# Tag Info

## Hot answers tagged pattern-recognition

58

How about $\dfrac{i^n + (-i)^n}{2}$? (Of course, that is arguably just trigonometry in disguise). Or as a recurrence: $a_n = -a_{n-2}$ with $(a_0,a_1)=(1,0)$. Or $\begin{bmatrix}1 & 0\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}^n\begin{bmatrix}1\\0\end{bmatrix}$? (Which can be viewed as a better-disguised version of either of the two ...

32

\begin{align} 3^{27}=3(3^{26})=3(9^{13})& =3(10-1)^{13} \\ & \equiv 3((-1)^{13}+13(-1)^{12}(10)+\binom{13}{2}(-1)^{11}(10^2)) \pmod{1000} \\ & \equiv 3(-1+130-7800) \pmod{1000} \\ & \equiv 987 \pmod{1000} \\ \end{align} Edit: The same method (using binomial theorem) can easily be applied to $3^n$, even for large $n$. \begin{align} ...

29

Whether this is simplest will depend on exactly what you mean, but the following is a pretty simple description. It's certainly simpler than anything involving trig functions. $$a_n=\begin{cases} 0 & \text{if n is odd} \\ 1 & \text{if n is divisible by 4} \\ -1 & \text{otherwise} \end{cases}$$

28

"Guess the next term" is very often not a proper mathematics problem, and I would agree with your teacher that it is not a proper mathematics problem in this case (but a diverting puzzle among friends). However, the sequence has surprising properties about the limit proportions of 1s and 2s that would be "true" mathematics problems for any mathematician and ...

24

You can look up integer sequences at OEIS: http://oeis.org/A056805 So your sequence is "Numbers $n$ such that $6*10^n+1$ is prime". I assume you're looking for a formula, but if there was a closed-form expression for these numbers, we could find arbitrarily large prime numbers! The largest known prime has 12978189 digits and right now there is a 250,000 ...

17

If you have a sequence and have a linear recursion formula generating the sequence, then you can easily transform it into a closed form solution using one of many methods available. In your case let us start with the simplest recursion: $a_0=1,a_1=0,a_n=-a_{n-2},n\ge 2$. The easiest (if you have access to a computer) way to obtain a closed form solution is ...

15

The specific sequence is not essential. You are asking how to construct a function with period 4. Linear combinations of shifts of one $n$-periodic function can be used to write down any other $n$-periodic function, so they are all equally good in that sense. The trouble is to get at a sequence with period 4 without basing it on another one already known ...

8

First of all notice that $f(x):=x^n+(x-1)^n=0$ is equivalent to $(\frac{x-1}{x})^n+1=0$. The roots $f$ are therefore of the form $x=1/(1-y)$ where $y$ are the roots of $y^n+1=0$, so the Galois group is the same as for the polynomial $x^n+1$. I'm not sure what "Galois group of a polynomial" is when the polynomial is reducible, so let me determine the Galois ...

6

There will be a pattern to the last three digits of a power of 3, in general. However, that pattern may not necessarily show itself within the first 27 terms. However, here's something you can do instead to solve your problem: \begin{align} \text{ last 3 digits of } 3^{27} &= \text{ last 3 digits of } (3^3)^9\\ &= \text{ last 3 digits of } 27^9\\ ... 6 There are some interesting patterns, so I would say that it is proper mathematics. http://en.wikipedia.org/wiki/Look-and-say_sequence http://www.ams.org/journals/era/1997-03-11/S1079-6762-97-00026-7/home.html 6 According to Ian Stewart's Galois Theory, Conway proved in 1985 that if L(n) is the length of the n'th term in this sequence, L(n) satisfies a 72-term recurrence relation, which implies that L(n) is proportional to \alpha^n (for large n) where \alpha \cong 1.303577 is the smallest solution of a polynomial with integer coefficients of degre 71. He gives ... 5 The easiest way to find the answer here is to divide out by the first terms, producing:\begin{array} &&&&&&1\\ &&&&1&&1\\ &&&1&&2&&1\\ &&1&&3&&3&&1\\ &1&&4&&6&&4&&1\\ \end{array}$$...which should look ... 5 This seems to be an example of the Look-and-say sequence, which is in itself interesting; its variant, the Kolkoski sequence, leads to several difficult problems. Specifically, each term of the sequence after the initial one describes the previous term of the sequence by listing the number of times each symbol appears in the term. Thus, since the first term ... 5 We have \gcd(A, B)=B if and only if B divides A. So the desired number of pairs would be$$ 2\left(\sum_{i=1}^{N} \lfloor N/i \rfloor\right)-N $$We need to subtract N, otherwise the pairs (i, i) would be counted twice. Here \lfloor N/i \rfloor accounts for the number of multiples of i up to N. Added. The above formula counts (A, B) and ... 4 Updated: let A(n,k) be the number in the table with row n and column k, where 0\le n and 0\le k\le n.$$A(n,k)=A(n,n-k)A(n,0)=A(n,n)=0$$The table would look like this: (top left is row 0 and column 0)$$\begin{array}{c} 0\\ 0& 0\\ 0& 1& 0\\ 0& 1& 1& 0\\ 0& 3& 2& 3& 0\\ 0& 5& ...

4

We can look at a regular $n$-gon of "radius" $r$, i.e., the convex hull of $r$ times the $n$-th roots of unity. Connecting the vertices of the polygon to the origin gives you $n$ isosceles triangles of area $\frac{r^2}{2}\sin \frac{2\pi}{n}$. The total area of the $n$-gon is thus $$A = \frac{nr^2}{2} \sin \frac{2\pi}{n}.$$ To calculate the length $l$ of ...

4

Your questions is rather broad, as we can have all types of patterns that can include logic patterns, number patterns, and even word patterns. For number patterns, there are all sorts of things (list not exhaustive) to learn, investigate and explore, such as: $\bullet$ Arithmetic Sequences $\bullet$ Geometric Sequences See for example, the Number ...

4

When you have a finite set $S$, a map $f:\ S\to S$ and an initial value $a_0\in S$ then the sequence $(a_n)_{n\geq0}$ recursively defined by $$a_{n+1}:=f(a_n)\qquad(n\geq0)\tag{1}$$ will eventually become periodic: After at most $|S|$ steps the recursion will produce a number which you have seen before, and from then on the procedure will repeat ...

4

Try looking at the sequence this way: 1 4 7 10 13 5 3 1 -1 -3 The top part of the sequence increases by 3 each time and the bottom part of the sequence decreases by 2 each time. The next 3 terms are therefore 16, -5 and 19.

4

The first forward difference is defined by $\Delta f = f(n+1)-f(n)$. According to the intended pattern, this difference increases by $8$ every term. In the first instance, you get $$\Delta f (1)=13-3=10$$ $$\Delta f(2)=31-13=10+8.$$ It's easy to see that $\Delta f(3)=10+8+8$ and so on, so we get that $\Delta f(n)=2+8n$. Note also that $$\Delta f(1)+\Delta ... 4 HINT: If the n th \displaystyle T_n=n^r \displaystyle T_{m+1}-T_m=(m+1)^r-m^r=\sum_{k=0}^{r-1}\binom mkm^k Observe that the difference of order O(r-1) If we set T'_m= T_{m+1}-T_m, T'_{s+1}-T'_s  will be of order O(r-2) and so on Reference : Finite Difference I, II Finite Sum of Power? 4 If the teacher's objection to the sequence was that the process of counting and quoting numbers is 'not real math', this is completely wrong. Counting is fundamental to math - there is nothing about the definition of this sequence that makes it less basic, mathematical or rigorous than say, the Fibonacci sequence. If his objection was that the presentation ... 4 Your first expression should be (A+B)^n=\sum_{k=0}^{n}{n \choose k}A^{n-k}B^k starting from k=0. Your second expression may have problems with k=n when evaluating {n-1 \choose n}. So I will try to help with a slightly altered version of your question:$$\sum_{k=0}^{n-1}n!{n-1\choose k}A^{k+1}B^{n+k} = n!A B^n \sum_{k=0}^{n-1}{n-1\choose ...

3

It's a well established 'empirical fact' that there is no significant autocorrelation in market returns. That is, knowing that the market moved up in the last period provides no information about what it will do in the next period. So the answer to your question (as backed up by a lot of data) is that it is still 50/50 whether the market goes up or down ...

3

If (and this is a big if) you are considering a symmetric $\pm1$ random walk or a driftless Brownian motion starting from $x=40$ and you are wondering about the probability $u(x)$ that it hits $x_1=100$ before hitting $x_0=-100$, then indeed the answer is $$u(x)=\frac{x-x_0}{x_1-x_0}=70\%.$$ Briefly, an elementary method in the random walk case is to ...

3

The left side of the triangle, as you note, is $(n+1)!$; the right side of the triangle is $2^n n!$. Just like last time, the trick is to divide out by the common factor $n!$ and search from there. Doing so gives the triangle: \begin{array}{cccccccccccc} &&&&&2\\ &&&&3&&4\\ ...

3

There is no one right answer to this type of questions. The next element can be anything you want. That is one of the typical IQ questions of the type what is the next in a sequence, by inferring it from the previous values. The answer is, the next item can be anything you want, the book can come up with some rules to justify the next element, but so can ...

3

The sequence is of the kind $x_{n+1}=f(x_n)$, where $f$ is most easily described by referring to the decimal digit representation of $x_n$. While such relying on the fact that we have ten fingers and not on the "proper" numbers themselves may be frowned upon as did your teacher, I don't think it is always justified. For one, the well-known divisibility rules ...

3

If the word has weight 0, 8, 12, 16, or 24, then you can shuffle the columns of a generator matrix to get that word. It likely wouldn't be one of the four Golay generators: Greedy, Icosahedron, Nonresidue, or Polynomial. But it doesn't have to be.

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