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Smallest a(n) such that C(a(n),n) >= 2^n.
1

%I #27 Mar 25 2023 07:42:38

%S 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,

%T 35,37,38,39,41,42,43,45,46,47,48,50,51,52,54,55,56,58,59,60,61,63,64,

%U 65,67,68,69,71,72,73,74,76,77,78,80,81,82,84,85,86,87,89

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

%C 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.

%C 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

%H Chai Wah Wu, <a href="/A254058/b254058.txt">Table of n, a(n) for n = 1..10000</a>

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

%o (Python)

%o def A254058(n):

%o ....b, a1, a2, t = 1, 0, n, 2**n

%o ....while b < t:

%o ........a2 += 1

%o ........a1 += 1

%o ........b = (b*a2)//a1

%o ....return a2 # _Chai Wah Wu_, Jan 30 2015

%o (PARI) a(n) = x=1; while(binomial(x, n) < 2^n, x++); x; \\ _Michel Marcus_, Jan 28 2015

%K nonn

%O 1,1

%A _Domotor Palvolgyi_, Jan 24 2015