A265739 - OEIS

%I #19 Nov 07 2022 09:51:29

%S 1,2,6,7,14,21,28,106,113,226,339,452,565,678,791,904,1017,1130,1243,

%T 1356,1469,1582,1695,1808,1921,33102,33215,66317,99532,165849,265381,

%U 364913,729826,1360120,1725033,3450066,5175099,25510582,27235615

%N Numbers k such that there exists at least one integer in the interval [Pi*k - 1/k, Pi*k + 1/k].

%C Conjecture: the sequence is infinite.

%C See the reference for a similar problem with Fibonacci numbers.

%C For k > 1, the interval [Pi*k - 1/k, Pi*k + 1/k] contains exactly one integer.

%C The corresponding integers in the interval [Pi*k - 1/k, Pi*k + 1/k] are 3, 4, 6, 19, 22, 44, 66, 88, ... (see A265735).

%C The sequence is infinite by Dirichlet's approximation theorem. In other words, the irrationality measure of Pi is at least 2 so this sequence is infinite. - _Charles R Greathouse IV_, Nov 07 2022

%H Takao Komatsu, <a href="http://www.fq.math.ca/Scanned/41-1/komatsu.pdf">The interval associated with a Fibonacci number</a>, The Fibonacci Quarterly, Volume 41, Number 1, February 2003.

%e For k=1, there exist two integers, 3 and 4, in the interval [1*Pi - 1/1, 1*Pi + 1/1] = [2.14159..., 4.14159...];

%e for k=2, the number 6 is in the interval [2*Pi - 1/2, 2*Pi + 1/2] = [5.783185..., 6.783185...].

%e for k=6, the number 19 is in the interval [6*Pi - 1/6, 6*Pi + 1/6] = [18.682889..., 19.016223...].

%p # program gives the interval [a,b], the first integer in [a,b] and n

%p nn:=10^9:

%p for n from 1 to nn do:

%p x1:=evalhf(Pi*n-1/n):y1:=evalhf(Pi*n+1/n):

%p x:=floor(x1):y:=floor(y1):

%p for j from x+1 to y do:

%p printf("%g %g %d %d\n",x1,y1,j,n):

%p od:

%p od:

%Y Cf. A000796, A265735.

%K nonn

%O 1,2

%A _Michel Lagneau_, Dec 15 2015