OFFSET
1,1
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 k are 1, 2, 6, 7, 14, 21, 28,...(see A265739).
We observe two properties:
(1) a(n) = m*a(n-m+1) for some n, m=2,3,4.
Examples:
m = 2 => a(7)=2*a(6), a(11)=2*a(10), a(15)=2*a(14), a(20)=2*a(19), a(25)=2*a(24), a(30)=2*a(29),...
m = 3 => a(16)=3*a(14), a(21)=3*a(19), a(26)=3*a(24), a(31)=3*a(29),...
m = 4 => a(4)=4*a(1), a(32)=4*a(29), ...
But, for m=5, the formula (1) is not valid. We find a(13)=5*a(9), a(18)=5*a(10), a(23)=5*a(11), ...
(2) a(n+2) = a(n) + a(n+1) for n = 4, 9, 26, 27, 28, 29, 35, ...
For k > 1, the integer satisfying the definition is such that ceiling(Pi*k - 1/k) = floor(Pi*k + 1/k). - Stefano Spezia, Apr 26 2023
LINKS
Stefano Spezia, Table of n, a(n) for n = 1..52
Takao Komatsu, The interval associated with a Fibonacci number, The Fibonacci Quarterly, Volume 41, Number 1, February 2003.
EXAMPLE
For k=1 there exists two integers a(1)=3 and a(2)=4 in the interval [1*Pi -1/1, 1*Pi + 1/1] = [2.14159...,4.14159...];
for k=2, the number a(3)=6 is in the interval [2*Pi-1/2, 2*Pi+1/2] = [5.783185..., 6.783185...];
for k=6, the number a(4)= 19 is in the interval [6*Pi-1/6, 6*Pi+1/6] = [18.682889..., 19.016223...].
MAPLE
*** the program gives the interval [a, b], a(n) and k ***
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:
MATHEMATICA
kmax=10^9; Flatten[Table[Range[Ceiling[Pi k-1/k], Floor[Pi k+1/k]], {k, kmax}]] (* or limiting memory usage *)
a = {3, 4}; kmax = 10^9; For[k = 1, k <= kmax, k++,
If[(nw = Ceiling[Pi k - 1/k]) == Floor[Pi k + 1/k],
AppendTo[a, nw]]]; a (* Stefano Spezia, Apr 26 2023 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Dec 15 2015
STATUS
approved