We are asked to find the nth term (general term) of a sequence that begins 7, 11, 19, 23, 31, 35 . . .
First a general note that there are an infinite number of sequences that begin with these six numbers.
We observe an alternating pattern: starting with 7 we add 4, add 8, add 4, add 8, then add 4. Assuming this pattern continues the sequence would continue 43,47,55 . . . One way to describe such a sequence is to define it in a piece-wise manner.
a_{n}={ (6n+1, " if " n "odd"),(6n-1," if " n "even"):}
So a1=6(1)+1=7, a2=6(2)-1=11, a3=6(3)+1=19 etc...
We can write this in closed form as:
a_{n}=6n + (-1)^(n-1)
Where did the 6 come from? Consider the interleaved sequences: one sequence is 7,19,31 . . . and the other is 11,23,35 . . . For the first sequence we have n=1 a(1)=7; n=3 a(3)=19; n=5 a(5)=31—the underlying model for (1,7), (3,19), (5,31) is linear with a slope of 6. Likewise we have (2,11), (4,23), (6,35) which also is linear with a slope of 6. The first sequence is given by 6n+1 and the second sequence by 6n-1.
http://mathworld.wolfram.com/Sequence.html
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