Can you find a function F with the following characteristics ?

Question

F(1) = 3
F(3) = 9
F(6) = 27
F(10) = 81
F(15) = 243
etc...
and the pattern repeats to infinity.

Derivations add to the worthiness of the answer.

Thanks Pascal.
The position of best answer has been filled folks. Others need not apply.

Answer 1

F(x) = 3^(√(2x+1/4) - 1/2)

The pattern is that the nth triangular number is mapped to 3^n. The nth triangular number is n(n+1)/2, so evaluating F at that point yields:

F(n(n+1)/2)
3^(√(2n(n+1)/2+1/4) - 1/2)
3^(√(n(n+1)+1/4) - 1/2)
3^(√(n²+n+1/4) - 1/2)
3^(√((n+1/2)²) - 1/2)
3^(n+1/2 - 1/2)
3^n

eta: the derivation is simple: we want a function that maps n(n+1)/2 to 3^n. Given x=n(n+1)/2, we have:

x = n(n+1)/2
2x = n²+n
2x+1/4 = n²+n+1/4 = (n+1/2)²
n+1/2=√(2x+1/4)
n = √(2x+1/4) - 1/2
3^n = 3^(√(2x+1/4) - 1/2)

Which is the function I just gave.