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a(n) = binomial(n, floor(log(n))).
1

%I #28 Sep 22 2022 13:39:22

%S 1,1,3,4,5,6,7,28,36,45,55,66,78,91,105,120,136,153,171,190,1330,1540,

%T 1771,2024,2300,2600,2925,3276,3654,4060,4495,4960,5456,5984,6545,

%U 7140,7770,8436,9139,9880,10660,11480,12341,13244,14190,15180,16215,17296,18424,19600,20825

%N a(n) = binomial(n, floor(log(n))).

%H Mathematics Stack Exchange, <a href="https://math.stackexchange.com/questions/363555/how-to-compute-the-asymptotic-growth-of-binomn-log-n">How to compute asymptotic growth of binomial(n, log(n))</a>.

%p a:= n-> binomial(n, ilog(n)):

%p seq(a(n), n=1..60); # _Alois P. Heinz_, Jul 31 2022

%t a[n_] := Binomial[n, Floor[Log[n]]]; Array[a, 50] (* _Amiram Eldar_, Jul 31 2022 *)

%o (Python)

%o from numpy import log

%o from math import comb

%o for n in range(1, 50):

%o x = comb(n, floor(log(n)))

%o print("{}, ".format(x), end='')

%o (PARI) a(n) = binomial(n, floor(log(n))); \\ _Michel Marcus_, Jul 31 2022

%Y Cf. A000195, A007318.

%Y Cf. A001405, A051033, A051036, A051052, A051053.

%K nonn

%O 1,3

%A _Christoph B. Kassir_, Jul 14 2022