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Theorem. If $2^n-1$ is prime then $n$ is prime. Proof. Suppose that $2^n-1$ is prime, and write $n=st$ where $s,t$ are positive integers. Since $$x^s-1=(x-1)(x^{s-1}+x^{s-2}+\cdots+1)\ ,$$ we can substitute $x=2^t$ to see that $2^t-1$ is a factor of $2^n-1$. Since $2^n-1$ is prime there are only two possibilities, $$2^t-1=1\quad\hbox{or}\quad ...


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 ...


All numbers of this kind are divisible by $11$. Consider prime $\mathit{abcdefg}$ where each letter is a digit. The number $\mathit{gfedcbaabcdefg}$ is divisible by $11$ because the alternating sum of digits is always zero: $$g - f + e - d + c - b + a - a + b - c + d - e + f - g = 0.$$


As John's answer notes, every palindromic number with an even number of digits is divisible by $11$, because the alternating sum of its digits is zero (and thus a multiple of $11$). For example, for $81233218$, we have: $$8-1+2-3+3-2+1-8 = 0,$$ and so $81233218$ is divisible by $11$. The reason why this divisibility rule works is most easily seen using a ...


Note that for $m\ge 2$ $$2^{2m}-1=(2^m)^2-1=(2^m-1)(2^m+1)$$ is not a prime number.


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


\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} ...


What could possibly be easier than $\Re(i^n)$, $n=0,1,\ldots$?


"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 ...


As has been pointed out in the comments, this is not a special property of the primes. Rather it's true that whenever you reverse a number and append the result to itself, the result is always divisible by 11. To see why this is true, consider the number $A = 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10^0 a_0$, where $a_0, \ldots, a_{n-1}$ are all ...


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 ...


$$\frac{1}{2} \left((-1)^{(n-1) n/2}+(-1)^{n (n+1)/2}\right)$$


I'm saying that $\color{red}{\text{435}}$ is the answer to the question. Why? Consider the polynomial $$\large p(x)=-\frac{3x^5}{2}+\frac{55x^4}{2}-\frac{375x^3}{2}+\frac{1175x^2}{2}-786x+525$$ Here is a table of values of $p(x)$ on consecutive $x$. $\begin{array}{|l|c|c|c|c|c|c|}\hline x & 1 & 2& 3& 4& 5&6\\ \hline ...


The ratio between Fibonacci numbers soon settles down to a number close to $1.618$. This number is called the Golden Ratio. You get an extra digit every time the Fibonacci numbers have increased by a factor 10. $1.618^4=6.854$ and $1.618^5=11.09$ Once the ratio settles down, you get at least one extra digit every five numbers. Sometimes the extra digit ...


In this case: consider sequence $$ a_n = 15 \cdot p_n, $$ where $p_n$ is $n$-th prime number. $a_n$: $\color{gray}{30, 45, 75, 105,} 165, 195, 255, 285, 345, \color{red}{435}, ...$


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 ...


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 ...


It's known as the look-and-say sequence.


Let $p(n)$ be the $n$th term in the sequence. Clearly, this sequence follows the formula: $$\begin{align} p(x) &= \frac{600631 x^{19}}{121645100408832000}-\frac{791723 x^{18}}{800296713216000}+\frac{196988587 x^{17}}{2134124568576000}\\ &-\frac{41785811 x^{16}}{7846046208000}+\frac{8219611 x^{15}}{38626689024}-\frac{49026370303 ...


Another approach but it's just mere an intuitive one.


Set $g_n=p_{n+1}-p_n$, where $p_n$ is the series of prime numbers, with $p_1=2$. Then $$ p_1+\sum_{i=1}^n g_i=\sum_{i=1}^n g_i+2=p_{n+1}. $$ So the conjecture is obviously true, but not useful.


This is just the sequence of positive integers expressed in base two: $a_n$ is the base-two representation of the positive integer $n$. There's not really much more to be said.


If you can multiply a 3-digit number by $3$ without a calculator, then you can answer the question without a calculator. Just start with $1$, multiply by $3$ $27$ times, keeping only the last three digits. $1,3,9,27,81,243,729,187$, and so on.


Nice observation! Here is an explanation: The $n$-th Fibonacci number $F_n$ is asymptotically equal to $\varphi^n/\sqrt5$, where $\varphi=(1+\sqrt5)/2$. This implies that the number of digits in $F_n$, which is essentially $\log_{10} F_n$, is asymptotically equal to $n\,\log_{10}\varphi\approx0.2090\,n$. As a consequence, there are either $4$ or $5$ ...


Hint $\ $ If in radix $b,\,$ we have $\,n = f(b) = f_0\! + f_1 b +\,\cdots+ f_n b^n,\,$ then reversing the digits of $\,n\,$ yields the integer $\,\bar n = \bar f(b)\,$ for $\,\bar f(b) = b^n f(b^{-1})$ the reversed polynomial. Appending yields $$ {\rm mod}\,\ b\!+\!1\!:\,\ b\equiv -1\,\Rightarrow\, b^{n+1} f(b) + b^n f(b^{-1})\,\equiv\, (-1)^n ...


Note that $1+3+4+9=17$ and $1+7+1+0=9$.


The expected answer is probably something like this: Each column is the difference of the previous column A B C D -------------- 87 2 89 4 6 2 95 6 12 2 107 8 20 2 ?127? 10 30 157 However, we can fit any number in the unknown spot with the proper function. Each ...


Yes, the formula $2^n-1$ may only yield a prime if $n$ is prime, therefore the only even value of $n$ which yields a prime is $2$, which yields the Mersenne prime $2^2-1=3$. However for some prime values of $n$ the result is not prime, for example $2^{11}-1=23\cdot 89$. In general $a^x-1$ is composite if $x$ is composite. This can be seen very easily by ...


Every number obtained by this construction is divisible by $11$. Here is a simple test for divisibility by $11$: if the number is $a_r\cdots a_0$, it is divisible by $11$ iff $\sum_{i=0}^r (-1)^ia_i = 0$. This follows from the fact that $10^{2i+1}=-1\bmod{11}$ and $10^{2i} = 1\bmod{11}$. Thus, $a_r\cdots a_0 = \sum 10^ia_i = \sum (-1)^i a_i\bmod{11}$. Now, ...

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