

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
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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).


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*n1/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 A226965
Adjacent sequences: A265736 A265737 A265738 * A265740 A265741 A265742


KEYWORD

nonn


AUTHOR

Michel Lagneau, Dec 15 2015


STATUS

approved



