

A254058


Smallest a(n) such that C(a(n),n) >= 2^n.


1



2, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 18, 20, 21, 22, 24, 25, 26, 28, 29, 30, 32, 33, 34, 35, 37, 38, 39, 41, 42, 43, 45, 46, 47, 48, 50, 51, 52, 54, 55, 56, 58, 59, 60, 61, 63, 64, 65, 67, 68, 69, 71, 72, 73, 74, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 89
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OFFSET

1,1


COMMENTS

Taking logarithms and using approximation for binomial coefficients, we obtain that n < log C(a(n), n) ~ a(n)H(n/a(n)), so we need to solve x=H(x) with x=n/a(n), which gives a(n) ~ 1.29n.
a(n) ~ r * n, where r = 1.293815373340415493316601653303657352145361654147... is the root of the equation r^r = 2*(r1)^(r1).  Vaclav Kotesovec, Jan 29 2015


LINKS

Chai Wah Wu, Table of n, a(n) for n = 1..10000


MATHEMATICA

f[n_] := Block[{k = n}, While[ Binomial[k, n] < 2^n, k++]; k]; Array[f, 67] (* Robert G. Wilson v, Jan 28 2015 *)


PROG

(Python)
a=n
while binomial(a, n)<2^n:
a=a+1
(Python)
from __future__ import division
def A254058(n):
....b, a1, a2, t = 1, 0, n, 2**n
....while b < t:
........a2 += 1
........a1 += 1
........b = (b*a2)//a1
....return a2 # Chai Wah Wu, Jan 30 2015
(PARI) a(n) = x=1; while(binomial(x, n) < 2^n, x++); x; \\ Michel Marcus, Jan 28 2015


CROSSREFS

Sequence in context: A014132 A184008 A183862 * A346128 A276220 A288204
Adjacent sequences: A254055 A254056 A254057 * A254059 A254060 A254061


KEYWORD

nonn


AUTHOR

Domotor Palvolgyi, Jan 24 2015


STATUS

approved



