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4

After some research and thinking I have realised the following : We can observe that : $$(2K-1)*(2K+1) < (2K)^2$$ Now, if we give values to K ( from 1 to n ) and the multiply all of them together we get : $$1 * 3^2 * 5^2 * ... * (2n-1)^2 * (2n+1) < 2^2 * 4^2 * ... * (2n)^2$$ Now by dividing with $2n+1$ and the right side plus taking the square ...

2

May this help a bit? For positive integers, those relations hold: $$(2n)!! = 2^n n!$$ $$(2n-1)!! = \frac{(2n)!}{2^n n!}$$ But I guess that your case need the general $n$ not only integers.. EDIT DUE TO THE COMMENTS BELOW Using the relationships above and the Stirling approximations below, we find: $$\frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{(2^n n!)(2^n ... 2 \lim\limits_{n\to \infty}\dfrac{(2n-1)!!}{(2n)!!}=0\quad and \quad\lim\limits_{n\to \infty}\dfrac{(2n+1)!!}{(2n)!!}=\infty.~ Where it really gets interesting, however, is when we attempt to evaluate their product. Their polar-opposite tendencies will cancel each other out, yielding ~\lim\limits_{n\to ... 5 You can notice that$$\frac{(2n-1)!!}{(2n)!!} = \prod_{k=1}^n \frac{2k-1}{2k} = \prod_{k=1}^n \left(1-\frac{1}{2k}\right).$$So you want in fact calculate the infinite product$$\lim_{n\to\infty} \prod_{k=1}^n \left(1-\frac{1}{2k}\right) = \prod_{k=1}^\infty \left(1-\frac{1}{2k}\right)$$Now you can use this fact: How to prove \prod_{i=1}^{\infty} (1-a_n) ... 2 n! is the number of permutations of S=\{1, \dots, n \}, while n^n is the number of functions from S to itself. Permutations are functions, so clearly n! \leq n^n. 2 It sounds like want to show that S = \{n \in \mathbb{N} : n! > n^n\} cannot have a smallest element. So let n be the smallest number in S, and (by what you say you have shown) n>1. Then we have that both n! > n^n and (n-1)! \le (n-1)^{n-1}, and we want to derive a contradiction. Hint: What is n!/(n-1)! and what can you say about ... 0 Note that we have$$\begin{align} n!&=n\cdot (n-1)\cdot (n-2)\cdot (n-3)\cdots3\cdot 2\cdot 1\\\\ &=n\cdot n\left(1-\frac1n\right)\cdot n\left(1-\frac2n\right)\cdot n\left(1-\frac3n\right)\cdots n\frac3n\cdot n\frac2n\cdot n\frac1n\\\\ &=n^n\cdot\left(1-\frac1n\right)\cdot\left(1-\frac2n\right)\cdot\left(1-\frac3n\right)\cdots \frac3n\cdot ...

0

Hint: prove using induction. That is, prove the statement for $n = 1$, then assume the statement holds for some $n$ and prove that the statement holds for $n + 1$. Edit: Specifically, note that for $n = 1$, we have $n^n = 1$ and $n! = 1$, so this is true. Then we assume that the statement holds for some integer $n \geq 1$ (that is, we assume $n^n \geq n!$, ...

0

a. 5!*5! If the kind of hat is wrong, we just get permutations. b. D5 * D5 (Where D5 is the number of derangements on 5 objects = [5!/e]) This derangement number is determined with the inclusion/exclusion method. c. 10*9*8*7*6. Without the correct hat/person limitations, the attendant can give the five fedoras to 10, 9, 8, 7, 6 persons respectively ...

1

With "arbitrarily large" I assume a >> $10^{12}$ or something like that? Which means that b is large as well? I would start by factoring b; if b has a factor $p^k$ with p ≤ a then we may be able to prove that a! = 0 (modulo $p^k$). Even if that is not the case, calculating a! modulo x on a current computer will be a lot faster if x < $2^{64}$. Various ...

0

There are certain cases you can evaluate easily however for most cases you would need to evaluate it recursively. Apply the mod after each recursive step will help with the storage of the value. Special cases: The answer will be zero if there are sufficient terms to cover the prime factorization of $b$. E.g. If $b=72$ you need three factors of 2 and two ...

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n! becomes large already for small values of n so storage quickly becomes an issue if we were to calculate and store the whole number before doing modulus. The idea by Mose Wintner in the comments avoids having to store the whole number which is really nice, although we can also do it pairwise and in any order, say for instance $$(1 \cdot 2) (3 \cdot 4) (5 ... 0 In general, the inverse of the \Gamma function is not known to possess a closed form expression, if that is what you're asking. The best you might hope for is an infinite series/product expansion, and/or a continued fraction representation. 3 The vector you want to build up can be seen as a number of 50 digit in base 10 (= length of the alphabet). The answer is 10^{50}. This is called "sequence with repetition", no permutation. Similarly, it can be seen as the number of words of 50 letters given an alphabet of 10 letters. 0 The factorial is n! = n(n-1)(n-2)\cdots 2\cdot 1 The double factoral is n!! = n (n-2)(n-4) \cdots  (terminates with 2 or 1) The triple factorial is n!!! = n(n-3)(n-6)\cdots  (terminates with 3,2, or 1) And so forth. n\underbrace{!!\ldots!!}_{m} = \prod_{k=0}^{\lceil n/m-1\rceil} (n-km) Once the distinction between (n!)! and n!! is ... 0 To solve a multifactorial, we can start by counting the number of ! we need to write to represent the factorial. We'll call this number "k". We'll call the number we are to factorialize "x". Now, we set up a sequence where a(n) = x - kn. Our next step is to take all a(n) from n=1 up to, but not including, the nearest rounded-down integer to n = x/k, and ... 0 n!! is the product of positive integers \le n in steps of 2, n!!! is the product of positive integers \le n in steps of 3, and so on 2 The double factorial is reasonably standard and has the meaning you state. n!! is the product of all the numbers down from n that are of the same parity. I think your statement that 20!!=20(20-2)(20-4)\dots (20-18) captures it within the range of a high school student. You might include an example where n is odd. The triple factorial is much less ... 4 Note that for n\ge 2 the top is equal to n\cdot n^{n-2}. The bottom is (1)(2)\left[(3)(4)\cdots(n)\right]. The product [(3)(4)\cdots(n)] consists of n-2 terms, all \le n. So the bottom is \le (1)(2)n^{n-2}. It follows that for n\ge 2 we have$$\frac{n^{n-1}}{n!}\ge \frac{n\cdot n^{n-2}}{(1)(2)n^{n-2}}=\frac{n}{2}.$$Since ... 0$$\lim_{n \to\infty} \left(\frac{n+1}{n}\right)^{n-1} = e$$so it diverges. 1 You could try to use the Stirling formel which states that for large n$$n! \sim \sqrt{2\pi n} (\frac{n}{e})^n$$So you would get :$$\frac{n^{n-1}}{n!} \sim n^{n-1}(\frac{e}{n})^n (2\pi n)^{-\frac{1}{2}} \sim \frac{1}{\sqrt{2\pi}n^{\frac{3}{2}}} e^n \rightarrow \infty$$Thus the sequence diverges. 1 Using logarithms makes this kind of problems simpler. Considering$$A=\frac{\ n^2 2^n}{\ n!}\log(A)=2\log(n)+n\log(2)-\log(n!)$$Now, Stirling approximation$$\log(n!)=\log(\sqrt{2\pi})+(n+\frac 12)\log(n)-n+\frac 1 {12n}+\cdots$$So$$\log(A)=-\log(\sqrt{2\pi})+(1+\log(2))n-(n-\frac 32)\log(n)-\frac 1 {12n}+\cdots$$from which you can conclude that ... 3 Just transform everything into exponential like$$\frac{e^{2\ln(n)}\cdot e^{n\ln(2)}}{e^{\ln(n!)}}$$Now use the fact that \ln(n!) \approx n\ln(n) - n and substitute using powers properties obtaining$$e^{n\ln(2) + 2\ln(n) - n\ln(n) + n}$$namely$$e^{n\cdot[\ln(2) - \ln(n) + 1] + 2\ln(n)}$$as n\to\infty you can easily see that what remains is ... 1 Stirling's formula states that$$\lim_{n \rightarrow \infty} \frac{n!}{\sqrt{2\pi n} (\frac{n}{e})^n} = 1$$so$$ \lim_{n \rightarrow \infty} \frac{n^22^n}{n!} = \lim_{n \rightarrow \infty} \frac{n^22^n}{\sqrt{2\pi n}(\frac{n}{e})^n} $$and$$\lim_{n \rightarrow \infty} \frac{n^22^n}{\sqrt{2\pi n}(\frac{n}{e})^n} = \lim_{n \rightarrow \infty} ...

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You add a zero every time that you multiply by $10$. Since the only prime factors of $10$ are $2$ and $5$, then clearly the trailing number of zeros in a number is the minimum of the two exponents in the prime factorization of that number. To relate this to the formula you found, note that when computing a factorial, you will add a zero to the end every ...

2

About de Polignac's formula: You found yourself how many factors 5 the number n! has: The number of factors 5 is n/5 + n/25 + n/125... If you take other prime numbers, then you get very similar results: The number of factors 2 is n/2 + n/4 + n/8 ..., then number of factors 103 is n/103 + n/103^2 + n/103^3 ... and so on. $s_p(n)$ is the formula you found, ...

0

$n!!$ often denotes the semi-factorial, i.e. the product of every other integer up to $n$.

3

Well i can only assume what you are looking for. By the term in your question i think you might mean a superfactorial. See this Wikipedia Article on Factorials - it might help. If you mean a product of factorials it would be a notation like $$sf(n)=\prod^n_{k=1}k!$$

0

A double factorial is something else. In your case, it would be something like $$1!2!3!\dotsm n! = \prod_{k=1}^n k!$$ That's about it.

1

$n!=2^{25} \times 3^{13} \times 5^6 \times 7^4 \times 11^2 \times 13^2 \times 17 \times 19 \times 23$ Count the prime factors: For $2^n!$ there are $2^{n-1}$ multiples of 2, $2^{n-2}$ multiples of 4, $2^{n-3}$ multiples of 8, and so on the $1 + 2 + 4 + .... + 2^{n-1} = 2^n - 1$ powers of 2. I.E. $2^{2^n - 1}|2^n!$ but $2^{2^n}$ does not. Now 25 = 15 + ...

10

You know $n$ is between $23$ and $28$, inclusive, because there is a $23$ in the prime factorization of $n!$ but no $29$. There are $4$ factors of $7$, so these must be from $7$, $14$, $21$, and $28$. Therefore $n=28$.

0

Define $\gamma (f_n) := \log f_{n + 1} - \log f_n$. Observe that $f_n \equiv g_n$ if and only if $\gamma (f_n) = \gamma (g_n)$ and $f_1 = g_1$. We have $\gamma (n!) = \log (n + 1)$. We are going to try to express $n!$ in closed form as $h_n$. Let $$h_n = c_n \left ( \frac {n} {e} \right ) ^ n.$$ We have $\gamma (h_n) = \log (n + 1) + \log \frac {c_{n + ... 2 You can't break apart the product like this$$\lim \sqrt[n] {n!} = (\lim \sqrt[n]{n})(\lim \sqrt[n]{n-1}) \cdots (\lim \sqrt[n]{1})$$simply because the number of terms also goes toward infinity. To illustrate this, take the related example$\lim_{n \to \infty}\sqrt[n]{n^n}, or written out: $$1,\quad \sqrt2\cdot \sqrt2,\quad \sqrt[3]3\cdot \sqrt[3]3 \cdot ... 1 Your first proof is correct. The second one is not, because you cannot use "algebra of limits" as you did: in the product you write the number of terms is not fixed but is becoming larger and larger (is equal to n); so you cannot say say that the limit of the product is equal to the product of the limits. The same kind of things happen with infinite ... 1 As @ADG notes in his answer,$$ \sum_{k=1}^{n}\binom{n}{k}(−1)^{n+k}k^n = \sum_{k=0}^{n}\binom{n}{k}(−1)^{n+k}k^n = \sum_{k=0}^{n}\binom{n}{k}(−1)^{k}(n-k)^n, $$since \binom{n}{n-k}=\binom{n}{k}. Now, the formular is a special case of$$ \sum_{k=0}^{n}(-1)^k\binom{n}{k}P(n-k) = n! a_n, $$where P is a polynomial of degree \leq n, and a_n is the ... 0 Rewrite it as the ratio of two products. The numerator will obviously be none other than (3k)! while the denominator will be$$\begin{align}\prod_{j=0}^{k-1}(3j+1) \quad&=\quad 3^k~\prod_{j=0}^{k-1}\bigg(j+\dfrac13\bigg) \quad=\quad \dfrac{3^k}{\Gamma\bigg(\dfrac13\bigg)}\cdot\Gamma\bigg(\dfrac13\bigg)\cdot\prod_{j=0}^{k-1}\bigg(\dfrac13+j\bigg) ... 2 By considering the contour integral $$\oint_C d\zeta \frac{\zeta^{-z}}{(1+\zeta)^2}$$ about a keyhole contour and using the residue theorem, we may derive the relation $$\left (1-e^{-i 2 \pi z} \right) \int_0^{\infty} dx \frac{x^{-z}}{(1+x)^2} = i 2 \pi \left [\frac{d}{d\zeta} e^{-z \log{\zeta}} \right ]_{\zeta=e^{i \pi}} = i 2 \pi (-z e^{-i \pi} ) e^{-i ... 1 Your idea is a special case of what is known as rising factorials and falling factorials. The rising factorial is defined as$$ x^{(n)} = x(x+1)\cdots(x+n-1), $$and the falling factorial is$$ x_{(n)} = x(x-1)\cdots(x-n+1), $$where n is a non-negative integer and x is an arbitrary (complex) number. These notations are used in combinatorics and ... 1 Sure, you could define (-n)¡=(-n)\cdot(-n+1)\cdot(-n+2)\dots\cdot(-2)\cdot(-1), when n is a nonnegative integer. Note that you can alternatively write this as (-1)^n n!, so you don't really need to introduce a new notation for it. I don't know of any applications of this concept, though. Generally, because you can just write it in terms of the ... 1 Put "wiki factorial" into a search engine of your choice. The gamma function is the generalisation of factorial which is discontinuous at negative integers for the reason you appear to have spotted. 0 Evaluating a double sigma. How do I start with the problem ? By evaluating a simple sigma! ;-) Could you prove that the inner sum evaluates to \dfrac1{(m+1)^2} ? My bet is that it telescopes, due to some clever algebraic manipulations of binomial coefficients. Afterwards, see Basel problem and the Riemann \zeta function. 2 Binomial Coefficient or Combination Wikipedia words it quite nicely: \binom{n}{k} is read as "n choose k", because there are \binom{n}{k} ways to choose k elements, disregarding their order, from a set of n elements. I'd bet that most calculators use this efficient method,$$\binom{n}{k} = \prod\limits_{i=1}^k \frac{n+1-i}{i}$$So for ... 3 {50}\choose{4} is read "50 choose 4" and is the number of ways to choose 4 things from 50 things, where order doesn't matter. Most graphing calculators will have a button for this, if not there is the formula:$${{n}\choose{k}} =\frac{n!}{k!(n-k)!}$where$n!=n\times (n-1)\times\cdots\times 1$. 6 The key part of this problem is to realize$(1+{1\over n})^n\geq 2$. We use the binomial theorem here:$(1+{1\over n})^n\geq 1+n\cdot1\cdot{1\over n}=2$Now we apply our induction step:$(\frac{n+1}{2})^{n+1}=(\frac{n}{2})^{n}\cdot({1+{1\over n}})^n\cdot({n+1\over2})\geq n!\cdot2\cdot ({n+1\over2})=(n+1)!$0 If you are using all the primes not greater than$n$then$p_1^{\frac{2n}{p_1}}\cdots p_k^{\frac{2n}{p_k}}$should be greater than$n!$. We know that$n!$is equal to$\prod p_i^{[\frac{n}{p_i}]+[\frac{n}{p_i^2}]+...}$($[x]$denotes the integer part of$x$). So we only have to compare the exponents. The exponent of every prime in the factorial is not ... 6 Hint: Bertrand's postulate(in actuality a theorem) states that for every prime$p$, there exists another prime number between$p$and$2p$. This means that$\forall n>1, n!$will always have a single power of some prime number(s). For any positive integer$n\ge 2$, there exists a prime$p$such that$\frac{n}{2}<p\le n$This implies that$p\mid n!$. ... 0 I will encourage you to learn combinatorics basics. Let me explain the answer. let$X=\{x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10}\}$be the 10 different items you want to choose$5$items from. This will be the total no. of subsets of$X$with size$5$. Let$A$denotes the set of all possible subsets with size$5$. Then there is natural bijection from$A$... 1 Well, it's$\frac{10!}{5!}$, if you're counting each order you select balls as a different "way." For each of those$\frac{10!}{5!}$sets of$5$balls, there's$5!$ways to order it, so one way to see the correct answer is to divide by the number of same re-orderings, and get$\frac{10!}{5!5!}$, which is correct. (I always found this explanation hand-wavey, ... 1 Here is a copy of an answer I gave to a similar post (I'll edit it to fit your question a bit better): The formula for permutations is$\frac{n!}{(n-k)!}$The formula for combinations is$\frac{n!}{k!(n-k)!}$(Here I assume you understand the permutation formula a bit). To derive the second from the first let's make an example... say we are finding the ... 0 There are$\dfrac{n!}{k!(n-k)!}$ways to choose$k$things from$n\$ things.

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