OFFSET
1,1
COMMENTS
Suppose that r = (r(n)) is a sequence satisfying (i) 1 = r(1) > r(2) > r(3) > ... and (ii) r(n) -> 0. For x in (0,1], let n(1) be the index n such that r(n+1) , x <= r(n), and let L(1) = r(n(1))-r(n(1)+1). Let n(2) be the index n such that r(n(1)+1) < x <= r(n(1)+1) + L(1)r(n), and let L(2) = (r(n(2))-r(r(n)+1)L(1).
Continue inductively to obtain the sequence (n(1), n(2), n(3), ... ), the r-nested interval sequence of x. Taking r = (1/n!) gives the factorial-nested interval sequence of x.
Conversely, given a sequence s= (n(1),n(2),n(3),...) of positive integers, the number x having satisfying NI(x) = s is the sum of left-endpoints of nested intervals (r(n(k)+1), r(n(k))]; i.e., x = sum{L(k)r(n(k+1)+1), k >=1}, where L(0) = 1.
Guide to related sequences:
x factorial-nested interval sequence
1/e A269970
e-2 A269971
1/pi A269972
pi-3 A269973
sqrt(1/2) A269974
sqrt(2)-1 A269975
sqrt(1/3) A269976
sqrt(3)-1 A269977
1/tau A269978
A269979 (1,2,3,4,5,6,7,...)
A269980 (1,3,5,7,9,11,...)
A269981 (2,4,6,8,10,13,...)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 08 2016
STATUS
approved