# Is there any pattern in this number sequence?

I have a feeling that it's obvious but I can't find it. I was wondering if anybody else could decode this cryptic thing.

$[1,2,4,7,10,14,19,24,30,37]$

This is very important, I hope someone can see something that I can't.

Also if possible, try to construct an algorithm that can actually predict the one, because I doubt I would have much luck with that either.

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Well... – J. M. Oct 6 '12 at 15:13
maybe there's a zero at the beginning, then this would work: +1,+1,+2,+3,+3,+4,+5,+5,+6,+7,+7,+8,+9,+9... so all odd numbers twice :) rather random. could you add some more numbers? – Alex Oct 6 '12 at 15:30
This would depend critically on where you found this sequence or how it arised. – hjg Oct 6 '12 at 15:55
Why not simply try OEIS or wolfram-alpha wolframalpha.com/input/…... – Gottfried Helms Oct 6 '12 at 18:05
Since there are 10 numbers, you can find a polynomial $P(x)$ of degree at most $9$ so that $a_n=P(n)$. Then, there also exists uncountably many functions $f(x)$ so that $a_n=f(n)$.... Is this the type of pattern you are looking for? ;) – N. S. Oct 6 '12 at 19:19

Compute the first and second differences of the given data: $$\matrix{ 1&&2&&4&&7&&10&&14&&19&&24&&30&&37 \cr &1&&2&&3&&3&&4&&5&&5&&6&&7&\cr &&1&&1&&0&&1&&1&&0&&1&&1&&\cr}$$ Here the third line seems periodic with period $3$ with mean ${2\over3}$. Therefore the given data $(y_k)_{0\leq k\leq9}$ can be produced by a function $k\mapsto f(k)$ of the form $$f(k)={1\over3} k^2 + a\ k+ b+ c\ \cos{2k\pi\over3}+ d\ \sin{2k\pi\over3}\ .$$ Now fix the undetermined coefficients $a$, $b$, $c$, $d$ such that for $k=0,\ldots,3$ the correct values $y_k$ are produced. The final result is $$y_k={1\over3}k^2 + k + {7\over9} + {2\over9} \cos{2k\pi\over3}\qquad(0\leq k\leq9)\ .$$
@aayush: The $y_k$ satisfy the linear inhomogeneous difference equation $y_{k+2}-2y_{k+1}+y_k=$ a given periodic function of period $3$. The general theory about such equations says that our "Ansatz" with undetermined $a$, $b$, $c$, $d$ works in this case. – Christian Blatter Oct 7 '12 at 9:18