cp's OEIS Frontend

A022922 Number of integers m such that 5^n < 2^m < 5^(n+1).

%I A022922 #18 Apr 09 2021 03:50:30
%S A022922 2,2,2,3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,2,3,2,2,
%T A022922 3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,2,3,2,2,3,2,2,3,2,2,
%U A022922 3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,2,3,2,2,3,2,2,3,2,2
%N A022922 Number of integers m such that 5^n < 2^m < 5^(n+1).
%C A022922 Let A(x) be the counting function of terms a(n) = 3 for n <= x. Then lim A(x)/x = 3*log(2)/log(5) - 1 = 0.29202967... as x goes to infinity. - _Vladimir Shevelev_, Mar 21 2013
%H A022922 Amiram Eldar, <a href="/A022922/b022922.txt">Table of n, a(n) for n = 0..10000</a>
%F A022922 Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = log_2(5) (A020858). - _Amiram Eldar_, Apr 09 2021
%e A022922 Contribution from _M. F. Hasler_, Mar 21 2013: (Start)
%e A022922 a(0)=2 because 5^0 = 1 < 2 = 2^1 < 2^2 = 4 < 5 = 5^1,
%e A022922 a(1)=2 because 5^1 = 5 < 8 = 2^3 < 2^4 = 16 < 25 = 5^2,
%e A022922 a(2)=2 because 5^2 = 25 < 32 = 2^5 < 2^6 = 64 < 125 = 5^3,
%e A022922 a(3)=3 because 5^3 = 125 < 128 = 2^7 < 2^8 < 2^9 = 512 < 625 = 5^4. (End)
%t A022922 Join[{2}, Differences @ Table[Floor[n*Log2[5]], {n, 100}]] (* _Amiram Eldar_, Apr 09 2021 *)
%Y A022922 First differences of A061785 (except for the first term).
%Y A022922 Cf. A020858.
%K A022922 nonn
%O A022922 0,1
%A A022922 _Clark Kimberling_
%E A022922 Definition clarified by _M. F. Hasler_, Mar 21 2013