# Tag Info

57

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. $$a_n=\begin{cases} 0 & \text{if n is odd} \\ 1 & \text{if n is divisible by 4} \\ -1 & \text{otherwise} \end{cases}$$

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 ...

15

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\\ ... 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 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}$$...which should look ... 5 We have \gcd(A, B)=B if and only if B divides A. So the desired number of pairs would be$$ 2\left(\sum_{i=1}^{N} \lfloor N/i \rfloor\right)-N $$We need to subtract N, otherwise the pairs (i, i) would be counted twice. Here \lfloor N/i \rfloor accounts for the number of multiples of i up to N. Added. The above formula counts (A, B) and ... 4 We can look at a regular n-gon of "radius" r, i.e., the convex hull of r times the n-th roots of unity. Connecting the vertices of the polygon to the origin gives you n isosceles triangles of area \frac{r^2}{2}\sin \frac{2\pi}{n}. The total area of the n-gon is thus$$A = \frac{nr^2}{2} \sin \frac{2\pi}{n}.$$To calculate the length l of ... 4 Your first expression should be (A+B)^n=\sum_{k=0}^{n}{n \choose k}A^{n-k}B^k starting from k=0. Your second expression may have problems with k=n when evaluating {n-1 \choose n}. So I will try to help with a slightly altered version of your question:$$\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 ...

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 $$\Delta f(1)+\Delta ... 4 Try looking at the sequence this way: 1 4 7 10 13 5 3 1 -1 -3 The top part of the sequence increases by 3 each time and the bottom part of the sequence decreases by 2 each time. The next 3 terms are therefore 16, -5 and 19. 4 Updated: let A(n,k) be the number in the table with row n and column k, where 0\le n and 0\le k\le n.$$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

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 ...

3

In a sense there must be patterns to be found in rainbows, because they can be described mathematically and math is all about patterns. However, the mathematics of rainbows is quite complex, and if you are looking for patterns that are simple and elegant, then probably the best example of such to be found in a rainbow is the pattern we see. That is, the ...

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.

3

I think that |n mod 4−2|−1 is a great solution. Here is some that not require absolute value: (1 - (n mod 4))((n+1) mod 2) The logic behind it is: n mod 2 gives 0, 1, 0, 1.. because we want all odd numbers to be 0 we use n+1 mod 2. Than we use (1 - (n mod 4)) to make 0 input to output 1 and 2 input to output -1. n mod 4 is just to limit the numbers between ...

3

You're interested in the special case of the divisor function $\sigma_0(n)$, which returns the number of divisors of $n$. The numbers with more divisors than any smaller number are called highly composite and they are all representable as a product of primorials. Equivalently, the exponents of their prime factors must be constant or decreasing. I believe the ...

3

As mentioned, a slightly non-standard use of double-factorial gives $$a_n=\frac{1}{n!!}$$ for even $n$. However, I usually see double factorial used with odd $n$. For even $n$, $$a_n=\frac{1}{2^{n/2}(n/2)!}$$ works as well. For odd $n$, without double factorial, $$a_n=\frac{2^{(n-1)/2}(\frac{n-1}{2})!}{n!}$$ so it is easy to see why $1/n!!$ is ...

3

For any integer $n$,there are $2$ solutions to $x^2 = x \pmod {2^n}$ and $x^2 = x \pmod {5^n}$, which are $0$ and $1$. Hence by the chinese remainder theorem, there are $4$ solutions to $x^2 = x \pmod {10^n}$. Two of those are the obvious $0$ and $1$, there is one solution which is $0$ mod $5^n$ and $1$ mod $2^n$, and the last one is $0$ mod $2^n$ and $1$ ...

3

Exponentiation by squaring reduces the number of multiplications from 26 to 7: First square repeatedly to create powers by powers of 2: $$3^1=3 \qquad 3^2=9 \qquad 3^4=81 \qquad 3^8 \equiv 561 \pmod{1000} \qquad 3^{16} \equiv 721 \pmod{1000}$$ Then, since $27=1+2+8+16$, $$3^3=3^1 3^2 = 27 \qquad 3^{11} = 3^3 3^8 \equiv 147 \pmod{1000} \qquad 3^{27} = ... 3 The left side of the triangle, as you note, is (n+1)!; the right side of the triangle is 2^n n!. Just like last time, the trick is to divide out by the common factor n! and search from there. Doing so gives the triangle:$$\begin{array}{cccccccccccc} &&&&&2\\ &&&&3&&4\\ ...

3

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 ...

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