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

1,2

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

Links

Table of n, a(n) for n=1..39.

Takao Komatsu, The interval associated with a Fibonacci number, The Fibonacci Quarterly, Volume 41, Number 1, February 2003.

Example

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...].

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:

Crossrefs

Cf. A000796, A265735.

Sequence in context: A087376 A199974 A176279 * A281167 A210660 A344343

Adjacent sequences: A265736 A265737 A265738 * A265740 A265741 A265742

Keyword

nonn

Author

Michel Lagneau, Dec 15 2015

Status

approved