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I am giving this question to see how much time people take to solve such problems and to know how the idea struck their minds . I solved but it took a lot of time(12 minutes) for the solution to strike my mind .I want to check how much time generally people take for these questions .

Find the next number in the sequence

$83 , 121 , 16 , 49 , 169 ,256 , ......$

Answerers please also share how much time it took to solve this problem and also the possible ways you tried to finally arrive at the answer . Maths is fun , sequence is more fun

..

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$169 = (2+5+6)^2$ –  Daniel Fischer Aug 20 '13 at 19:48
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It is a sequence not a series. –  Mhenni Benghorbal Aug 20 '13 at 19:49
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The sequence continues $$83,121,16,49,169,256,169,256,\ldots$$ –  DonAntonio Aug 20 '13 at 19:51
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@HarishKayarohanam Don't know how long it took. Not very long, the squares were kind of a giveaway - although I was tempted to answer $42$. –  Daniel Fischer Aug 20 '13 at 19:57
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I have the nagging suspicion that the posers of these questions don't know that. –  Daniel Fischer Aug 20 '13 at 20:11

3 Answers 3

It is trivial not to observe that the sequence is the polynomial $$(263 x^5)/120-(1121 x^4)/24+(8915 x^3)/24-(31915 x^2)/24+(41197 x)/20-974$$ evaluated at $1,2,3,4,5,6$. Substituting $x=7$, we obtain $384$ as the next element.

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Clearly, this is the only correct answer! ;) –  mrf Aug 20 '13 at 19:55
    
How did you arrive at this polynomial ? –  Harish Kayarohanam Aug 20 '13 at 19:56
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@HarishKayarohanam, in case you're confused, this answer is a joke; the purpose is to show that there are infinitely many sequences you can make which fit your initial terms. For example, I could make my sequence exactly like yours for the first 6 terms, and then suddenly switch to $-34102495073$ as the next number. –  George V. Williams Aug 20 '13 at 20:06
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@HarishKayarohanam It is great to see that this answer served a purpose then :) –  Lord Soth Aug 20 '13 at 20:16
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@MartinArgerami If one continues a finite sequence of integers via polynomial interpolation (of minimal degree), the sequence will continue to consist of integers. You can see this by noting that the same continuation is obtained by working with repeated differences. –  Hagen von Eitzen Aug 20 '13 at 20:36

There is no unique solution to such a problem. For any arbitrary choice of $t$ the sequence $$a_n := 720t-974+(\tfrac{41197}{20}-1764t)n+(1624t-\tfrac{31915}{24})n^2+(\tfrac{8915}{24}-735t)n^3+\cdots$$ $$\cdots + (175t-\tfrac{1121}{24})n^4+(\tfrac{263}{120}-21t)n^5+tn^6$$ satisfies $a_1=83,$ $a_2=121,$ $a_3=16,$ $a_4=49,$ $a_5=169,$ and $a_6=256$.

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As it seems that the "correct" answer was quickly found by many commentators, often within very few minutes, but has not been posted yet, here goes:

For $n\in\mathbb N$ let $Q(n)$ denote the digit sum of $n$, that is $$ Q(n)=\begin{cases}n&\text{if }0\le n\le 9\\r+Q(q)&\text{if }n=10k+r\text{ with }k\ge1,0\le r\le 9\end{cases}$$. Then the sequence obeys the recursion $a_{n+1}=Q(a_n)^2$. Hence the seuence continues $169,256,160, 256,\ldots$

The exact behaviour depends on the starting value $a_1$, but the sequence is bounded from above: For $a_n<10^{k+1}$ we have $a_{n+1}\le 81k^2<10^k$ as soon as $k\ge3$, hence the sequence is decreasing until it falls and stays below $1000$. Therefore the sequence must be eventually periodic. The given sequence is presented just immediately before the first repetition. The only periods of length one are $1$ and $81$. The only period of length two is seen in the problem statement: $169,256$. By inspection of all starting values $\le 999$, there are no longer periods. Also, there is no starting value below $1000$ with a longer pre-period than $83$ (though there are many with equal length).

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