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$$3,8,17,32,57,\ldots$$ How do we find the $n^{th}$ term? I have been roaming aimlessly for a few minutes now. A definite pattern is in the differences of differences. Hints will be apreciated.

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up vote 1 down vote accepted

$$3,8,17,32,57,\cdots$$

Taking the differences of consecutive terms $$5,9,15,25,\cdots$$

Taking the differences of consecutive terms $$4,6,10,\cdots$$

Taking the differences of consecutive terms $$2,4,\cdots,\text{ whose } n\text{ term } =2^n$$

The $n$ th term of $4,6,10,\cdots$ will be $2^n+(n$ th term of $2,2,2,\cdots)$ i.e., $2^n+2$

The $n$ th term of $5,9,15,25\cdots$ will be $2^n+2+(n$th term of $1,3,5,\cdots)$ i.e., $2^n+2+(2n-1)$

The $n$ th term of $3,8,17,32,57,\cdots$ will be $(2^n+2+2n-1)+(n$ th term of $-2,-1,2,7\cdots)$

Now the $n$ th term of $-2,-1,2,7\cdots$ can be found to be $(n-1)^2-2$ by similar difference method.

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From OEIS, the sequence is $$2^n + n^2$$

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I was hovering along the same direction almost.Thanks for the answer. – rah4927 Jun 9 '13 at 16:55

Let p(n) be the nth prime. Let o(n) be the nth term in OEIS sequence A190525, the number of n-step one-sided prudent walks. The given sequence is clearly:

p(1) + o(1), p(3) + o(2), p(5)+ o(3), p(7)+ o(4), p(9)+o(5),..., p(2n-1)+o(n).

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This can be anything you want, you only have to choose a particular pattern.

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I had forseen this answer even before I ever posted this question. – rah4927 Jun 9 '13 at 17:32

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