Simple C++ recursion code isn't working how I expected it to

Question

#include <iostream>
using namespace std;

//a function using recursion to add up numbers in a sequence
int add(int a) {
if (a < 3) {
return a + add(a+1);
} else {
return 0;
// return 0 because otherwise it adds whatever other number to the total
}

}

//how I expected it to run
//1 + 2 (a = 1)
//1 + 2 + 3(a = 2)
//1 + 2 + 3 + 4(a = 3)

int main()
{
cout << add(1);
}

Assigning 1 as a parameter returns 3, it seems as though the function only runs once where 1 is the parameter(return 1 + 2)
Why isn't the function running a second time with 2 as a parameter?(where the total would be return 3(total so far) + 3(current parameter + 1)
If I misunderstood something please let me know

edit
The code returns the desired output if i change the if statement to (a < 4), why?

Answer 1

if you just want to do this
1
2
3
4
5
6
7
8
etc..
then you dont need to say a + add(a+1) . you can just say add(a+1). and instead of returning 0 when you hit the limit, return a

Answer 2

it gets added. otherwise a would not increase in value

Answer 3

huh, fair enough.. but that makes it more complicated...but ok.. still addition, not multiplication (see code)

Answer 4

https://code.sololearn.com/cwDxz9xvTF6W/?ref=app

Answer 5

This is what happen :
add(1)
1<3 true
return 1+add(2) <--1+2=3-- //// output 3
add(2) |
2<3 true |
return 2+add(3) <-2+0=2--
add(3) |
3<3 false return 0 >--0---
Hope you'll understand, if you don't understand, I can try with sentence 😂
Ask me if you want anything !

Answer 6

I may be messing up a bit, thanks for the answers
but I have a slight feeling, true or not that the (+ add(a+1) isn't added to the total but only executed, so my code is run as:
return a (a = 1)
return a (a = 2)
= 3

Answer 7

the a + add(a+!) was me making the lesson in recursion more simple by adding instead of multiplying, also I removed the factorial element completely