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Finding a formula for a sequence with recurring digits

  1. Jul 26, 2012 #1
    how do i determine the formula for the sequence below?

    1,1,2,2,2,2,3,3,3,3,3,4,4,4,4,4,4,4,4,...,n,n,n,n,n,...,n,...

    need some instructions. thanks in advance
     
  2. jcsd
  3. Jul 26, 2012 #2
    Ok, there are subsequences of equal entries (increasing by one) in a row, and the numbers of equal entries in each of them form the following sequence:

    2, 4, 5, 8, ...

    I can't see any regularity. Sorry.
     
  4. Jul 26, 2012 #3

    haruspex

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    I can't see a pattern here: you have two 1s, four 2s, five 3s and eight 4s. Can you confirm that you've written it out correctly?
     
  5. Jul 27, 2012 #4
    Seems like one too many 2's with n-1 repeated F(n+1) times for n = 2,3,4,5,... .
     
    Last edited: Jul 27, 2012
  6. Jul 28, 2012 #5
    sorry, i've typed it incorrectly, it should be:

    1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,...

    two 1s, four 2s, six 3s, eight 4s,...
     
  7. Jul 28, 2012 #6

    haruspex

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    If you don't mind the use of [] (integer part of) it's [1/2 +√(n-3/4)]
     
  8. Jul 28, 2012 #7
    So, there are [itex]2 n[/itex] n's. The total number of elements no smaller than n is:
    [tex]
    \sum_{k = 1}^{n}{2 k} = n(n + 1)
    [/tex]

    Thus, all the elements with an index k that satisfies:
    [tex]
    (n - 1) n < k \le n (n + 1)
    [/tex]
    are equal to [itex]x_k = n[/itex] (why?). Can you find n, given k from the above inequalities?
    [tex]
    n^2 + n - k \ge 0 \Rightarrow n \ge -\frac{1}{2} + \sqrt{k + \frac{1}{4}}
    [/tex]
    [tex]
    n^2 - n - k < 0 \Rightarrow n < \frac{1}{2} + \sqrt{k + \frac{1}{4}}
    [/tex]

    If you know the [STRIKE]floor[/STRIKE] floor function, which is the same as integer part for positive integers, you should be able to deduce:
    [tex]
    x_k = \mathrm{Ceiling} \left[ \sqrt{k + \frac{1}{4}} - \frac{1}{2}\right]
    [/tex]
     
    Last edited: Jul 28, 2012
  9. Jul 28, 2012 #8
    [itex] a(n) = round(\sqrt{n})[/itex] seems to work, not that I could prove it.

    This is where a(1) is the first entry in the sequence.
     
  10. Jul 28, 2012 #9
    Good intuition abacus...

    A000194 n appears 2n times; also nearest integer to square root of n
    http://oeis.org/A000194

    If one desires that digits are repeated 2n - 1 times, that's an easy one: floor [sqrt n]
     
  11. Jul 28, 2012 #10
    The definition of ceiling and round functions is:
    [tex]
    \mathrm{Ceiling}(x) \equiv n, \ n \le x < n + 1
    [/tex]
    [tex]
    \mathrm{Round}(x) \equiv m, \ \vert x - m \vert < 1/2
    [/tex]
    (the last definition is ambiguous if the argument is half-integer, but there can never happen for acabus's argument)

    We want to know for what x, and y, [itex]\mathrm{Ceiling}(x) \stackrel{?}{=} \mathrm{Round}(y) = n[/itex]. From the above definitions, this happens when:
    [tex]
    n \le x < n + 1, \ n - 1/2 < y < n + 1/2
    [/tex]
    or, if we rewrite these inequalities for n, we get:
    [tex]
    x - 1 < n \le x, \ y - 1/2 < n < y + 1/2
    [/tex]
    Then, a necessary and sufficient (why?) condition is:
    [tex]
    x - 1 < y + 1/2, \ y - 1/2 < x
    [/tex]

    In our case, [itex]x = \sqrt{k + 1/4} - 1/2[/itex], and [itex]y= \sqrt{k}[/itex], so we have:
    [tex]
    \sqrt{k + 1/4} - 1/2 - 1 < \sqrt{k} + 1/2 \Leftrightarrow \sqrt{k + 1/4} - \sqrt{k} < 2
    [/tex]
    [tex]
    \sqrt{k} - 1/2 < \sqrt{k + 1/4} - 1/2 \Leftrightarrow \sqrt{k + 1/4} - \sqrt{k} > 0
    [/tex]
    The second inequality is surely true, but the first is proven as follows:
    [tex]
    \begin{array}{l}
    \sqrt{k + 1/4} - \sqrt{k} \stackrel{?}{<} 2 \\

    0 < \sqrt{k + 1/4} \stackrel{?}{<} \sqrt{k} + 2 \\

    k + 1/4 \stackrel{?}{<} k + 4 \sqrt{k} + 4 \\

    0 \stackrel{\surd}{<} \sqrt{k} + 15/16
    \end{array}
    [/tex]
    This is surely true. Therefore, both the formulas give the same output for all positive integers.
     
  12. Jul 28, 2012 #11
    Thanks all for your responses.
     
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