A153725 Least number m such that floor((3^n-m)/(2^n-m)) > floor(3^n/2^n).

1, 2, 2, 3, 2, 4, 7, 4, 8, 7, 12, 9, 17, 4, 8, 16, 99, 20, 39, 235, 49, 97, 194, 885, 1106, 439, 2059, 968, 4034, 5268, 3070, 1163, 2325, 4649, 9297, 18593, 16210, 4452, 8903, 67524, 68757, 49124, 98248, 39360, 288234, 17763, 35526, 567677, 1135354

Offset: 1

Author

Hieronymus Fischer, Jan 06 2009

Keywords

• nonn
• easy

Comments

Provided A002379(n) = floor((3^n-1)/(2^n-1)) holds (which is proved only for 1 < n <= 305000), then a(n) > 1.

Examples

			a(5)=2, since floor((3^5-1)/(2^5-1)) = floor(242/31) = 7 = floor(243/32) = floor(3^5/2^5), but floor((3^5-2)/(2^5-2)) = floor(241/30) = 8 > 7.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..8005

Crossrefs

Cf. A002379, A081464, A153701, A137994, A153717, A154130.

Programs

Mathematica

Table[n3 = 3^n; n2 = 2^n; m = 1;
While[Floor[(n3 - m)/(n2 - m)] <= Floor[n3/n2], m++]; m, {n,1,50}] (* Robert Price, Mar 27 2019 *)

PARI

a(n) = my(f = floor(3^n/2^n)); ceil(((f + 1)*(2^n) - 3^n)/f) \\ David A. Corneth, Mar 27 2019

Formula

a(n) = ceiling(((f + 1)*(2^n) - 3^n)/f) where f = floor(3^n/2^n). - David A. Corneth, Mar 27 2019