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0

Here is a closed form $$ \mathcal{I}=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)x^k= 2 \arcsin \left( x/2 \right) \left(\pi - \arcsin \left( x/2\right) \right) .$$ Now just plug in $x=1$ and the result follows.


1

This is more a comment that an answer As Jack D'Aurizio commented, I suppose that the summation starts at $k=1$ and not $k=0$ as written in the post. Using a CAS, the result obtained is $$\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\left[\Gamma\left(\frac{k}{2}\right)\right]^2=\frac{5 \pi ^2}{18}$$ which matches the value you obtained using Wolfram Alpha. ...


2

(This is more of a comment that an answer but, ...) Consider the even and odd $k$ in separate sums: Note: You probably do not want $\Gamma(0)$ in the sum, so I'll start at $k=1$. $$\begin{array}\\ S &=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\left[\Gamma\left(\frac{k}{2}\right)\right]^2\\ ...


10

You can use the identity given by the Euler Beta function $$\int_{0}^{1}x^{a-1} (1-x)^{b-1} \,dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ to state: $$S=\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k!}\Gamma(k/2)^2=\sum_{k=1}^{+\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}\left(x(1-x)\right)^{k/2-1}\,dx $$ and by switching the series and the integral: $$ S = ...


-2

To form factor of zeros at the end, we need $5$. Thus, if $Z(n)$ is the number of zeros of $n!$ then $$ Z(100) = \frac{100}{5} + \frac{100}{5^2} = 20 + 4 = 24 \ \text{zeros} $$


0

The fact that there is a special notation for the factorial suggests that there is no simpler formula for it other than the definition. Note that there is no special notation for $\sum_{i=1}^n i$, since it can be written as $n(n+1)/2$. (I'm not sure $\binom{n+1}{2}$ counts as special notation in this context.) If you need a formula, you could use $n! = ...


0

You could compute the "recursive" version of the function with a calculator that can handle recursives. Smart Math Calculator is an example of such calculator. For your factorial example you could type this in Smart Math Calculator: factorial(n)=n*factorial(n-1); n>1 factorial(n)=1; n<=1 Then to compute 10! you can use the defined function above in ...


1

It is known as factorial and denoted as $n!$. The case $10!$ can be reduced: \begin{eqnarray} 1 \cdot 2 \cdot 3 \cdots \cdot 8 \cdot 9 \cdot 10 &=& 10 \Big(5-4\Big) \Big(5-3\Big) \cdots \Big(5+4\Big)\\ &=& 50 \Big(25-16\Big) \Big(25-9\Big) \Big(25-4\Big) \Big(25-1\Big)\\ &=& 50 \Big(15-6\Big) \Big(15+6\Big) \Big(20-4\Big) ...


1

If you take the natural log we get $$ \ln(a_n)=(n+1)\ln(n+1)+\ln(n!)-( (n+1)!-n! )\ln(2)$$ $$= (n+1)\ln(n+1)+\sum_{1}^{n}\ln(k)-n\,n!\ln(2) $$ $$ = (n+1)\ln(n+1)+n\ln(n)-n+1-n\,n!\ln(2) \sim_{ \infty} 2n\ln(n)-n-nn!\,\ln(2) $$ $$ = 2n\ln(n)-(n+1)!\,\ln(2) \to -\infty $$ $$ \implies a_n \rightarrow_{n\to\infty} 0 $$ Note: $$ \sum_{1}^{n}\ln(k) ...


0

The basic sums of distinct factorial numbers are: $0! = 1!$ $0! + 1! = 2!$ There are no equal sums of distinct factorial numbers which don't boil down to one of those two. Proof: consider the numbers in the factorial number system. The factorials are $0_{10}!=10_!$, $1_{10}!=10_!$, $2_{10}!=100_!$, $3_{10}!=1000_!$, $\ldots$. Because $n > 2 \implies ...


1

The correct formula for the number of trailing zero digits in $n!$ is $$ \frac{n-\sigma_5(n)}{4} $$ where $\sigma_5(n)$ is the sum of the base-$5$ digits of $n$. So the formula given to you is only correct if the sum of the base-$5$ digits of $n$ is $4$. Since $$ \begin{array}{l} 100_\text{ten}=400_\text{five}\\ 200_\text{ten}=1300_\text{five}\\ ...


0

As explained in this answer, for a prime $p$, the number of factors of $p$ in $n!$ is $$ \frac{n-\sigma_p(n)}{p-1} $$ So as an estimate, we can use $n/4$ for the number of factors of $5$ that divide $n!$. For $30$ zeros, we would try $n=120$ ($440_\text{five}$). $$ \frac{120-8}{5-1}=28 $$ Since no factors of $5$ are added until $n=125$ ($1000_\text{five}$), ...


9

The number of zeros at the and of $n!$ is just the number of factors of $5$ in $n!$ (since the number of factors of $2$ in $n!$ is always larger than that). So, we need all $n$ such that the number $1$ up to $n$ have a total of $30$ factor of $5$. A first guess would be $30\cdot 5=150$, but then we forget that the multiples of $5^2=25$ have (at least) two ...


5

Your conjecture is true. We have the following celebrated result in math olympiads: Theorem. (Lifting The Exponent) If $p$ is an odd prime, $n\in\mathbb N$, $a,b\in\mathbb Z$ such that $p\mid a-b$ but $p\nmid a,b$, then $\nu_p(a^n-b^n)=\nu_p(n)+\nu_p(a-b)$, where $\nu_p(x)$ denotes the number of factors $p$ in the prime factorisation of the integer ...


0

So to answer your question in short: Part I: Yes Part II: Not Applicable Why? Here it is below, have fun... Our objective is to count the number of factors that are necessarily divisible by powers 5. and check there are at least as many powers of 2 dividing this expression. We begin by factoring the expression $3^{n!} - 1$ through the use of the ...


1

Just that you can check much higher; you know how to find $b_n.$ Then calculate $$ 3^{n!} - 1 \pmod {2^{b_n + 2}}, $$ $$ 3^{n!} - 1 \pmod {5^{b_n + 2}}. $$ I don't see much reason that (1) should hold forever, but up to 100 would be impressive. There are topics sort of like this where there is agreement for an initial set but later disagreement. The ...


2

We have four groups of three arranged around the circle. First choose the four seats of the managing directors. This can be done in $3$ ways. Next, distribute the four companies around the four quarters of the circle (around the four chosen seats). That can be done in $4!=24$ ways. Next, for each company, choose who sits to the left an who sits to the right ...


2

There are three choices for the first digit. Once this choice has been made, there are four digits left. These can be arranged in $4!$ ways. Then $3\times 4!$ gives the correct answer. Added Later: From what I can tell, you made two mistakes. First of all, after the choice of the first digit, you said there were three numbers left when in fact there are ...


0

EDIT: the full algorithm: The full algorithm can be summarized as follow: First check if $N<p$. If it is, then the answer is $N$. Otherwise proceed to next step. Calculate $M=\lfloor\frac{N}{p}\rfloor$. As explained in the explanation part, now we need to find the number of $k$ such that $kp$ is a valid $n$, and that $0\leq k\leq M$. This is equivalent ...


0

It is maximized when $x$ is as large as possible, so $x=a/b$. The ratio $g(k,j+1)/g(k,j)=(b/a)(k+n-j)$. So long as the ratio is more than 1, $g(k,j)$ is increasing with $j$, but when the ratio dips below 1, $g(k,j)$ starts decreasing. so it is maximized when $j=k+n-a/b$ - or else at an endpoint $j=2$ or $j=k+n$


6

Note that $$ \begin{align} \frac{x}{1-x} &=\sum_{n=1}^\infty x^n\\ &=\sum_{n=1}^\infty\sum_{k=1}^n\frac{a_k}{(n-k+1)!}x^n\\ &=\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{a_k}{(n-k+1)!}x^n\\ &=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{a_k}{(n+1)!}x^{n+k}\\ &=\frac{e^x-1}{x}\sum_{k=1}^\infty a_kx^k\tag{1} \end{align} $$ Therefore, $$ ...


4

Put $\displaystyle f(x)=\sum_{k\geq 1} a_k x^{k-1}$, $\displaystyle g(x)=\sum_{l\geq 0} \frac{x^l}{(l+1)!}=\frac{\exp(x)-1}{x}$, and $\displaystyle h(x)=f(x)g(x)=\sum_{n\geq 0}c_n x^n$. The coefficient $c_n$ is equal to $1$ if $n=0$, and for $n\geq 1$: $$c_n=\sum_{k-1+l=n, k\geq 1, l\geq 0}\frac{a_k}{(l+1)!}=\sum_{k=1}^{n}\frac{a_k}{(n+1-k)!}=1$$ Hence: ...


0

Show that $\frac{a_{n+1}}{a_n}$ converge to some value $>1$, keep in mind the definition of $e = \lim_{n \to \infty} (1+\frac{1}{n})^n$. Hence the series diverges.


4

Note that $$ (n!)^2 = \prod_{i=1}^n i(n+1-i) \ge \prod_{i=1}^n n = n^n $$ Hence $n! \ge n^{n/2}$. This gives $$ (2n)! \ge (2n)^n = 2^n n^n $$ For $n \ge 2$ we have $2^n \ge n^2$, hence $(2n)!\ge n^{n+2}$.


1

Several problems: Using $P(26,10)$ includes arrangements where a and z are not 10 apart. There are 24 letters excluding a and z. Once the letters between a and z are chosen, there are $26-(10+2)=14 \neq 16$ remaining letters. The number of letters to the left of a. It is possible to have anywhere between $0$ and $26-(10+2)$ letters to the left of a. a ...


1

There are $24!$ permutations of the letters b through y. For each such permutation, if a is to the left of z, it can appear in any of 15 positions (from before the first letter to before the fifteenth letter), and z appears 10 positions later. Similarly, if z is first, it can appear in any of 15 positions and a appears 10 positions later. So $30\cdot 24!$.


1

This is a very amazing problem. As I said earlier, I did not find any closed form for the integral and then I only performed numerical integrations. What is surprizing if that $y(a)=\displaystyle\int_{0}^{a} \Gamma(1+x)~ dx$ looks very much to $x(a)=\Gamma(1+a)$ (plot the two curves as a function of $a$). Taking into account Lucian's comment, I performed a ...


1

The factorial function is only defined on the positive integers, so those don't make sense. However, there is a generalization of the factorial called the Gamma function which you might want to check out.


4

You just need to do it once and from that, get a recursion relation: $$\begin{align*} I_n &= \int_{x=0}^\infty x^n e^{-x} \, dx \\ &= \lim_{t \to \infty} \Bigl[ -x^n e^{-x} \Bigr]_{x=0}^t + n \int_{x=0}^\infty x^{n-1} e^{-x} \, dx \\ &= \lim_{t \to \infty} - t^n e^{-t} + n I_{n-1} \\ &= n I_{n-1} . \end{align*}$$ Then it is trivial to ...


3

The asymptotic expansion for $\Gamma(z)$ is given by $$\Gamma(z) \sim \sqrt{\frac{2\pi}{z}}\left(\frac{z}{e}\right)^z\left(1 + \frac{1}{12z} + \cdots\right)$$ Therefore we have $$\begin{align}\Gamma\left(n+\frac{1}{2}\right) &\sim \sqrt{\frac{2\pi}{\left(n+\frac{1}{2}\right)}}\left(\frac{n+\frac{1}{2}}{e}\right)^{n+\frac{1}{2}} \\&= ...


0

It's easier if you redefine $t = \log \log n$, then RHS becomes $e^{e^{kt}}$. Now apply Stirling's approximation to the LHS and take logarithm. You'll see that lhs is $t \log t -t +O(\log t)$, which is of course $O(e^{kt})$.


5

The largest factorial primes are discovered only recently. From an announcement: On 30 August 2013, PrimeGrid’s PRPNet found the 2nd largest known Factorial prime: 147855!-1 The prime is 700,177 digits long. The discovery was made by Pietari Snow (Lumiukko) of Finland using an Intel(R) Core(TM) i7 CPU 940 @ 2.93GHz with 6 GB RAM running Linux. ...


1

In addition to Michael Hardy's answer, which precludes all of n+1 to n+50, you can also prove that n is in fact composite as well using Wilson's Theorem which states that a number n is prime if and only if (n - 1)! ≡ (n-1) mod n. Here, n is (51! + 1), so (n-1)! = 51!!. Therefore, (51! + 1) is prime iff 51!! ≡ 51! mod (51! + 1). However, we now that (51!+1) ...


8

Just looking at the heuristics of the problem: If you pick a random integer $x$, it will be a prime number with a probability about $1 / \ln x$. Now the number $n! + 1$ is not a random integer. We know that $n! + 1$ is not divisible by any prime number $p ≤ n$. A random large integer is not divisible by any prime $p ≤ n$ with probability ...


35

$n! + 1$ is prime for $n = 0, 1, 2, 3, 11, 27, 37, 41, 73, 77, 116, 154, 320, 340, 399, 427, 872, 1477, 6380, 26951, 110059, 150209, \dots$, no other factorial primes are known as of May 2014. See here for more info on factorial primes.


21

First you need to review the definition of the factorial: $$m! = \prod_{i = 1}^m i = 1 \times 2 \times 3 \times \cdots \times m.$$ This means that $m!$ is divisible by 2, by 3, by 4, by every number up to $m$. Therefore $51!$ is divisible by 2, by 3, by 4 and by every number up to 51 (and a few others greater than 51, but you don't need to worry about those ...


4

The number $51!$ has as non-trivial factors every natural number preceding $51$. Thus any $51! + 2$, $51! + 3$, etc. will be divisible by $2,3,4,\cdots$ respectively. $$n=51! + 1$$ $$n+a = 51! + (1+a) = (\frac{51!}{1+a}+1)\cdot (1+a)$$ If $1\leq a \leq 50$ then $\frac{51!}{1+a}$ is an integer and so $a+1$ divides $n+a$. (and they are obviously not equal, ...


2

The offset by $1$ on $51!$ may make it a little confusing. For any integer $2 \leq n \leq 51$, you will have that $51! + n$ is divisible by $n$ because $51!$ is divisible by $n$. This is an equivalent statement to your problem. Now you just need to show $51!$ is divisible by $n$ and that $n$ does not equal $51! + n$.


1

Hints: 1) use the inclusion-exclusion principle, 2) If there are more girls than boys on the committee of five, you have three choices: $(0,5), (1,4), (2,3)$. Then you choose $j$ from the set of boys and $5-j$ from the set of girls.


0

The simplest way would be; let $$ \color{fuchsia}{P_n=\frac{x^n}{n!}=} \color{maroon}{\frac x1.\frac x2.\frac x3\cdots\frac x{x-1}.\frac xx.\frac x{x+1}\cdots\frac x{n-1}.\frac xn}$$ Then $$\color{maroon}{0}\color{red}{<}\color{fuchsia}{P_n}\color{red}{<}\color{maroon}{\frac11.\frac12\cdots\frac{x-1}{x-1}.\frac xx.}\color{green}{\frac x{x+1}.\frac ...


0

Here's one set of "completely" unique permutations $f_1, f_2, \ldots f_{12}$: $$ f_n(x) = n\cdot x \bmod 13 $$ So we have: n=1: 1 2 3 4 5 6 7 8 9 10 11 12 n=2: 2 4 6 8 10 12 1 3 5 7 9 11 n=3: 3 6 9 12 2 5 8 11 1 4 7 10 n=4: 4 8 12 3 7 11 2 6 10 1 5 9 n=5: 5 10 2 7 12 4 9 1 6 11 3 8 n=6: 6 12 5 11 4 10 ...



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