A265739 Numbers k such that there exists at least one integer in the interval [Pi*k - 1/k, Pi*k + 1/k].

1, 2, 6, 7, 14, 21, 28, 106, 113, 226, 339, 452, 565, 678, 791, 904, 1017, 1130, 1243, 1356, 1469, 1582, 1695, 1808, 1921, 33102, 33215, 66317, 99532, 165849, 265381, 364913, 729826, 1360120, 1725033, 3450066, 5175099, 25510582, 27235615

Offset: 1

Author

Michel Lagneau, Dec 15 2015

Keywords

• nonn

Comments

Conjecture: the sequence is infinite.

See the reference for a similar problem with Fibonacci numbers.

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

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).

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

Examples

			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...];
for k=2, the number 6 is in the interval [2*Pi - 1/2, 2*Pi + 1/2] = [5.783185..., 6.783185...].
for k=6, the number 19 is in the interval [6*Pi - 1/6, 6*Pi + 1/6] = [18.682889..., 19.016223...].
		

Links

• Takao Komatsu, The interval associated with a Fibonacci number, The Fibonacci Quarterly, Volume 41, Number 1, February 2003.
• Eric Weisstein's World of Mathematics, Pi Continued Fraction

Crossrefs

Cf. A000796, A265735. Contains A002486 (without the first two terms) as a subsequence.

Programs

Maple

# program gives the interval [a,b], the first integer in [a,b] and n
nn:=10^9:
for n from 1 to nn do:
x1:=evalhf(Pi*n-1/n):y1:=evalhf(Pi*n+1/n):
x:=floor(x1):y:=floor(y1):
for j from x+1 to y do:
printf("%g %g %d %d\n",x1,y1,j,n):
od:
od: