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%I #5 Mar 10 2016 12:16:24
%S 1,6,9,4,4,6,6,4,9,0,6,6,4,3,3,8,5,1,0,7,6,1,9,5,9,6,5,9,7,9,6,9,0,2,
%T 4,1,7,5,8,6,9,5,6,0,5,1,9,2,6,9,9,2,4,2,8,4,1,9,7,8,2,4,9,4,2,4,3,0,
%U 9,5,1,0,1,3,4,7,0,6,1,8,1,0,9,9,6,3
%N Decimal expansion of the number having (2,4,6,8,10,...) = A005843 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 = 0.16944664906643385107619596597969024...
%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. A000142, A005843, A269970.
%K nonn,cons,easy
%O 0,2
%A _Clark Kimberling_, Mar 08 2016
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