A269979 Decimal expansion of the number having (1,2,3,4,...) = A000027 as its factorial-nested interval sequence.

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

Offset

0,1

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.

See A269970 for a guide to related sequences.

Links

Table of n, a(n) for n=0..85.

Example

x = 0.59045235435205548168124328101350...

Mathematica

r[n_] := 1/n!; n[k_] := k; Table[n[k], {k, 1, 1000}];
len[1] = r[n[1]] - r[n[1] + 1];
len[k_] := len[k - 1]*(r[n[k]] - r[n[k] + 1])
sum = r[n[1] + 1] + Sum[len[i]*r[n[i + 1] + 1], {i, 1, 300}];
g = N[sum, 150]
RealDigits[g, 10, 100][[1]]

Crossrefs

Cf. A000027, A000142, A269970.

Sequence in context: A144408 A371323 A019636 * A102521 A198615 A195372

Adjacent sequences: A269976 A269977 A269978 * A269980 A269981 A269982

Keyword

nonn,cons,easy

Author

Clark Kimberling, Mar 08 2016

Status

approved