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160

$\frac{680}{340}=2$, $\frac{1428}{680}=2.1$, and $\frac{3141.6}{1428}=2.2$, so we can expect that the person posing the question intended the next ratio to be $2.3$; this makes the next number $$3141.6\cdot2.3=7,225.68\;.$$

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Although 7225.68 is the obvious answer, as mentioned in other solutions, it should be noted that there are an uncountably infinite number of "correct" answers which can be attained from 4th-degree polynomials. Just solve a linear systems of equations of $$p(x) = ax^4 + bx^3 + cx^2 + dx + e$$ and p(0) = 340,\;\; ...

67

Let's try too : $|n \bmod 4-2|-1$

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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 ... 62 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 ... 55 Internet search gave me 340, 680, 1428, 3141.6, 7225.68 as: \begin{align} 680/340 = 2 \\ 1428/680 = 2.1 \\ 3141.6/1428 = 2.2 \\ 7225.68/3.141.6 = 2.3 \end{align} Edit: Or it could be a number on this german webpage, which compares different types of ovens. All other 4 numbers can be found there, so good look finding a pattern! 55 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.$$52 Eyebrow raising indeed, though the pattern does not continue as you suggest. I get$$ 0, 1, 1, 2, 3, 5, 7, 10, 16, 23, 37, 55, 84, 125, 198 $$Remember that the the number of primes has a well known growth rate (https://en.wikipedia.org/wiki/Prime_number_theorem). Since the Fibonacci numbers are relatively spread out, using n/\log n to approximate the ... 47 As John's answer notes, every palindromic number with an even number of digits is divisible by 11, because the alternating sum of its digits is zero (and thus a multiple of 11). For example, for 81233218, we have:$$8-1+2-3+3-2+1-8 = 0,$$and so 81233218 is divisible by 11. The reason why this divisibility rule works is most easily seen using a ... 38 Note that for m\ge 2$$2^{2m}-1=(2^m)^2-1=(2^m-1)(2^m+1)$$is not a prime number. 36 Whether this is simplest will depend on exactly what you mean, but the following is a pretty simple description. It's certainly simpler than anything involving trig functions.$$a_n=\begin{cases} 0 & \text{if n is odd} \\ 1 & \text{if n is divisible by 4} \\ -1 & \text{otherwise} \end{cases}33 \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} ... 30 What could possibly be easier than \Re(i^n), n=0,1,\ldots? 26 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 ... 25 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 ... 23 I'm saying that 435 is the answer to the question. Why? Consider the polynomial p(x)=-\frac{3x^5}{2}+\frac{55x^4}{2}-\frac{375x^3}{2}+\frac{1175x^2}{2}-786x+525$$Here is a table of values of p(x) on consecutive x. \begin{array}{|l|c|c|c|c|c|c|}\hline x & 1 & 2& 3& 4& 5&6\\ \hline ... 22 This is a fun observation... but I think you have a mistake. I wrote a quick python script to generate the fibonacci numbers and primes and make the counts and this is what I get: Between 5 and 8 there is 1 prime: 7 Between 8 and 13 there is 1 prime: 11 Between 13 and 21 there are 2 primes: 17, 19 Between 21 and 34 there are 3 primes: 23, 29, 31 ... 22$$\frac{1}{2} \left((-1)^{(n-1) n/2}+(-1)^{n (n+1)/2}\right)$$20 In this case: consider sequence$$ 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}, ... 20 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 ... 18 Your conjectured fact is true. Of any 3 consecutive integers, one is divisible by 3. If 2 of the integers are a pair of twin primes neither of which is 3, then neither of the "end" numbers is divisible by 3. So the "middle" number must be divisible by 3. So the twin primes are 3k-1 and 3k+1 for some integer k. It follows that their ... 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 ... 16 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 ... 15 You can make sense of this pattern (which as yoann points out does not go on very far) as saying that the density of primes is a constant \phi^{-3}\approx0.236, since the gap length is F_{n+1}-F_n=F_{n-1} and the conjectured prime count in this gap is F_{n-4}\approx F_{n-1}\phi^{-3}. Unfortunately, this goes against the prime number theorem or various ... 14 It's known as the look-and-say sequence. 12 Let p(n) be the nth 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 ...

11

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

11

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.

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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.

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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.

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