Question 107351
n=1,  0
n=2,  3
n=3,  8
n=4, 15
n=5, 24
If I add 1 to each output I have a perfect square. 
The function is then is a perfect square minus 1. 
{{{f(n)=n^2-1}}}
The other pattern is that the difference between two output numbers is a consecutive odd number.
{{{f(n+1)-f(n)=((n+1)^2-1)-(n^2-1))}}}
{{{f(n+1)-f(n)=((n^2+2n+1-1)-n^2+1)}}}
{{{f(n+1)-f(n)= 2n+1}}}
That's just interesting more than anything else. 
Now back to the problem. 
The expression for the nth term is then, as above,
{{{f(n)=n^2-1}}}
and the value for the 43rd term is then,
{{{f(n)=n^2-1}}}
{{{f(43)=43^2-1}}}
{{{f(43)=1848}}}