Find the next number in the sequence

Question

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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Answer 1

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.

Answer 2

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 $a_6=256$.

Answer 3

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 $169,256,169,256,\ldots$

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).