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How can i calculate and print the sum of first N elements of below sequence using recursive function

How can i calculate and print the sum of first N elements of below sequence using recursive function
Example:
Sample Input N: 4
Sample Output: 7 + 12 + 17 + 22 = 58
I did some part of the code but it gives wrong result and I don't know where I'm making mistakes. That's why I need help!
def recur_sum(n):
if n <= 1:
return n
else:
for i in range(7,n+1):
return i + recur_sum(i-1)
num = int(input("Enter a number: "))
if num < 0:
print("Enter a positive number")
else:
print("The sum is",recur_sum(num))

As far as I understand your question, you want the sum of a sequence which each element of this sequence is increased by a step (such as 5), and it has an initial value (like 7), and you want the sum of first N (like 4) elements of this sequence.
At each recursion level, we add the current value to step , but when n == 1 we only return the current value (which is the Nth item of the sequence).
Do this:
def recur_sum(cur_val, step, n):
if n == 1:
return cur_val
return cur_val + recur_sum(cur_val+step,step,n-1)
num = int(input("Enter a number: "))
init_val = 7
step = 5
if num < 0:
print("Enter a positive number")
else:
print("The sum is",recur_sum(init_val, step, num))
Output:
The sum is 58

this returns a list of all sequential numbers starting at 7 each increment of 5. sum the return array.... change 5 and 2 to change increments for desired jumps/steps, and change the initial return value..
def recur_sum(n):
if n == 1:
return [7]
else:
return [5*n+2] + recur_sum(n-1)
num = int(input("Enter a number: "))
res = recur_sum(num)
print(sum(res))

A recursive function which returns all the elements up to n (where n is the input of your function) has already been proposed above.
In my understanding, you want a function with some recursive logic that return the sum of all the elements up to the n-th.
Your sequence is 7, 12, 17, 22, 27 and so forth. If we disect it:
it element sum sum is element is
1 7 7 1 * 7 + 0 * 5 1 * 7 + 0 * 5
2 12 19 2 * 7 + 1 * 5 1 * 7 + 1 * 5
3 17 36 3 * 7 + 3 * 5 1 * 7 + 2 * 5
4 22 58 4 * 7 + 6 * 5 1 * 7 + 3 * 5
5 27 85 5 * 7 + 10 * 5 1 * 7 + 4 * 5
If you want at any cost to implement a recursive solution, if is evident that at each step you need to increase the rolling sum by it * 7 + (it - 1) * 5 (where 7 is your start point, while 5 is your step).
You can implement a recursive solution as follows:
def recursive(n, step = 5, start = 7, counter = 1):
if n > 0:
this_element = start + (counter - 1) * step
if counter == n:
return this_element
else:
return this_element + recursive(n, step = step, start = start, counter = counter + 1)
else:
return 0
for i in range(1, 10):
print(recursive(i))
OUTPUT
7
19
36
58
85
117
154
196
243
From the table above you can see though that maybe a recursive solution is overkilling here given that the sum of elements up to n-th step has solution:
def my_sum(n, step = 5, start = 7):
return n * start + int(step * (n - 1) * n / 2)
for i in range(1, 10):
print(my_sum(i))
OUTPUT
7
19
36
58
85
117
154
196
243

python Program is saying 2 numbers are different when they are the same

Thanks for the help in advance.
I compare 2 variables which hold numbers with the operation !=
the program takes input like this
4
1 2 3 4
Then the program should return the highest product exluding a square product. In this case the result should be 12(4x3) not 16(4x4) but it is incorrectly returning 16.
The code is here
def max_pairwise_product():
n = int(input())
a = input().split()
maxno = 0
for integer1 in a:
product = int(integer1) * n
if n != integer1:
if product > maxno:
maxno = product
return maxno
print(max_pairwise_product())

This is because your comparing an int to a str. I have edited your code minimally to fix this issue, but you have better to edit your list input before hand.
def max_pairwise_product():
n = int(input())
a = input().split() #If I were you I would do: a = [int(x) for x in input().split()]
maxno = 0
for integer1 in a:
product = int(integer1) * n
if n != int(integer1): #This is were I have changed
if product > maxno:
maxno = product
return maxno
print(max_pairwise_product())

You compare string vs. integer which is always False:
for integer1 in a: # integer1 is a string
product = int(integer1) * n
if n != integer1: # n is a number
Fix:
def max_pairwise_product():
_ = input() # not used
a = "1 2 3 4".split() # input() removed and fixed input given
a = sorted(set(map(int,a))) # convert all to intergers and sort unique numbers
maxno = 0
for idx,number in enumerate(a):
for number2 in a[idx+1:]:
maxno = max( maxno, number*number2 )
return maxno
print(max_pairwise_product())
Output:
12
The loops are optimized leveraging the fact that they numbers are sorted and unique - thats why the second loop can be shorter - no need to recompute already computed results.
Its good if you get inputs like '1 1 1 1 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6' because it will only compute for '1 2 3 4 5 6' and never compute any paring twice.

A test interview question I could not figure out

So I wrote a piece of code in pycharm
to solve this problem:
pick any 5 positive integers that add up to 100
and by addition,subtraction or just using one of the five values
you should be able to make every number up to 100
for example
1,22,2,3,4
for 1 I could give in 1
for 2 i could give in 2
so on
for 21 I could give 22 - 1
for 25 I could give (22 + 2) - 1
li = [1, 1, 1, 1, 1]
lists_of_li_that_pass_T1 = []
while True:
if sum(li) == 100:
list_of_li_that_pass_T1.append(li)
if li[-1] != 100:
li[-1] += 1
else:
li[-1] = 1
if li[-2] != 100:
li[-2] += 1
else:
li[-2] = 1
if li[-3] != 100:
li[-3] += 1
else:
li[-3] = 1
if li[-4] != 100:
li[-4] += 1
else:
li[-4] = 1
if li[-5] != 100:
li[-5] += 1
else:
break
else:
if li[-1] != 100:
li[-1] += 1
else:
li[-1] = 1
if li[-2] != 100:
li[-2] += 1
else:
li[-2] = 1
if li[-3] != 100:
li[-3] += 1
else:
li[-3] = 1
if li[-4] != 100:
li[-4] += 1
else:
li[-4] = 1
if li[-5] != 100:
li[-5] += 1
else:
break
this should give me all the number combinations that add up to 100 out of the total 1*10 ** 10
but its not working please help me fix it so it prints all of the sets of integers
I also can't think of what I would do next to get the perfect sets that solve the problem

After #JohnY comments, I assume that the question is:
Find a set of 5 integers meeting the following requirements:
their sum is 100
any number in the [1, 100] range can be constructed using at most once the elements of the set and only additions and substractions
A brute force way is certainly possible, but proving that any number can be constructed that way would be tedious. But a divide and conquer strategy is possible: to construct all numbers up to n with a set of m numbers u0..., um-1, it is enough to build all numbers up to (n+2)/3 with u0..., um-2 and use um-1 = 2*n/3. Any number in the ((n+2)/3, um-1) range can be written as um-1-x with x in the [1, (n+2)/3] range, and any number in the (um-1, n] range as um-1+y with y in the same low range.
So we can use here u4 = 66 and find a way to build numbers up to 34 with 4 numbers.
Let us iterate: u3 = 24 and build numbers up to 12 with 3 numbers.
One more step u2 = 8 and build numbers up to 4 with 2 numbers.
Ok: u0 = 1 and u1 = 3 give immediately:
1 = u0
2 = 3 - 1 = u1 - u0
3 = u1
4 = 3 + 1 = u1 + u0
Done.
Mathematical disgression:
In fact u0 = 1 and u1 = 3 can build all numbers up to 4, so we can use u2 = 9 to build all numbers up to 9+4 = 13. We can prove easily that the sequence ui = 3i verifies sum(ui for i in [0, m-1]) = 1 + 3 + ... + 3m-1 = (3m - 1)/(3 - 1) = (um - 1) / 2.
So we could use u0=1, u1=3, u2=9, u3=27 to build all numbers up to 40, and finally set u4 = 60.
In fact, u0 and u1 can only be 1 and 3 and u2 can be 8 or 9. Then if u2 == 8, u3 can be in the [22, 25] range, and if u2 == 9, u3 can be in the [21, 27] range. The high limit is given by the 3i sequence, and the low limit is given by the requirement to build numbers up to 12 with 3 numbers, and up to 34 with 4 ones.
No code was used, but I think that way much quicker and less error prone. It is now possible to use Python to show that all numbers up to 100 can be constructed from one of those sets using the divide and conquer strategy.

How do I solve this: "Step forward and backward problem"?

The Problem Statement:
Sanjay is addicted to alcohol. Every night he drinks 4 bottles of vodka. He is going to his home. At first, he takes a step forward (which is 5m) but beacuse he is drunk, after his each step in forward direction, his body gets imbalanced and he takes a step backward (which is 3m).
Each step takes 1 min to complete. The distance from the bar to home is n metres. Calculate the time taken by him to reach his home.
Input Format:
single line containing one integer n.
Constraints:
0 <= n < 10^18
Output Format
single integer describing the time taken by him to reach home.
from math import *
n = int(input())
x = 0
m = 0
n = n % 1000000007
n = n % 1000000007
while x < n:
x += 5
m += 1
if x >= n:
break
x -= 3
m += 1
print(m)
But the time limit is exceeding in the last test case i.e. for n = 10^18 like numbers
Sample Input 0
11
Sample Output 0
7

The time taken is simply n/2 * 2
He advances 2 meters each "cycle" 5 forward 3 back
So we see how many "cycles" go into n (n / 2m) this will result
In the number of "cycles" taken to reach his house
Then we simply multiply by the amount of time taken per cycle (2 minutes)
to get the total time taken (t = n/2 * 2).

Try reducing the problem. Let time_taken(dist) be the function that tells us how long it takes to get home. Then the following hold:
time_taken(1) == 1
time_taken(2) == 1
time_taken(3) == 1
time_taken(4) == 1
time_taken(5) == 1
time_taken(6) == 1 * 2 + time_taken(4) (since 5-3 = 2)
== 1 * 2 + 1
time_taken(7) == 1 * 2 + time_taken(5)
== 1 * 2 + 1
time_taken(11) == 1 * 2 + time_taken(9)
== 2 * 2 + time_taken(7)
== 3 * 2 + time_taken(5)
== 3 * 2 + 1
time_taken(26) == 1 * 2 + time_taken(24)
== 2 * 2 + time_taken(22)
== ...
== 11 * 2 + time_taken(4)
== 11 * 2 + 1
if n > 5:
time_taken(n) == 1 * 2 + time_taken(n - 2)
== 2 * 2 + time_taken(n - 4)
== ...
== (formula here) * 2 + time_taken(4 or 5)

Python times table using while

Why in python a times table program doesn't work like this?
n = int(input("Type a number: "))
count = 10
while count < 0:
print(f"{count} x {n} = {n * count}")
count = count - 1
But this works correctly:
n = int(input("Type a number: "))
for i in range(1, 11):
print(f"{i} x {n} = {n * i}")
The result is (or should be), for example:
1 x 5 = 5
2 x 5 = 10
3 x 5 = 15
4 x 5 = 20
5 x 5 = 25
6 x 5 = 30
7 x 5 = 35
8 x 5 = 40
9 x 5 = 45
10 x 5 = 50
As a beginner I need to understand... and is it possible to use while in that situation?
This is python 3.x

You want to start count from 1 so that your loop increments rather than decrements. Your while loop conditional check is also incorrect as count is never less than 0. Try this:
n = int(input("Type a number: "))
count = 1
while count <= 10:
print(f"{count} x {n} = {n * count}")
count = count + 1

You need to set the condition in the while to be true to go through the loop. Thus we will say do while count is greater than or equal to zero.
n = int(input("Type a number: "))
count = 10
while count >= 0:
print(f"{count} x {n} = {n * count}")
count = count - 1