Here is a Python solution for the problem:

```python
def tribonacci(n):
if n==1:
return 0
elif n==2 or n==3:
return 1
else:
return tribonacci(n-1) + tribonacci(n-2) + tribonacci(n-3)

n = int(input())
for i in range(1, n+1):
print(tribonacci(i), end=' ')
```

This program works by defining a recursive function `tribonacci(n)` that calculates the nth number in the Tribonacci series. The base cases are defined for n=1, n=2, and n=3. For n3, the function calls itself with the arguments n-1, n-2, and n-3, and returns their sum.

The main part of the program reads an integer n from the input, then prints the first n numbers in the Tribonacci series by calling the `tribonacci(i)` function in a loop from 1 to n. The numbers are printed on the same line, separated by spaces.

Please note that this solution has a high time complexity due to the repeated calculation of the same Tribonacci numbers. For larger inputs, a more efficient solution would be to use dynamic programming to store the calculated Tribonacci numbers and reuse them when needed.

Question

Here is a Python solution for the problem:

```python
def tribonacci(n):
    if n==1:
        return 0
    elif n==2 or n==3:
        return 1
    else:
        return tribonacci(n-1) + tribonacci(n-2) + tribonacci(n-3)

n = int(input())
for i in range(1, n+1):
    print(tribonacci(i), end=' ')
```

This program works by defining a recursive function `tribonacci(n)` that calculates the nth number in the Tribonacci series. The base cases are defined for n=1, n=2, and n=3. For n>3, the function calls itself with the arguments n-1, n-2, and n-3, and returns their sum.

The main part of the program reads an integer n from the input, then prints the first n numbers in the Tribonacci series by calling the `tribonacci(i)` function in a loop from 1 to n. The numbers are printed on the same line, separated by spaces.

Please note that this solution has a high time complexity due to the repeated calculation of the same Tribonacci numbers. For larger inputs, a more efficient solution would be to use dynamic programming to store the calculated Tribonacci numbers and reuse them when needed.

Knowee AI · Accepted Answer

Here is a Python solution for the problem:

```python
def tribonacci(n):
    if n==1:
        return 0
    elif n==2 or n==3:
        return 1
    else:
        return tribonacci(n-1) + tribonacci(n-2) + tribonacci(n-3)

n = int(input())
for i in range(1, n+1):
    print(tribonacci(i), end=' ')
```

This program works by defining a recursive function `tribonacci(n)` that calculates the nth number in the Tribonacci series. The base cases are defined for n=1, n=2, and n=3. For n>3, the function calls itself with the arguments n-1, n-2, and n-3, and returns their sum.

The main part of the program reads an integer n from the input, then prints the first n numbers in the Tribonacci series by calling the `tribonacci(i)` function in a loop from 1 to n. The numbers are printed on the same line, separated by spaces.

Please note that this solution has a high time complexity due to the repeated calculation of the same Tribonacci numbers. For larger inputs, a more efficient solution would be to use dynamic programming to store the calculated Tribonacci numbers and reuse them when needed.

Question

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