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/...
1
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0
answers
323
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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 ...
- 6,565
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 ...
- 6,565
2
votes
1
answer
221
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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
votes
0
answers
251
views
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
votes
0
answers
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
votes
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 ...
- 1,790
1
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1
answer
83
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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
votes
1
answer
61
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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
votes
0
answers
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
answer
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
697
views
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
votes
2
answers
602
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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 ...
- 1,068
3
votes
1
answer
164
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Combinatorial interpretation for $\binom{n}{3}- \lfloor \frac{n}{3} \rfloor$
P2, RMO 2003, India
For any natural number $n\gt7$, prove that $\binom{n}{7}-\lfloor \frac{n}{7} \rfloor$ is divisible by $7$.
My algebraic solution :
$$ \binom{n}{7} = \dfrac{n(n-1)(n-2)(n-3)(n-4)(n-...
- 10.6k
5
votes
2
answers
942
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Relation between Pascal's Triangle and Euler's Number
My friends and myself were discussing Pascal's Triangle, specifically the following property of it.
First, consider the Pascal's Triangle -
$$1\\ 1\ 1\\ 1\ 2\ 1\\ 1\ 3\ 3\ 1\\ 1\ 4\ 6\ 4\ 1\\ 1\ 5\ ...
- 256
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votes
1
answer
36
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On powers of binomial coefficient
Consider $(1+n)^k$. Where both n, k are natural numbers.
We have binomial expansion $\sum_{i=0}^{k} \binom{k}{i}n^i $
Then we have for each i-th term, certain powers of n(at least $>=i$).
As i ...
0
votes
1
answer
45
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A proof for sum including binomial [duplicate]
I was trying to prove some equation and reached this summation $\sum \limits_{j=0}^{m} {2j\choose{j}} {2m-2j \choose {m-j}} = 4^m $
I tried pascal identity and other known binomial identities but ...
user411780
0
votes
0
answers
48
views
Amount of Compositions of a Number n into k parts when the components are limited to the range [1;m] with m<n
The number of compositions of a number n into k parts is given by the binomial coefficent ${n-1 \choose k-1}$.
Is there a closed formula to this question, when the summands of the composition are ...
- 9
0
votes
1
answer
54
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Identity on Theory of Numbers
Can anyone help me identify this identity? Or is there a known principle regarding this?
$k\binom{k}{k}-(k-1)\binom{k}{1}+(k-2)\binom{k}{2}-(k-3)\binom{k}{3}+\ldots +(-1)^{k-1}\binom{k}{k-1}$
Any ...
- 13
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votes
1
answer
720
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How many non-negative integer solutions exist for: $x+y+z=48$ where, $x<y<z$?
I want to find the number of non-negative integral solutions to the following:
$x+y+z=48$ where, $x<y<z$.
The answer is apparently 192 and the solution provided is $$\frac{\dbinom{50}{2}-\...
- 123
2
votes
1
answer
77
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Digits patterns in power number
Definition
Given positive integers $a,b$, with $a>1$, let $D(a,b)$ be the sum of the base-$a$ digits of $b$.
In other words
Rearranging, we get: $b = r_{l} a^l + ... + r_2 a^2 + r_1 a^1 + r_0 a^...
- 2,697
4
votes
2
answers
6k
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$n$ choose $k$ where $n$ is negative
I saw (in the book $A~ Walk~ Through~ Combinatorics$) that $\sum_{n \geq 0}{-3 \choose n} = \sum_{n \geq 0}{n+2 \choose 2}(-1)^n$, which confuses me. It seems that it can be derived directly from ...
- 43
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0
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221
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What is the growth rate of the products of binomial coefficients?
Claim: Experimental data seems to suggest that
$$
{n \choose 1^a b}{n \choose 2^a b}{n \choose 3^a b}\cdots {n \choose m^a b}
\sim \exp\bigg(\frac{2n^{1 + \frac{1}{a}}}{ab+3b}\bigg)
$$
where $a$ and ...
- 20.8k
3
votes
2
answers
1k
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Number of six digit numbers divisible by $3$ but none of the digits is $3$
Find number of six digit numbers divisible by $3$ but none of the digits is $3$
My try:
Let the six digits are $a,b,c,d,e,f$ such that
$$a+b+c+d+e+f=3p$$
where $1 \le p \le 18$
Now since $a \ge 1$...
- 10.2k
3
votes
2
answers
204
views
Prove that $x^n=\sum_{k=1}^{n}\sum_{j=1}^{k}(-1)^{k-j}\binom{k}{j}\binom{x}{k}j^n$
Prove that for every $x,n \in \mathbb{N}$ holds
$$x^n=\sum_{k=1}^{n}\sum_{j=1}^{k}(-1)^{k-j}\binom{k}{j}\binom{x}{k}j^n$$
This is so called MacMillan Double Binomial Sum, see Mathworld - Power, ...
- 558
2
votes
1
answer
219
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Are there any power identities which don't belong to this list?
The problem of finding expansions of monomials, binomials etc. is classical and there is a lot of beautiful solutions have been found already, the most prominent examples are
Binomial Theorem, ...
- 558
3
votes
3
answers
200
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A conjecture on the sum of binomial coefficients
I am looking for a proof, disproof or counter example of the following claim.
Let $C = \{k_1, k_2, \ldots, \}$ be a strictly increasing infinite sequence of positive integers which have a certain ...
- 20.8k
2
votes
2
answers
279
views
Concerning the identity in sums of Binomial coefficients
Let be the following identity
$$\sum_{k=1}^{n}\binom{k}{2}=\sum_{k=0}^{n-1}\binom{k+1}{2}=\sum_{k=1}^{n}k(n-k)=\sum_{k=0}^{n-1}k(n-k)=\frac16(n+1)(n-1)n$$
As we can see the partial sums of binomial ...
- 558