[BEGINNER] reccurence! and sequence!

[ag0aep6g via Digitalmars-d-learn]

On 06/26/2017 11:51 AM, helxi wrote:
>> auto tri = sequence!((a,n) => n*(n+1)/2)();
> 
> /** okay, it's a triangular number array
> * I understand n is the index number, the nth term
> * However where does this 'a' go?
> */

`a` is a tuple of the run-time arguments you pass to `sequence`. In this 
example, no arguments are passed (empty parens at the end of the call), 
so `a` is empty.

>> auto odds = sequence!("a[0] + n * a[1]")(1, 2);
> 
> /** okay, this is a range of odd numbers
> / where and how do I plug (1, 2) into ""a[0] + n * a[1]"

a[0] = 1
a[1] = 2

>> sequence!("n+2")(1).take(10).writeln;
> //[2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

`a` isn't used in the string lambda, and it's not considered an element 
of the range. So n starts at 0 and this just prints 0+2, 1+2, 2+2, etc.

>> recurrence!("n+2")(1).take(10).writeln;
> //[1, 3, 4, 5, 6, 7, 8, 9, 10, 11]

`a` is still not used in the string lambda, but `recurrence` uses the 
values in `a` as the first elements of the range. `n` is incremented 
accordingly (to 1), so this prints:

1 = a[0],
3 = (n = 1) + 2,
4 = (n = 2) + 2,
etc.

Another difference between `sequence` and `recurrence` is that `a` 
always refers to the same initial value(s) in `sequence`, while in 
`recurrence` it gets updated and refers to the previous element(s) of 
the range:

----
sequence!((a, n) => a[0] + 1)(1).take(10).writeln;
     // [2, 2, 2, 2, 2, 2, 2, 2, 2, 2]
     // because a[0] is always 1

recurrence!((a, n) => a[0] + 1)(1).take(10).writeln;
     // [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
     // because a[0] refers to the previous value
----