This is (probably) a series with a recurrence relation:
a(n)=a(n-3)+a(n-2)+a(n-1)
Try
0=-1+0+1
1=0+1+0
2=1+0+1
...
37=6+11+20
Some call this tribonacci numbers, from
Fibonacci numbers where
a(n)=a(n-2)+a(n-1)
-1,0,1,0,1,2,3,6,11,20,37
a(n)=a(n-3)+a(n-2)+a(n-1)
Try
0=-1+0+1
1=0+1+0
2=1+0+1
...
37=6+11+20
Some call this tribonacci numbers, from
Fibonacci numbers where
a(n)=a(n-2)+a(n-1)
2+3+6=11, ...
So the next number is:
11+20+37 = 68
20+37+68 = ?
...
Looking at the given sequence: -1, 0, 1, 0, 1, 2, 3, 6, 11, 20, 37
We observe that the numbers seem to switch between incrementing and decrementing. Let's break down the sequence and identify the pattern:
Step 1: Starting from -1, we see an increment of 1 in the next number, resulting in 0.
Step 2: The next number decreases by 1, resulting in -1.
Step 3: Now, the pattern seems to repeat with an increment of 1, resulting in 0 again.
Step 4: The next number decreases by 1, resulting in -1.
So, we have identified the repeating pattern of incrementing and decrementing values: 1, -1, 1, -1.
Now, to find the next three numbers in the pattern, we continue this same pattern:
Step 5: Incrementing by 1 from -1, we get 0.
Step 6: The next number will be -1 since we continue with the decrementing pattern.
Step 7: Incrementing by 1 from -1, we get 0 again.
Therefore, the next three numbers in the pattern are: 0, -1, 0.
Hence, the pattern rule for the given sequence is an alternating pattern of incrementing by 1 and decrementing by 1.