|
%I #5 Mar 10 2016 12:16:15
%S 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,
%T 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,
%U 3,6,0,0,1,4,3,3,2,3,4,6,3,9,4,6,6,0
%N Decimal expansion of the number having (1,3,5,7,9,...) = A005408 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 = 0.5209201496534874757622819891187433754...
%t r[n_] := 1/n!; n[k_] := 2 k -1; 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, A005408, A269970, A269979.
%K nonn,cons,easy
%O 0,1
%A _Clark Kimberling_, Mar 08 2016
|
