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| Integers only (Posted on 2014-03-31) |
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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?
Generalised approach
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 | Comment 6 of 7 | 
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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.
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Posted by Harry
on 2014-03-31 21:53:28 |
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