The clue is that the difference between consecutive terms increases by 2 each time, starting at 3. So we have the progressive difference between the first term and each of the others is 7-4=3, 12-4=8, 19-4=15.
3, 8 and 15 are each one less than the squares of the numbers 2, 3 and 4, because 2×2-1=3, 3×3-1=8, 4×4-1=15.
These correspond to the differences for n=1, 2, 3.
So we can write 7=4+(1+1)²-1, 12=4+(2+1)²-1, 19=4+(3+1)²-1.
So the (n+1)th term=4+(n+1)²-1, and the nth term is 4+(n-1+1)²-1=3+n².
Let’s test the formula by putting n=1 for the 1st term: 3+1²=4.
2nd term is 3+2²=7; 3rd term is 3+3²=12; 4th term is 3+4²=19. The formula works! So the nth term is 3+n² or n²+3.