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A254058 Smallest a(n) such that C(a(n),n) >= 2^n.

%I A254058 #29 May 22 2025 10:21:42
%S A254058 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 A254058 35,37,38,39,41,42,43,45,46,47,48,50,51,52,54,55,56,58,59,60,61,63,64,
%U A254058 65,67,68,69,71,72,73,74,76,77,78,80,81,82,84,85,86,87,89
%N A254058 Smallest a(n) such that C(a(n),n) >= 2^n.
%C A254058 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 A254058 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 A254058 Chai Wah Wu, <a href="/A254058/b254058.txt">Table of n, a(n) for n = 1..10000</a>
%t A254058 f[n_] := Block[{k = n}, While[ Binomial[k, n] < 2^n, k++]; k]; Array[f, 67] (* _Robert G. Wilson v_, Jan 28 2015 *)
%o A254058 (Python)
%o A254058 def A254058(n):
%o A254058     b, a1, a2, t = 1, 0, n, 2**n
%o A254058     while b < t:
%o A254058         a2 += 1
%o A254058         a1 += 1
%o A254058         b = (b*a2)//a1
%o A254058     return a2 # _Chai Wah Wu_, Jan 30 2015
%o A254058 (PARI) a(n) = x=1; while(binomial(x, n) < 2^n, x++); x; \\ _Michel Marcus_, Jan 28 2015
%K A254058 nonn
%O A254058 1,1
%A A254058 _Domotor Palvolgyi_, Jan 24 2015