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A215894 a(n) = floor(2^n / n^k), where k is the largest integer such that 2^n >= n^k.

%I A215894 #34 Sep 08 2022 08:46:03
%S A215894 1,2,1,1,1,2,4,6,1,1,2,3,5,9,1,1,2,4,6,10,17,1,2,3,5,9,15,26,1,2,4,6,
%T A215894 11,18,31,1,2,4,6,11,19,32,1,2,3,5,9,16,28,49,1,2,4,7,13,22,38,1,1,3,
%U A215894 5,9,16,27,47,1,2,3,5,10,17,30,51,1,2,3,5,10
%N A215894 a(n) = floor(2^n / n^k), where k is the largest integer such that 2^n >= n^k.
%C A215894 a(n) < n.
%C A215894 n such that a(n) = n-1: 2, 3, 996, 3389, 149462.
%F A215894 a(n) = modlg(2^n, n) = floor(2^n / n^floor(n*logn(2))), where logn is the logarithm base n.
%F A215894 In the base-b representation of k, modlg(k,b) is the most significant digit: k = c0 + c1*b + c2*b^2 + ... + cn*b^n, cn = modlg(k,b), c0 = k mod b. - _Alex Ratushnyak_, Aug 30 2012
%e A215894 a(2) = floor(2^2 / 2^2) = 1,
%e A215894 a(3) = floor(2^3 / 3) = 2,
%e A215894 a(4)..a(9) are floor(2^n / n^2),
%e A215894 a(10)..a(15) are floor(2^n / n^3),
%e A215894 a(16)..a(22) are floor(2^n / n^4), and so on.
%t A215894 Table[Floor[2^n/n^Floor[n Log[n, 2]]], {n, 2, 64}] (* _Alonso del Arte_, Aug 26 2012 *)
%o A215894 (Python)
%o A215894 import math
%o A215894 def modiv(a,b):
%o A215894     return a - b*(a//b)
%o A215894 def modlg(a,b):
%o A215894     return a // b**int(math.log(a,b))
%o A215894 for n in range(2,100):
%o A215894     a = 2**n
%o A215894     print(modlg(a,n), end=',')
%o A215894 (Magma) [Floor(2^n div n^Floor(n *Log(n,2))): n in [2..100]]; // _Vincenzo Librandi_, Jan 09 2019
%Y A215894 Cf. A215892, A000799, A060505.
%K A215894 nonn
%O A215894 2,2
%A A215894 _Alex Ratushnyak_, Aug 25 2012