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Problem: Simplify the following equation:
begin {equation}
sum limits_ {k = 1} ^ n dbinom {n} {k} begin {Bmatrix} n \ k end {Bmatrix} k!
end {equation}
One solution: I am aware of the following relationship:
begin {equation}
sum limits_ {t = 1} ^ nt ^ n = sum limits_ {k = 1} ^ n dbinom {n + 1} {k + 1} begin {Bmatrix} n \ k end {Bmatrix } k!
end {equation}
Now, I struggle to get a somewhat similar (ie net) expression for the given equation.
Thank you for your help!
For $ p $ a prime number, $ x $ a positive real number greater than $ p $ and k $ an even positive integer, say a maximum subset $ A $ prime numbers not exceeding $ x $containing $ p $ and such as the distance between two consecutive elements of $ A $ Is at least k $ to the $ (k, p, x) $property -gap. Now for such a set $ A $ let $ N_ {A} (x) $ the set of integers not exceeding $ x $ whose elements are products of elements of $ A $. Finally leave $ h_ {A} (x) $ to be the cardinal of $ mathbb {N} cap ([2,x] setminus N_ {A} (x)) $ .
What is the best upper limit for $ inf {h_ {A} (x), a text {a} (k, p, x) text {-gap property} } $ in terms of k $, $ p $ and $ x $?
I have this question on Sierpinski's triangle:
What percentage of the figure is blue?
Graphic[
Table[
If[
EvenQ[Binomial[col, k]],
{EdgeForm @ Thin, Blue, Rectangle[{col, k}]}
{EdgeForm @ Thin, Yellow, Rectangle[{col, k}]}
],
{col, 0, 7, 1},
{k, 0, col, 1}
]]
create a collection of six characters, each character representing a new stage of the Sierpinski triangle. Combination table[] with the previous batch of code:
Table[
Graphics[
Table[
If[
EvenQ[Binomial[col, k]],
{Blue, rectangle[{col, k}]}
{Yellow, rectangle[{col, k}]}
],
{col, 0, n, 1},
{k, 0, col, 1}
]],
{n, {7, 15, 31, 63, 127, 255}}
]
more elegantly:
Table[
Graphics[
Table[
If[
EvenQ[Binomial[col, k]],
{Blue, rectangle[{col, k}]}
{Yellow, rectangle[{col, k}]}
],
{col, 0, 2 ^ n - 1, 1}, (* This ensures that we have a
of-2 number of columns. *)
{k, 0, col, 1}
]],
{n, 3, 8, 1}
]
Think of some counting operations:
Table[
If[
EvenQ[Binomial[col, k]],
1
0
],
{col, 0, 7, 1},
{k, 0, col, 1}
]
count[] the 1 in the previous list, start with Flatten[]in the list:
Flatten[
Table[
If[
EvenQ[Binomial[col, k]],
1
0
],
{col, 0, 7, 1},
{k, 0, col, 1}
]]count[
Flatten[
Table[
If[
EvenQ[Binomial[col, k]],
1
0
],
{col, 0, 7, 1},
{k, 0, col, 1}
]],
1
]
To find the percentage of evens, divide that number by the length[] from the same list:
100. Count[
Flatten[
Table[
If[
EvenQ[Binomial[col, k]],
1
0
],
{col, 0, 2 ^ 9 - 1, 1}, (*
Change the polynomial *)
{k, 0, col, 1}
]],
1
]/Length[
Flatten[
Table[
If[
EvenQ[Binomial[col, k]],
1
0
],
{col, 0, 2 ^ 9 - 1, 1},
{k, 0, col, 1}
]]]
What changes do I need to make to give the percentage of entries that are even?
How can I generate a table where the left column shows the number of columns in the triangle while the right column gives the percentage of entries identical to those in the image?
I am very new at Mathematica, so I spend a lot of time on it.
The context : I am working on a telematic application for an insurance company. Drivers must drive properly and the AI calculates their overall score in relation to their journey score. The score may vary (increase or decrease). There were frauds in the past when people could see their trip score and report that trip as "did not drive", so we decided to delete the trip score.
The problem Can people not evaluate their driving skills effectively because there is nothing to compare or any indication that the trip has been good except in the details, but the information is too specific ex: fast acceleration, hard braking, etc. Do not give a big picture to the user.
Proposal: I've tried the metaphor of traffic lights (red, yellow, green), but it's not accessible to visually impaired users. I've also tried the stars (1, 2, 3) but it's not going well with the brand. Finally, I've tried the metaphor of the gauge, but it's too complicated to understand at first glance (low gauge = bad or good?).
I'm not aware, do you have any suggestions?
Let's say there is a group of 5 numbers or more. In this group of numbers, the largest number contributes approximately. 63% in total of the group. How to increase or decrease or do nothing of such number, so that the largest number of the new series of figures now represents 25% of the group total. For simplicity, a group of 6-digit samples consists of – 535904.48, 51345976.47, 334394997.88, 101100599.81, 35426627.55, and 45416.58.
Any idea or suggestion on how to get a new set of numbers that satisfies the requirement of the greatest number not contributing to more than 25% of the new total should be highly appreciated.
For any positive integer k $ and $ l $, does the equation
$$ ( sum_ {i = 1} ^ k frac {1} {p_i}) ( sum_ {j = 1} ^ l frac {1} {q_j}) = 1 $$
have solutions in distinct primes, that is to say $ p_1, p_2, points, p_k, q_1, q_2, points, q_l $ are distinct?
I need clarification for the factorization when the variable 'd' is included in $ (a + ib) ^ n $, such as $ (da + dib) ^ n $.
Yes $ (a + ib) ^ 2 $ = $ 1 * a ^ 2 + 2iab – 1 * b ^ 2 $,
while what:
$ (da + dib) ^ 2 $
= ??
For the first, I was able to determine that $ (da + dib) ^ 2 $ is either
Yes $ (a + bi) ^ 4 $ = $ a ^ 4 + 4a ^ 3bi – 6a ^ 2b ^ 2 – 4ab ^ 3i + b ^ 4 $
while what:
$ (da + dib) ^ 4 $ = ??
Yes $ (a + bi) $ 8 = ## EQU1 ##
while what: $ (da + dib) ^ $ 8 = ??
Today, I have a very difficult problem and yet I have no idea how to solve it.
Someone who can help?
https://www.namepros.com/threads/00567-com-sold-for-60-500-icrm-com-for-8-988.1113567/
Well, the field is 18, but I have not seen good indicators – in fact, zero. OK, what is the justification? I guess it's rare, but it's hard to believe that 00567.com is rare.
How can I generate random numbers from the tics of a Geiger counter?
Suppose I have a file with the timestamps of each event.
0
325
456
920
....
The time intervals between two events can be described by a Poisson distribution.
What is the next step to get real random numbers distributed uniformly from this?
A solution on Fourmilab compares the time interval between two successive events and sets 0 and 1 by testing whether the second interval is longer.
This will create one bit from two normal distributed values.
Is there a standard way to more efficiently create random numbers distributed evenly across time slots?
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