To help pass time on a long bus ride, a group of math teachers created a sequence of numbers, with each teacher saying one term in the sequence. The first and second teachers each said a non-negative integer, and every teacher after that said the sum of all of the previous terms in the sequence.
For example, if the first teacher said the number \(2\) and the second teacher said the number \(8\), then
the third teacher would say the sum of the first and second terms, which is \(2+8=10\), and
the fourth teacher would say the sum of the first, second, and third terms, which is \(2+8+10=20\).
How many possible sequences could the teachers have said if the first teacher said the number \(3\) and another teacher said the number \(3072\)?
We know how to construct the sequence, and we know that the first term is \(3\), but where is the term whose value is \(3072\)?
Could \(3072\) be the second term?
If the first two terms are \(3\) and \(3072\), then we can calculate the next few terms.
The third term would be \(3+3072=3075\).
The fourth term would be \(3+3072+3075=3075 + 3075=2(3075)=6150\).
The fifth term would be \(3+3072+3075+6150=6150+6150=2(6150)=12\,300\).
We see that we can determine any term beyond the third term by summing all of the previous terms, or we can simply double the term immediately before the required term, since that term is the sum of all the preceding terms. (This also means that if any term after the third term is known, then the preceding term is half the value of that term.)
Therefore, there is one possible sequence with \(3072\) as the second term. The first \(6\) terms of this sequence are \(3,~3072,~3075,~6150,~12\,300,~24\,600\).
Could \(3072\) be the third term?
Yes, since the third term is the sum of the first two terms, and the first term is \(3\), then the second term would be \(3072-3=3069\) and the first \(6\) terms of this sequence are \(3,~ 3069,~ 3072,~ 6144,~ 12\,288,~ 24\,576\).
Could \(3072\) be the fourth term?
Yes, since the fourth term is even, then we can determine the third term to be half of the fourth term, which is \(3072\div 2=1536\), then the second term would be \(1536-3=1533\). The first \(6\) terms of this sequence are \(3,~ 1533,~ 1536,~ 3072,~ 6144,~ 12\,288\).
Could \(3072\) be the fifth term?
To get from the fifth term to the third term we would divide by \(2\) twice, or we could divide by \(4\). If the resulting third term is a non-negative integer greater than or equal to \(3\), then the sequence exists. The third term would be \(3072\div 4=768\), and the second term would be \(768-3=765\). Thus the sequence exists and the first \(6\) terms are \(3,~ 765,~ 768,~ 1536,~ 3072,~ 6144\).
We could continue in this way until we discover all possible sequences that are formed according to the given rules with first term \(3\) and \(3072\) somewhere in the sequence. However, if we look at the prime factorization of \(3072\) we see that the highest power of \(2\) that divides \(3072\) is \(1024\) (or \(2^{10}\)), since \(3072=2^{10}\times 3\). In fact, dividing \(3072\) by \(1024\) would produce a third term that would be \(3\). The second term would then be \(0\), a non-negative integer, and the resulting sequence would be \(3,~0,~3,~6,~12,~24,~48,~96,~192,~384,~768,~1536,~3072,~6144,\ldots\).
If we divide \(3072\) by any integral power of \(2\) from \(2^0=1\) to \(2^{10}=1024\), the resulting third term would be an integer greater than or equal to \(3\), and \(3072\) would appear in each of these sequences. There are \(11\) such sequences. The number \(3072\) would appear somewhere from term \(3\) to term \(13\) in the acceptable sequence. However, \(3072\) can also appear as the second term, so there are a total of \(12\) possible sequences.
Could \(3072\) be the fourteenth term? From the fourteenth term to the third term we would need to divide \(3072\) by \(2^{11}\). The resulting third term would be \(\frac{3}{2}\). This would mean the second term is not an integer and so the sequence is not possible.
Therefore, there are a total of \(12\) such sequences.