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 Integers only (Posted on 2014-03-31)
What is the smallest pair (a1,a2) of integers (a1, less than a2) to provide a list of 24 integers using
a recurrent function a(n)=(a(n-1)+a(n-2))/2?

 See The Solution Submitted by Ady TZIDON No Rating

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 Generalised approach | Comment 6 of 7 |
Assuming an proportional to mn, the  equation an = (an-1 + an-2)/2 yields

the auxiliary equation   m2 = m/2 + ½ with roots  1 and -1/2, giving the

general solution:   an = A(1)n + B(-1/2)n   where A and B are constants.

Using the first two terms gives  a1 = A – B/2  and  a2 = A + B/4, so that

A = (a1 + 2a2)/3,  B = 4(a2 – a1)/3  and the general solution is:

an = (a1 + 2a2)/3 + 4(a2 – a1)/(3*(-2)n)

Thus        a24     = (1/3)[a1 + 2a2 + 4(a2 – a1)/224]

= (1/3)[a1 + 2a2 + (a2 – a1)/222]

So for a24 to be an integer, (a2 – a1) must be divisible by 222, and for a1 and

a2 to be as close together as possible, we need a2 – a1 = 222, which gives

a24      = (a1 + 2a2 + 1)/3

= (a1 + 2(a1 + 222) + 1)/3

= (3a1 + 2*222 + 1)/3                                         (1)

Now, 22 = 1 (mod 3), so 2*222 = 2*(22)11 = 2*111 = 2 (mod 3) and this

makes the numerator in (1) a multiple of 3, showing that a2 – a1 = 222 is

a sufficient condition for a24 to be an integer.

If only positive integers are allowed for (a1, a2) then choose (1, 222 + 1).

If zero is allowed then (0, 222) is best, as Jer recommended.

If all integers are allowed then (-221, 221) have the smallest abs. values,

but (-2796202, -2796202 + 222) will also work as tomarken found.

 Posted by Harry on 2014-03-31 21:53:28

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