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Scary sequences

  1. Nov 6, 2013 #1

    adjacent

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    Gold Member

    1. The problem statement, all variables and given/known data
    I have two questions
    1.Today,in my test paper,I got this sequence. Find the nth term formula
    1,3,15,61,253
    I didn't know how to start.This is clearly not an arithmetic or geometric sequence.
    Any help?

    2. And is there any formula for finding the nth term of sequences with multiple level?
    What I mean as multiple level here is this.
    2,4,6,8,10 below the sequence is the difference between two numbers.
    1. 2 2 2 2

    This is a first level sequence.

    2,4,8,14,22
    1. 2 4 6 8
    2. 2 2 2

    This is a 2nd level sequence.

    Is there any formula for finding the nth term of n level sequences?

    2. Relevant equations
    a+d(n-1)
    ar(n-1)


    3. The attempt at a solution
    I really don't know
     
    Last edited: Nov 6, 2013
  2. jcsd
  3. Nov 6, 2013 #2

    jedishrfu

    Staff: Mentor

    What was the actual question?

    Did they want you to create a formula for the nth term or just predict the next number in the sequence?
     
  4. Nov 6, 2013 #3

    adjacent

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    Gold Member

    Create the formula for nth term.
     
  5. Nov 6, 2013 #4

    jedishrfu

    Staff: Mentor

    I don't think there is a generic formula that you can use here instead you must discern the pattern and write one yourself.

    I see a kind of powers of two pattern in your sequence but I'm not sure if there is a simpler solution. I noticed that 1 is 2^0 and 3 is 2^2-1 ...
     
  6. Nov 6, 2013 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    There is always formulas- generally, an infinite number of them, for any finite sequence of numbers.

    One way of getting such a formula is to use "Newton's difference formula" to get a polynomial. Given n (x, y) points, there exist a unique polynomial of degree n-1 and up. We can think of the sequence 1,3,15,61,253 as given by the function f(0)= 1, f(1)= 3, f(2)= 15, f(3)= 61, and f(4)= 253.
    The "first differences" are 3-1= 2, 15- 3= 12, 61- 15= 46, 253- 61= 192.
    The "second differences" are 12- 2= 10, 46- 12= 34, 192- 46= 144.
    The "third differences" are 34- 10= 24, 144- 34= 110.
    The "fourth difference" is 110- 24= 76.

    By "Newton's difference formula" these numbers are given by 1+ 2n+ (10/2)n(n+1)+(24/6)n(n+1)(n+2)+ (76/24)n(n+1)(n+ 2)(n+ 3).
     
  7. Nov 6, 2013 #6

    jedishrfu

    Staff: Mentor

    A small correction 192-46 = 146 and so 146-34 = 112 and 112-24 = 88

    I saw a curious pattern

    2^0, 2^2-1, 2^4-1, 2^6-3, 2^8-3, 2^10-5 so I figured the next number would be 2^12-5 ...
     
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