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To generate laundau's prime numbers using python

I want to generate list of primes of the format x2+1 but I am not getting any output.
My code:
def LPrime(n):
for i in range(1,n):
x = i**2+1;
for j in range(2,x):
if x % j != 0:
print(x,i)
Please tell me what am I doing wrong.
The expected output should be something like
2,1
5,2
and so on
Also I am not able to execute it properly, I am using Ubuntu 22.04
There is no output when I try to

If if x % j == 0: you have a divisor. and j will not be 1 or x. So you should mark p as False (there is no prime). else, it is a prime; print it.
def LPrime(n):
for i in range(0, n):
x = i**2+1
p = True
for j in range(2, x):
if x % j == 0:
p = False
if p:
print(f"Prime = {x},\t i = {i}")
Don't forget to call it
>>> LPrime(4)
Prime = 1, i = 0
Prime = 2, i = 1
Prime = 5, i = 2
If you want n primes, use a while loop:
def LPrime(n):
count = 0
i = 0
while count < n:
x = i**2+1
p = True
for j in range(2, x):
if x % j == 0:
p = False
if p:
print(f"Prime = {x},\t i = {i}")
count += 1
i += 1
>>> LPrime(4)
Prime = 1, i = 0
Prime = 2, i = 1
Prime = 5, i = 2
Prime = 17, i = 4

Function to sum and compare two sides of an array

For example this is array = [1,2,3,4,3,2,1]
i want computer to choose i=4 and sum both left of the four and right of the four and then compare if they are equal to each other, program prints i value to the screen.
Now i write it in here and it works as long as the i value is 0 but if i want to make this search for i>0 then i gets complicated.
this is my main program:
# importing "array" for array creations
import array as arr
# Function to find sum
# of array exlcuding the
# range which has [a, b]
def sumexcludingrange(li, a, b): # I FOUND THIS FUNCTION ONLINE BUT ALSO DIDN`T WORK EITHER.
sum = 0
add = True
# loop in li
for no in li:
# if no != a then add
if no != a and add == True:
sum = sum + no
# mark when a and b are found
elif no == a:
add = False
elif no == b:
add = True
# print sum
return sum
#lis = [1, 2, 4, 5, 6]
#a = 2
#b = 5
#sumexcludingrange(arr, 0, i-1)
def my(arr):
for i in range(len(arr)):
if i == 0: #this works when array is like [1,0,0,1] and i equals zero
sum_left = arr[0]
sum_right = sum(arr) - arr[0]
while sum_left == sum_right:
print(i)
break
elif i > 0 : # i 1 2 3 4 5 6 7 8 9..
#sumleft and sumright are not right.. what is the issue here?
sum_left = sumexcludingrange(arr, 0, i-1) #result is 16, sumexcludingrange function didn`t
#work.
print (sum_left)
#print(i)
sum_right == sum(arr) - sum_left
#print(sum_right)
#break
while sum_left == sum_right:
print(sum_left)
print(sum_right)
print("they are equal now")
break
else:
print("index cannot be a negative number.")
something = arr.array('i', [1,2,3,4,3,2,1]) # len(arr) = 7
my(something)
After seeing that i need another function to compare two sides of the array, i created a new file and write it in here and somehow it outputs this:
1
13
13
2
10
10
#array[i] array[i+1] .. + array[length-1]
array = [1,2,3,4,5,6,7] # 7 element so length=7,
for i in range(len(array)):
if i > 0 :
ln = len(array) + 1
sum_right = sum(array[i+1:ln])
sum_left = sum(array) - sum_right
print(i)
print(sum_right)
print(sum_right)
the way i think is this:
# array[i] i>0 i={1,2,3,4..}
sum_total = sum(arr) = array[0] + ..+ array[i-1] + array[i] + array[i+1] + ... + array[length-1]
sum_right = array[i+1] + .. + array[length-1]
sum_left = sum_total - sum_right
Is there a better way to accomplish that?

If I understand your question correctly, the solution could be something like this:
array = [1,2,3,4,5,6,7]
a1 = array[:4] # first 4 numbers
a2 = array[-4:] # last 4 numbers
total = sum(array)
diff = total-sum(a2)

Estimating in Python

I am trying to write a python function which will give me back the first value of a number k which will function 2 >= function 1
function 1(p) => 1 + 1/1! + 1/2! + 1/p!
function 2(k) => (1 + 1/k)^k
So I input use 2 for example in function 1. It would take estimate e at 2.5 but it would take k being 6 for it to get close which is 2.522.
I want to return the 6.
I get this far but I am not sure where to go from there.
for x in range(p):
factorial(x):
if x == 0:
return 0
else:
return x * factorial(x-1)
result = 1 + 1/factorial(p)

I think you need two separate functions, and no loops are needed.
def factorial(x):
if x in {0, 1}:
return 1 # This needs to be 1, otherwise you multiply everything by 0
else:
return x * factorial(x-1)
def summ(p):
if p == 0:
return 1
elif p == 1:
return 2
else:
return 1/factorial(p) + summ(p-1)
Regarding the rest of the question, I think this will help
def f2(k):
return 1 if k == 0 else (1 + 1/k)**k
max_iter = 10 # TODO: increase
k = 0
delta = 0.000001
while max_iter > 0:
f1_result = summ(k)
f2_result = f2(k)
check = f1_result <= f2_result
print("k={}: {:6.8f} <= {:6.8f} == {}".format(k, f1_result, f2_result, check))
k += 1
max_iter -= 1
# if check:
# break
# TODO: Check if difference is within acceptable delta value
Output
k=0: 1.00000000 <= 1.00000000 == True
k=1: 2.00000000 <= 2.00000000 == True
k=2: 2.50000000 <= 2.25000000 == False
k=3: 2.66666667 <= 2.37037037 == False
k=4: 2.70833333 <= 2.44140625 == False
k=5: 2.71666667 <= 2.48832000 == False
k=6: 2.71805556 <= 2.52162637 == False
k=7: 2.71825397 <= 2.54649970 == False
k=8: 2.71827877 <= 2.56578451 == False
k=9: 2.71828153 <= 2.58117479 == False
From larger numbers, this check still fails at k=996 and throws a recursion error.
k=996: 2.71828183 <= 2.71691848 == False
Traceback (most recent call last):
File "python", line 22, in <module>
File "python", line 13, in summ
File "python", line 5, in factorial
File "python", line 5, in factorial
File "python", line 5, in factorial
[Previous line repeated 992 more times]
RecursionError: maximum recursion depth exceeded
Therefore, it would help if you wrote a factorial function using a loop rather than recursion.
Edit
I think I understand what you are trying to get now
in1 = int(input("value 1:"))
v1 = summ(in1)
v2 = 0
max_iter = 50 # Recursion execution limit
while f2(v2) <= v1 and max_iter > 0:
v2 += 1
max_iter -= 1
print(v2)
Output
value 1: <enter 2>
6

sum of factors of a given number

i have written this code which finds factors of a number .after thinking and trying so much i could not get the sum of the numbers I get in output.I wish to get the sum of these numbers as output recursively.here's my code:
def p(n,c):
s = 0
if c >= n:
return n
if n % c == 0:
s += c
print(s,end=',')
return p(n,c+1)
n = int(input('enter no:'))
c = 1
print(p(n,c))

Given the comments, it appears that this might be what you want:
sum([n for n in xrange(1,24) if 24 % n == 0])
To make it a bit more generic:
def sum_of_factors(x):
return sum([n for n in xrange(1,x) if x % n == 0])
EDIT: here's a recursive version:
def sum_of_factors(x, y=1):
if (y >= x):
return 0
if (x % y == 0):
return y + sum_of_factors(x, y + 1)
return sum_of_factors(x, y + 1)
>>> sum_of_factors(24)
36

Is this the output you are looking for?
Use global variable,
s = 0
def p(n,c):
global s
if c >= n:
return n
if n % c == 0:
s += c
print(s,end=',')
return p(n,c+1)
n = int(input('enter no:'))
c = 1
print(p(n,c))
Output
enter no:1,3,6,10,16,24,36,24

Collatz conjecture sequence

The Collatz conjecture
what i am trying to do:
Write a function called collatz_sequence that takes a starting integer and returns the sequence of integers, including the starting point, for that number. Return the sequence in the form of a list. Create your function so that if the user inputs any integer less than 1, it returns the empty list [].
background on collatz conjecture:
Take any natural number n. If n is even, divide it by 2 to get n / 2, if n is odd multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1.
What I have so far:
def collatz_sequence(x):
seq = [x]
if x < 1:
return []
while x > 1:
if x % 2 == 0:
x= x/2
else:
x= 3*x+1
return seq
When I run this with a number less than 1 i get the empty set which is right. But when i run it with a number above 1 I only get that number i.e. collatz_sequence(6) returns [6]. I need this to return the whole sequence of numbers so 6 should return 6,3,10,5,16,8,4,2,1 in a list.

You forgot to append the x values to the seq list:
def collatz_sequence(x):
seq = [x]
if x < 1:
return []
while x > 1:
if x % 2 == 0:
x = x / 2
else:
x = 3 * x + 1
seq.append(x) # Added line
return seq
Verification:
~/tmp$ python collatz.py
[6, 3, 10, 5, 16, 8, 4, 2, 1]

def collatz_sequence(x):
seq = [x]
while seq[-1] > 1:
if x % 2 == 0:
seq.append(x/2)
else:
seq.append(3*x+1)
x = seq[-1]
return seq
Here's some code that produces what you're looking for. The check for 1 is built into while statement, and it iteratively appends to the list seq.
>>> collatz_sequence(6)
[6, 3, 10, 5, 16, 8, 4, 2, 1]
Note, this is going to be very slow for large lists of numbers. A cache won't solve the speed issue, and you won't be able to use this in a brute-force solution of the project euler problem, it will take forever (as it does every calculation, every single iteration.)

Here's another way of doing it:
while True:
x=int(input('ENTER NO.:'))
print ('----------------')
while x>0:
if x%2==0:
x = x/2
elif x>1:
x = 3*x + 1
else:
break
print (x)
This will ask the user for a number again and again to be put in it until he quits

def collatz(x):
while x !=1:
print(int(x))
if x%2 == 0:
x = x/2
else:
x = 3*x+1
this is what i propose..

seq = []
x = (int(input("Add number:")))
if (x != 1):
print ("Number can't be 1")
while x > 1:
if x % 2 == 0:
x=x/2
else:
x = 3 * x + 1
seq.append (x)
print seq

This gives all the steps of a single number. It has worked with a 50-digit number in 0,3 second.
collatz = []
def collatz_sequence(x):
while x != 1:
if x % 2 == 0:
x /= 2
else:
x = (3*x + 1)/2
collatz.append(int(x))
print(collatz)
collatz_sequence()

Recursion:
def collatz(n):
if n == 1: return [n]
elif n % 2 == 0: return [n] + collatz(int(n/2))
else: return [n] + collatz(n*3+1)
print(collatz(27))

steps=0
c0 = int(input("enter the value of c0="))
while c0>1:
if c0 % 2 ==0 :
c0 = c0/2
print(int(c0))
steps +=1
else:
c0 = (3 * c0) + 1
print(int(c0))
steps +=1
print("steps= ", steps)

import numpy as np
from matplotlib.pyplot import step, xlim, ylim, show
def collatz_sequence(N):
seq = [N]
m = 0
maxN = 0
while seq[-1] > 1:
if N % 2 == 0:
k = N//2
seq.append(N//2)
if k > maxN:
maxN = k
else:
k = 3*N+1
seq.append(3*N+1)
if k > maxN:
maxN = k
N = seq[-1]
m = m + 1
print(seq)
x = np.arange(0, m+1)
y = np.array(seq)
xlim(0, m+1)
ylim(0, maxN*1.1)
step(x, y)
show()
def collatz_exec():
print('Enter an Integer')
N = int(input())
collatz_sequence(N)
This is how you can use it:
>>> from collatz_sequence import *
>>> collatz_exec()
Enter an Integer
21
[21, 64, 32, 16, 8, 4, 2, 1]
And a plot that shows the sequence:

seq = []
def collatz_sequence(x):
global seq
seq.append(x)
if x == 1:
return
if (x % 2) == 0:
collatz_sequence(x / 2)
else:
collatz_sequence((x * 3) + 1)
collatz_sequence(217)
print seq

def collataz(number):
while number > 1:
if number % 2 == 0 :
number = number //2
print(number)
elif number % 2 ==1 :
number = 3 * number + 1
print(number)
if number == 1 :
break
print('enter any number...!')
number=int(input())
collataz(number)