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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 (list; graph; refs; listen; history; text; internal format)
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*(r-1)^(r-1). - 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

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Last modified November 28 21:48 EST 2022. Contains 358421 sequences. (Running on oeis4.)