *)
Join[{1}, Table[Ceiling[-LambertW[-1, -4*Log[2]/(Pi*n^2)]/(4*Log[2]) * (1 + 2*Log[2]/(3*Log[2] + (2 + Log[2])*LambertW[-1, -4*Log[2]/(Pi*n^2)] + 2*LambertW[-1, -4*Log[2]/(Pi*n^2)]^2))], {n, 2, 100}]] (* Vaclav Kotesovec, Jan 04 2025 *)

Crossrefs

Cf. A000984.

Sequence in context: A235224 A092139 A084558 * A163291 A156875 A066339

Adjacent sequences: A320062 A320063 A320064 * A320066 A320067 A320068

Keyword

nonn

Author

David Lewis, Oct 22 2018

Status

approved