All Questions
96
questions
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Summation of n-simplex numbers
Gauss proved that every positive integer is a sum of at most three triangular(2-simplex) numbers. I was thinking of an extension related to n-simplex.
Refer: https://upload.wikimedia.org/wikipedia/...
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321
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A conjecture on representing $\sum\limits_{k=0} ^m (-1)^ka^{m-k}b^k$ as sum of powers of $(a+b)$.
UPATE: I asked this question on MO here.
I was solving problem 1.2.52 in "An introduction to the theory of numbers by by Ivan Niven, Herbert S. Zuckerman, Hugh L. Montgomery"
Show that if ...
- 6,350
5
votes
1
answer
205
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Conjecture: $\binom{n}{k } \mod m =0$ for all $k=1,2,3,\dots,n-1$ only when $m $ is a prime number and $n$ is a power of $m$
While playing with Pascal's triangle, I observed that $\binom{4}{k } \mod 2 =0$ for $k=1,2,3$,and $\binom{8}{k } \mod 2 =0$ for $k=1,2,3,4,5,6,7$ This made me curious about the values of $n>1$ and ...
- 6,350
2
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1
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204
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Generalization of binomial coefficients to both non-integer arguments
It is known that binomial coefficients can be generalized to the following: for $s\in\mathbb R$ and $k\in\mathbb N$,
\begin{equation*}
\binom{s}{k} := \prod_{i=0}^{k-1} \frac{s-i}{k-i} = \frac{s(s-1)...
- 1,972
4
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0
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251
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Sum of even binomial coefficients modulo $p$, without complex numbers
Let $p$ be a prime where $-1$ is not a quadratic residue, (no solutions to $m^2 = -1$ in $p$).
I want to find an easily computable expression for
$$\sum_{k=0}^n {n \choose 2k} (-x)^k$$
modulo $p$. ...
- 4,284
0
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0
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100
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Show that $2n\choose n$ divisible by primes $p,$ such that $n<p<2n$? [duplicate]
Suppose on the contrary that $2n \choose n$ is not divisible by $p\in (n,2n)$. There exis $k$ and $0\ne r\lt p$ such that
$(2n)\cdots (n+1)=kp \,n!+r \, n!$. The second term on RHS is not divisible by ...
- 11.5k
-1
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1
answer
48
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Number of ones in the dyadic expansion of m [closed]
I was going through a paper where I stuck on a combinatorial argument as follows
I want help with the first assertion i.e proving the inequality $\alpha(m+l)\le\alpha(m)$.
As the author suggests it is ...
- 1,770
1
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1
answer
82
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prove that: there are exactly ${n-1 \choose k-1}$products that consist of $n-k$ factors
prove that: there are exactly ${n-1 \choose k-1}$products that consist of $n-k$ factors,so that all these factors are elements of $[k]$. Repetition of factors is allowed,
that is my attempt :
for ...
user998997
0
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1
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$\sum\limits_{k=1}^{n-1}\binom{n-1}{k-1}\frac{a^{n-k}b^k}{k}=\frac{(a+b)^n-a^n-b^n}{n}$
In "New properties for a composition of some generating functions for primes", properties of generating functions are used to form a primality test, speaking specifically in example 1 the ...
- 1,262
2
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0
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47
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A $\Bbb Z$-valued $\Bbb Q$-coefficient polynomial is a $\Bbb Z$-combination of $\binom{x}{k}$s: $q$-analog?
I am wondering what a $q$-analog of the following should be:
Proposition. If $f(x)\in\Bbb Q[x]$ is a polynomial for which $f(\Bbb Z)\subseteq\mathbb{Z}$ (that is, $f(x)$ is an integer for all integers ...
- 152k
5
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1
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186
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For how many $n$ is $2021^n$ + $2022^n$ + $2023^n$ + ... + $2029^n$ prime?
For how many $n$ is $2021^n$ + $2022^n$ + $2023^n$ + ... + $2029^n$ prime?
My first thought is set x = 2021 so we can create:
$x^n$ + $(x+1)^n$ + $(x+2)^n$ + ... + $(x+8)^n$
And then we expand each ...
- 199
2
votes
1
answer
108
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Parity of a Binomial Coefficient Using Lucas' Theorem
I'm trying to prove a property involving the parity of a binomial coefficient using Lucas' theorem. For this, let $r=\lfloor \log_2(k_s)\rfloor+1$, with $k_s\geq2$. We know that if $0\leq i\leq 2^{r}-...
- 453
1
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1
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53
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How to summarise pattern in binomial-type expansion into a single expression
I have a series of polynomials as follows:
\begin{eqnarray}
&1\\
&1+4x^3+x^6\\
&1+20x^3+48x^6+20x^9+x^{12}\\
&1+54x^3+405x^6+760x^9+405x^{12}+54x^{15}+ x.^{18}\\
& \vdots\tag{1}
\...
- 317
2
votes
1
answer
690
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binomial coefficients modulo $3$
Let $a(n)$ be the number of binomial coefficient's on the $n$-th row of Pascal's triangle which leave remainder $1$ upon division by $3$, and $b(n)$ be the number that leave remainder $2$. Show that $...
user765855
12
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2
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601
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If a power of 2 divides a number, under what conditions does it divide a binomial coefficient involving the number that it divides?
We have had many questions here about the divisibility of $\binom{n}{k}$, many of them dealing with divisibility by powers of primes, or expressions involving the $\textrm{gcd}(n,k)$ (I originally ...
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