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A265739
Numbers k such that there exists at least one integer in the interval [Pi*k - 1/k, Pi*k + 1/k].
1
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
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
Sequence in context: A199974 A176279 A376296 * A281167 A210660 A344343
KEYWORD
nonn
AUTHOR
Michel Lagneau, Dec 15 2015
STATUS
approved