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Yet another number pattern

  1. Jun 9, 2005 #1
    4
    5
    11
    34
    65
    111
    175
    260
    505
    540
    671

    what is the next line
     
  2. jcsd
  3. Jun 10, 2005 #2
    really no one got an idea ?
     
  4. Jun 11, 2005 #3
    Is the next number 802.
     
  5. Jun 11, 2005 #4
    ArielGenesis, maybe you can give any hint?
     
  6. Jun 13, 2005 #5
    my fault acually, but it is suppose to be
    1
    5
    15
    34
    65
    111
    175
    260
    369
    505
    671

    not much diffrent isn't it.
    no neveza, it's not 802
    the hint is that it involve listing a group of number and doing sum.
     
  7. Jun 14, 2005 #6

    1 = 1
    5 = 2 + 3
    15 = 4 + 5 + 6
    34 = 7 + 8 + 9 + 10
    65 = 11 + 12 + 13 + 14 + 15
    111 = 16 + 17 + 18 + 19 + 20 + 21
    175 = 22 + 23 + 24 + 24 + 26 + 27 + 28
    260 = 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36
    369 = 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45
    505 = 46 + 47 + 48 + 49 + 50 + 51 + 52 + 53 + 54 + 55
    671 = 56 + 57 + 58 + 59 + 60 + 61 + 62 + 63 + 64 + 65 + 66
    870 = 67 + 68 + 69 + 70 + 71 + 72 + 73 + 74 + 75 + 76 + 77 + 78
     
  8. Jun 14, 2005 #7
    4
    5
    11
    34
    65
    111
    175
    260
    505
    540
    671
    1381

    Is that anywhere close? You really don't want to know how my messed up mind found that answer though. It might make you go crazy.... :biggrin:

    Oopss! You posted another one...

    1
    5
    15
    34
    65
    111
    175
    260
    369
    505
    671
    870

    Looks like that should be the answer for this second set of numbers.
     
    Last edited: Jun 14, 2005
  9. Jun 14, 2005 #8
    Wow! We got the answers a different way; but I'm sure that they equate.

    I did it this way:

    C = 6 9 12 15 18 21 24 27 30 ??? [ Difference in #'s from B ]
    B = 4 10 19 31 46 64 85 109 136 166 ??? [ Difference in #'s from A]
    A=1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, ???

    In C, the ??? should, of course, be 33, which makes the ??? in B = 199, which makes the ??? in A = 870
     
  10. Jun 14, 2005 #9
    Rahmuss, your approach is systematic and works in a large number of cases of this kind of puzzle.
     
  11. Jun 15, 2005 #10
    sorry, it is my fault. and i yup, jimmy (as usual) got it right

    can we form a formula based on rahmuss asnwer. as i also had not notice it. and how could it happen to be 3 if he is going to make D = 3,3,3,3,3,3,3 [diffrence in #'s from c]
     
  12. Jun 15, 2005 #11
    One possible formula is this difference formula:
    [itex]A_n = 3 \times A_{n-1} - 3 \times A_{n-2} + A_{n - 3} + 3;[/itex]

    Another would be a cubic polynomial in n, but I haven't figured out the coefficients yet.
     
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