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Friday, 24 January 2025

A Challenging Sequence

Here is a sequence that appeared in a 1926 SAT examination where participants had about twenty seconds on average to solve each question: 750,21,264,183,210,,

Given the first five members of the sequence, one must find the next two members. At first there seems to be no pattern at all but the key to the solution is to look at the differences between the members of the sequence. 

With this insight, the solution is straightforward. Let's look at the differences between successive terms:21750=72926421=243183264=81210183=27

This gives the sequence:729,243,81,27=36,+35,34,+33
Clearly the remaining differences are -3^2 and +3 and the sequence is then as shown in Figure 1:


Figure 1

In sequence notation we could express the successive terms using this formula:an+1=an+(1)n+136n with a0=750

The terms of this sequence will approach a limit of 203.25 (permalink). 

It's easy to generate other sequences using this approach. For example:100,104,95,120,56,,

Once again, we are given the first five members of the sequence and must find the next two. Let's look at the differences:104100=495104=912095=2556120=64
These differences are all squares so let's write the difference in base index notation. We get:22,32,52,82
The bases (2, 3, 5 and 8) are Fibonacci numbers and so the progression must be 132 and 212. The sequence is thus:
100,104,95,120,56,225,216
The absolute value of the terms of this sequence will increase without bound, alternating between positive and negative values. The formula for this sequence is (permalink):an+1=an+(1)n(fibonacci(n+3))2 with a0=100
The general formula for these types of sequences is:an+1=an+f(n)
The approach to finding missing terms is to find what f(n) is:f(n)=an+1an
The form of this function should emerge when considering the initial differences between the terms:
f(0),f(1),f(2),f(3),

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