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A269980 Decimal expansion of the number having (1,3,5,7,9,...) = A005408 as its factorial-nested interval sequence. 1
5, 2, 0, 9, 2, 0, 1, 4, 9, 6, 5, 3, 4, 8, 7, 4, 7, 5, 7, 6, 2, 2, 8, 1, 9, 8, 9, 1, 1, 8, 7, 4, 3, 3, 7, 5, 4, 8, 1, 4, 5, 7, 9, 0, 7, 6, 5, 4, 9, 6, 8, 3, 6, 7, 1, 8, 3, 5, 7, 1, 7, 3, 6, 0, 5, 6, 5, 6, 3, 6, 0, 0, 1, 4, 3, 3, 2, 3, 4, 6, 3, 9, 4, 6, 6, 0 (list; constant; graph; refs; listen; history; text; internal format)
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
EXAMPLE
x = 0.5209201496534874757622819891187433754...
MATHEMATICA
r[n_] := 1/n!; n[k_] := 2 k -1; 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
Sequence in context: A213198 A021196 A257406 * A319231 A058512 A111560
KEYWORD
nonn,cons,easy
AUTHOR
Clark Kimberling, Mar 08 2016
STATUS
approved

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Last modified March 29 10:44 EDT 2024. Contains 371268 sequences. (Running on oeis4.)