A265735 Integers in the interval [Pi*k - 1/k, Pi*k + 1/k] for some k > 0.

3, 4, 6, 19, 22, 44, 66, 88, 333, 355, 710, 1065, 1420, 1775, 2130, 2485, 2840, 3195, 3550, 3905, 4260, 4615, 4970, 5325, 5680, 6035, 103993, 104348, 208341, 312689, 521030, 833719, 1146408, 2292816, 4272943, 5419351, 10838702, 16258053, 80143857, 85563208

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 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

Examples

			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...].
		

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.
• Eric Weisstein's World of Mathematics, Pi Continued Fraction

Crossrefs

Cf. A000796, A265739. Contains A002485 (without the first two terms) as a subsequence.

Programs

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 *)