Posers and Puzzles

Posers and Puzzles

  1. Subscriberjoe shmo
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    19 Nov '07 22:25
    like

    series, 6,8,11,15,20

    i have read a little bit about it

    a(1) =6 a(2) = 8 a(3)= [a(2) - a(1)+1] + a(2)

    a(4) = [a(3) - a(2) +1] + a(3)...........

    how is a(n) composed?
  2. Subscriberjoe shmo
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    19 Nov '07 22:49
    Originally posted by joe shmo
    like

    series, 6,8,11,15,20

    i have read a little bit about it

    a(1) =6 a(2) = 8 a(3)= [a(2) - a(1)+1] + a(2)

    a(4) = [a(3) - a(2) +1] + a(3)...........

    how is a(n) composed?
    would it be something like this for the above series

    a(n) = [ a(n-1) - a(n-2) + 1] + a(n-1)

    i see no way to move to a(n) without figuring out all previous a(n)'s

    Is there
  3. Standard memberPalynka
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    19 Nov '07 23:191 edit
    Originally posted by joe shmo
    like

    series, 6,8,11,15,20

    i have read a little bit about it

    a(1) =6 a(2) = 8 a(3)= [a(2) - a(1)+1] + a(2)

    a(4) = [a(3) - a(2) +1] + a(3)...........

    how is a(n) composed?
    a(n) = a(n-1) + n = a(n-2) + n-1 + n = 5 + 1 + 2 + ... + n

    So if your n is 100 you have 5 + (1 + 2 + 3 + ... + 100)

    If I understand your notation.
  4. Subscriberjoe shmo
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    19 Nov '07 23:421 edit
    Originally posted by Palynka
    a(n) = a(n-1) + n = a(n-2) + n-1 + n = 5 + 1 + 2 + ... + n

    So if your n is 100 you have 5 + (1 + 2 + 3 + ... + 100)

    If I understand your notation.
    so this is a simplified version of what i have above?

    Your version looks like an arithmatic sequence, is it? could you use the nth term formula in this instance?

    which is a(n) = a(1) + (n-1)d

    or would the be the sum of the first n terms S(n)= n/2[a(1)+a(n)]
  5. Subscriberjoe shmo
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    19 Nov '07 23:501 edit
    Originally posted by joe shmo
    so this is a simplified version of what i have above?

    Your version looks like an arithmatic sequence, is it? could you use the nth term formula in this instance?

    which is a(n) = a(1) + (n-1)d

    or would the be the sum of the first n terms S(n)= n/2[a(1)+a(n)]
    so S(n) = n/2[ a(1) + a(n) ] + 5 for the series 6,8,11,15.......
  6. Standard memberPalynka
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    19 Nov '07 23:572 edits
    Originally posted by joe shmo
    would it be something like this for the above series

    a(n) = [ a(n-1) - a(n-2) + 1] + a(n-1)

    i see no way to move to a(n) without figuring out all previous a(n)'s

    Is there
    To explain a little more, one way to mathematically do it is by recursive substitution (this is the brute force, no-shortcuts method).

    You have:
    (1) a(n) = [ a(n-1) - a(n-2) + 1] + a(n-1) = 2.a(n-1) - a(n-2) + 1

    But if this is valid for all n, then it is also valid for n-1:

    (2) a(n-1) = 2.a(n-2) - a(n-3) + 1

    Replace (2) for the a(n-1) (1) and you have:

    a(n) = 4.a(n-2) - 2.a(n-3) + 2 - a(n-2) + 1 = 3. a(n-2) - 2.a(n-3) + 2 + 1

    Do it again for a(n-2) as in (2), you have:

    a(n-2) = 2.a(n-3) - a(n-4) + 1
    => a(n) = 4.a(n-3) - 3.a(n-4) + 3 + 2 + 1

    Do you see a pattern here?

    a(n) = (t+1)a(n-t) - t.a(n-t-1) + t + t-1 + ... + 1

    Choose t = n-2 and you have:
    a(n) = (n-1)a(2) - (n-2)a(1) + n-2 + n-3 + ... + 1

    a(n) = 8n - 8 - 6n + 12 + n-2 + n-3 + ... + 1

    a(n) = 2n + 4 + n-2 + n-3 + ... + 1

    Since 2n = n + n-1 + 1, you can change the last term to:

    a(n) = 5 + n + n-1 + n-2 + n-3 + ... + 1

    Uf.

    Alternatively (which was what I did). You can look at your series and see a pattern:

    Value-> 6, 8, 11, 15, 20
    Rank-> 1, 2, 3 , 4 , 5

    Each number is the previous number plus its rank:
    (3) a(n) = a(n-1) + n
    (4) a(n-1) = a(n-2) + n-1
    Substitute (4) in (3) and you have:
    a(n) = a(n-2) + n + n-1
    ...
    a(n) = a(1) + n + n-1 + ... + 2

    Since a(1) = 6 = 5+1

    a(n) = 5 + 1 + 2 + ... + n-1 + n
  7. Subscriberjoe shmo
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    20 Nov '07 00:10
    thanks, I'll work on attempting to fully understand this for kicks, but I have to be honest it looks intimidating....😕

    haha..... thanks again
  8. Standard memberPalynka
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    20 Nov '07 12:06
    Originally posted by joe shmo
    thanks, I'll work on attempting to fully understand this for kicks, but I have to be honest it looks intimidating....😕

    haha..... thanks again
    Look at the alternative if it looks intimidating.

    Each term is the sum of the previous one plus its rank, i.e. a(5) = 20 = a(4) + 5; a(4) = 11 + 4 and so on.
  9. Standard memberwolfgang59
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    20 Nov '07 13:15
    a(n) = 5 +{n(n+1)/2}

    Its triangular numbers plus five

    pictorially its;

    00000
    0
    00
    000
    0000
    00000
    000000

    etcetera
  10. Subscriberjoe shmo
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    20 Nov '07 17:37
    Originally posted by wolfgang59
    a(n) = 5 +{n(n+1)/2}

    Its triangular numbers plus five

    pictorially its;

    00000
    0
    00
    000
    0000
    00000
    000000

    etcetera
    what about these triangular numbers plus five, I pulled the series from the "what you have is ....." thread the creator posted a problem with these numbers
  11. Subscriberjoe shmo
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    20 Nov '07 17:41
    Originally posted by Palynka
    Look at the alternative if it looks intimidating.

    Each term is the sum of the previous one plus its rank, i.e. a(5) = 20 = a(4) + 5; a(4) = 11 + 4 and so on.
    right.....but that quite easy to understand so i will try to put some effort into the theory.
  12. Standard membergenius
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    20 Nov '07 23:45
    Originally posted by joe shmo
    right.....but that quite easy to understand so i will try to put some effort into the theory.
    In all honesty, your answer is sufficient. Take the Fibonacci sequence, for instance, which takes the form 1,1,2,3,5,8...

    Normally you would write it as a recurrence relation,

    f[n]=f[n-2]+f[n-1].

    However, we can also write it as

    f[n]=((A^n)-(1-A)^n)/sqrt(5)

    where A=(1+sqrt(5))/2. Writing it in any other form would be unnecessary.
  13. Standard memberchessisvanity
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    20 Nov '07 23:491 edit
    ...........😞
  14. Subscriberjoe shmo
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    21 Nov '07 00:11
    Originally posted by genius
    In all honesty, your answer is sufficient. Take the Fibonacci sequence, for instance, which takes the form 1,1,2,3,5,8...

    Normally you would write it as a recurrence relation,

    f[n]=f[n-2]+f[n-1].

    However, we can also write it as

    f[n]=((A^n)-(1-A)^n)/sqrt(5)

    where A=(1+sqrt(5))/2. Writing it in any other form would be unnecessary.
    thanks for that, I was just going to ask about the Fibonacci sequence and I had a hunch that the golden ratio could be used to figure out
    a(n) is the ratio used as the average distance for n > or = to 3
  15. Joined
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    21 Nov '07 10:141 edit
    Originally posted by joe shmo
    thanks for that, I was just going to ask about the Fibonacci sequence and I had a hunch that the golden ratio could be used to figure out
    a(n) is the ratio used as the average distance for n > or = to 3
    There's a standard way of solving these sort of problems ('difference equations'😉. It's actually very similar to the method of solving ordinary differential equations (if you've come across those).

    For the Fibonnaci series: a(n+2) = a(n + 1) + a(n)

    Look for a solution in the form a(n) = x^n.

    x^(n + 2) = x^(n + 1) + x^n
    => x^n[x^2 - x - 1] = 0

    This gives you a quadratic equation. Using the usual quadratic formula:
    x = (1 +- sqrt(5))/2

    But the problem is linear - if you have two solutions you can add them together and you still have a solution. So the most general solution is:

    a(n) = A[(1 + sqrt(5))/2]^n + B[(1 - sqrt(5))/2]^n

    To find A and B you look at the first two terms. You know a(0) = 1 and a(1) = 1.
    => A + B = 1
    and (A - B)sqrt(5) = 1

    Solve for A and B and you've got the answer.
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