All Questions
96
questions
1
vote
3
answers
51
views
Show ($n+1$)$2^n$ = $\sum_{i\geq 0}^{} {n + 1\choose i}i$ algebraically. [duplicate]
Show ($n+1$)$2^n$ = $\sum_{i\geq 0}^{} {n + 1\choose i}i$ algebraically.
I know $2^n$ = $\sum_{i\geq 0}^{} {n\choose i}$. But how do I manipulate the $(n+1)$ to make it look like the right side?
- 87
0
votes
2
answers
420
views
Prove Prove $n2^{n - 1}$ = $\sum_{i\geq 0}^{} {n \choose i}i$ algebraically and using induction.
Suppose $n \in$ natural numbers.
Prove $n2^{n - 1}$ = $\sum_{i\geq 0}^{} {n \choose i}i$
I have proven it combinatorially. Just having troubles algebraically and using induction.
- 87
14
votes
1
answer
352
views
are there known cases where $\binom{n}{k}$ is a perfect prime power?
I was wondering about cases where $\binom{n}{k}=p^j$ with $p$ a prime (nontrivially, so that $ n-k>1$ and $n \neq p^j$.) I had the terrible idea of checking binomial expansions
$$(x+y)^n \equiv x^n+...
- 21k
8
votes
1
answer
172
views
Largest power of $p$ which divides $F_p=\binom{p^{n+1}}{p^n}-\binom{p^{n}}{p^{n-1}}$ [closed]
I would like to know your comments in order to obtain the largest power of the prime number $p$ which divides
$$
F_p=\binom{p^{n+1}}{p^n}-\binom{p^{n}}{p^{n-1}}.
$$
I proved the largest power that ...
- 649
2
votes
2
answers
114
views
How to check if $k$ divides $\binom {n - 1} {k - 1}$ cheaply?
I'm writing a computer algorithm to do binomial expansion in C#. You can view the code here; I am using the following identity to do the computation:
$$
\dbinom n k = \frac n k \dbinom {n - 1} {k - 1}...
- 353
2
votes
2
answers
329
views
sum of squares of diagonal binomial coefficients.
My question is if there exists a way to evaluate the sum
$$
{{s}\choose{s}}^{\!2} + {{s + 1}\choose{s}}^{\!2} + \ldots {{s+r}\choose{s}}^{\!2}.
$$
In other words, it's the sum of the squares of the ...
- 39
4
votes
2
answers
861
views
Sum involving the product of binomial coefficients
I wonder whether it is possible to calculate the folowing sum that involves the Binomial coefficients
$$\sum_{k=0}^n \binom{n}{k}^2 \binom{2k}{k} .$$
- 321
3
votes
2
answers
91
views
How to find the remainder of $\binom{2013}{101}$ when it is divided by 101
I start first from the definition of
$$\binom{n}r=\frac{n!}{r!(n-r)!} $$
then I used the Wilson's theorem for p is prime
$$(p-1)!\equiv-1\pmod p$$
now how we can continue??
- 571
-1
votes
1
answer
23
views
calculating ${Q + K - 1 \choose K - 1}$ *$k!$. with logic,K*product(K-i+Q) , for i=1..(K-1)
I need to calculate ${Q + K - 1 \choose K - 1}$ *$k!$.
I came across this small logic which calculates the above expression but could not get the logic as how this works ?
Here is the logic :
K*...
- 696
27
votes
2
answers
1k
views
A nice formula for the Thue–Morse sequence
The Thue–Morse sequence$^{[1]}$$\!^{[2]}$ $t_n$ is an infinite binary sequence constructed by starting with $t_0=0$ and successively appending the binary complement of the sequence obtained so far:
$$\...
- 47.3k
5
votes
0
answers
115
views
What's a good estimate of $\sum\limits_{k \mid n} {n \choose k}$?
Question: What's a good estimate of $\sum\limits_{k \mid n} {n \choose k}$?
We have found an OEIS page for that: https://oeis.org/A056045
$1, 3, 4, 11, 6, 42, 8, 107, 94, 308, 12, 1718, 14, 3538, ...
- 5,847
1
vote
1
answer
301
views
Anti diagonal elements of table forming pascal triangle
A function in $k$ and $n$ leads to the formation of this table. The elements in this table are rows of pascal triangle if we look at the anti diagonals elements of this table. They have also been ...
- 696
0
votes
1
answer
252
views
How to choose $n$ balls from the bags?
Given $4$ bags A, B, C and D.
Bag A contains 'a' number of balls.
Bag B contains 'b' number of balls.
Bag C contains 'c' number of balls.
Bag D contains 'd' number of balls.
I have another bag E ...
- 1,020
0
votes
1
answer
3k
views
number of subsets of a set with even sum using combinatorics or binomial
Let S={a1,a2,a3.......aN}.There are 2^N subsets of this set so if we don't consider the empty set we are left with 2^N-1.We do need to consider cases where it number of odd numbers may be zero and ...
2
votes
1
answer
831
views
Multinomial coefficients modulo a prime
Let $p$ be a prime and let $m \geq 1$. Lucas' theorem implies that the binomial coefficient ${p^m-1 \choose k}$ is not divisible by $p$ for any $0 \leq k \leq p^m-1$. I wonder if something similar ...
- 2,003