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0

If you graph this sequence of numbers, you will find an interesting curve: (0,1,2), (1,2,3), (2,3,4), -> first cycle (1,2,3), (2,3,4), (3,4,5), -> second cycle (2,3,4), ... Use Excel or squared paper.


1

Look up Kolmogorov complexity. One can fix a programming language and then define the answer to such pattern-continuation questions as the shortest program $P$ such that $P(0),P(1),\cdots$ agrees with the given terms. Now in general it is impossible to systematically find such a program, which means that we cannot hope to solve all except the simplest ...


4

First you write the nonnegative integers in base $3$ $0,1,2,10,11,12,20,21,22,100,101,102,110,111,112,120,\dots$ Then you add up the (base 3) digits in each number (presumably in base $10$) $0,1,2,1,2,3,2,3,4,1,2,3,2,3,4,3,\dots$ Since we start from zero, we express $2011$ in base $3$, getting $2202111_3$, then add the digits to get $9$.


3

The key part to this, in my opinion, is that "if the pattern continues, it will continue to be rainy." However, that doesn't mean that the inside beginning part of the pattern is definitely always increasing. Yes, it looks that way from the beginning, but what if it's actually some polynomial that has some twists and turns in the beginning, but then ...


8

Let $z_i=a_i + ib_i$. Then your recursion is $$ z_{n+1}=a_{n+1}+ib_{n+1}=\frac{1}{2}(a_n+ib_n)+\frac{1}{2}(1+i)\sqrt{\frac{a_n^2+b_n^2}{2}}=\frac{1}{2}z_n+\frac{1+i}{2\sqrt{2}}|z_n|=\frac{1}{2}z_n+\frac{1}{2}e^{i\pi/4}|z_n|. $$ Now, let $z_n=e^{i(\theta_n+\pi/4)}y_n$, where the $y_n$ and $\theta_n$ are real. Then $$ y_{n+1}e^{i\theta_{n+1}}=\frac{1}{2}y_{n}...


1

Geometrically: $(a_{n+1}, b_{n+1})= \frac 12 (a_n,b_n) + \frac 12 \sqrt{a_n^2+b_n^2} (\cos \pi/4, \sin \pi/4)$ So we draw a line between $(a_n,b_n)$ and connect it to a point equidistant from the origin on the line $x = y$, and find the mid-point. $r_n = \sqrt{a_n^2 + b_n^2}\\ \phi_n = \tan^{-1}(\frac{b_n}{a_n})$ $\phi_{n+1} = \frac{\frac{\pi}{4} + \...


0

Assuming you're not dealing with a large data set (read: millions of objects) and thus aren't concerned with performance, the following algorithm should work: Each object will be numbered, e.g. object $k$, and $k$ will be it's id number. It's data will be an ordered triple with each constrained to $[0,100]$ (are these integers? I think you intend for that.) ...


1

In general, the answer is no. For example, if this is your histogram, there are multiple shifts that give the same result.


0

$x=20$ Since in first riddle the numbers are 10 and 3 they are formed by $5×2$ & $5-2$ . In the second numbers are 6 and 1 they are formed by $3×2$ &$3-2$ . From here $x=20$ gives the numbers 20 and 8 and they are formed by $10×2$ & $10-2$.


4

I get $6$ for the fifth term: $T(n)=\frac{378-272n+75n^2-7n^3}{3}$ gives the general term. $58=T(1)$ $26=T(2)$ $16=T(3)$ $14=T(4)$ $6=T(5)$


10

The missing item is $42$ of course, and the sequence is: $$a_n=-\frac{41n^4}{12}+\frac{211n^3}{6}-\frac{1435n^2}{12}+\frac{1025n}{6}-83$$


3

General formula to find next term is~ $$x^3-1$$ $$1^3-1=0$$ $$2^3-1=7$$ $$3^4-1=26$$ $$4^3-1=63$$ $$5^3-1=124$$ Upcoming number in sequence is 215


-1

It must be zero, and the sequence is, of course, $$5,10,0,50,122,0,5,10,0,50,122,0,5,10,0,\dots$$


6

$2^2+1,3^2+1 , .., 7^2+1, 11^2+1$, seems like series of primes to me so I would guess 26



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