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

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, 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, 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

Offset: 0

Author

Clark Kimberling

Keywords

• nonn

Comments

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

Examples

			Contribution from _M. F. Hasler_, Mar 21 2013: (Start)
a(0)=2 because 5^0 = 1 < 2 = 2^1 < 2^2 = 4 < 5 = 5^1,
a(1)=2 because 5^1 = 5 < 8 = 2^3 < 2^4 = 16 < 25 = 5^2,
a(2)=2 because 5^2 = 25 < 32 = 2^5 < 2^6 = 64 < 125 = 5^3,
a(3)=3 because 5^3 = 125 < 128 = 2^7 < 2^8 < 2^9 = 512 < 625 = 5^4. (End)
		

Links

• Amiram Eldar, Table of n, a(n) for n = 0..10000

Crossrefs

First differences of A061785 (except for the first term).

Cf. A020858.

Programs

Mathematica

Join[{2}, Differences @ Table[Floor[n*Log2[5]], {n, 100}]] (* Amiram Eldar, Apr 09 2021 *)

Formula

Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = log_2(5) (A020858). - Amiram Eldar, Apr 09 2021

Extensions

Definition clarified by M. F. Hasler, Mar 21 2013