A269979 - OEIS

%I #4 Mar 10 2016 21:02:45

%S 5,9,0,4,5,2,3,5,4,3,5,2,0,5,5,4,8,1,6,8,1,2,4,3,2,8,1,0,1,3,5,0,2,4,

%T 2,7,9,7,1,0,4,3,5,7,7,1,7,7,7,3,5,0,0,6,3,9,6,4,8,3,9,1,0,6,9,9,8,9,

%U 2,0,1,5,6,0,3,8,8,8,9,8,9,4,6,7,7,0

%N Decimal expansion of the number having (1,2,3,4,...) = A000027 as its factorial-nested interval sequence.

%C 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.

%C 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.

%C See A269970 for a guide to related sequences.

%e x = 0.59045235435205548168124328101350...

%t r[n_] := 1/n!; n[k_] := k; Table[n[k], {k, 1, 1000}];

%t len[1] = r[n[1]] - r[n[1] + 1];

%t len[k_] := len[k - 1]*(r[n[k]] - r[n[k] + 1])

%t sum = r[n[1] + 1] + Sum[len[i]*r[n[i + 1] + 1], {i, 1, 300}];

%t g = N[sum, 150]

%t RealDigits[g, 10, 100][[1]]

%Y Cf. A000027, A000142, A269970.

%K nonn,cons,easy

%O 0,1

%A _Clark Kimberling_, Mar 08 2016