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A269980
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Decimal expansion of the number having (1,3,5,7,9,...) = A005408 as its factorial-nested interval sequence.
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1
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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
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OFFSET
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0,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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x = 0.5209201496534874757622819891187433754...
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MATHEMATICA
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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]]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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