OFFSET
1,2
COMMENTS
Conjecture: the sequence is infinite.
See the reference for a similar problem with Fibonacci numbers.
The corresponding integers in each interval [e*k - 1/k, e*k + 1/k] are 2, 3, 5, 8, 11, 19, 38, 87, 106, 193, 386, 1264, ...(see A265741).
For k > 1, the interval [e*k - 1/k, e*k + 1/k] contains exactly one integer.
We observe two properties:
(1) a(n) = m*a(n-m+1) for some n, m=2,3,4 and 5
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), ...
m = 5 => a(33)=5*a(29), ...
(ii) a(n+2) = a(n) + a(n+1) for n = 1, 3, 7, 11, 13, 16, 18, 21, 23, 26, 28, ...
LINKS
Takao Komatsu, The interval associated with a Fibonacci number, The Fibonacci Quarterly, Volume 41, Number 1, February 2003.
EXAMPLE
a(1) = 1 because there exist two integers (2 and 3) in the interval [1*e -1/1, 1*e + 1/1] = [1.71828..., 3.71828...];
a(2) = 2 because the number 5 belongs to the interval [2*e-1/2, 2*e+1/2] = [4.936564..., 5.936564...];
a(3) = 3 because the number 8 belongs to the interval [3*e-1/3, 3*e+1/3] = [7.821512..., 8.488179...].
MAPLE
*** the program gives the interval [a, b], the integer(s) between [a, b] and k ***
nn:=10^9:
e:=exp(1):
for n from 1 to nn do:
x1:=evalhf(e*n-1/n):y1:=evalhf(e*n+1/n):
x:=floor(x1):y:=floor(y1):
for j from x+1 to y do:
printf("%g %g %g %g\n", x1, y1, j, n):
od:
od:
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Michel Lagneau, Dec 15 2015
STATUS
approved