What is nth term of sequence 1,2,5,14,42,132,...?

Question

Is there a closed form exist for the following sequence to find the nth term?

1, 2, 5, 14, 42, 132, ...

I tried comparing the differences of subsequent terms which are - 1, 3, 9, 28, 90, which does not seem to follow any pattern.

Any help is much appreciated. Thanks.

Answer 1

There are several possibilities. One of them is that these are the Catalan numbers:

$$ 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440,\ldots $$

Answer 2

Assuming you created those first entries while working on a certain problem, chances are these are the Catalan numbers. However, you should of course go back to your problem and verify that this hypothesis is indeed correct, for there are an infinite number of formulas that create the beginning of your sequence. Here's one:

$$f(n)=1*(1-min(|n-1|,1))+2*(1-min(|n-2|,1))+5*(1-min(|n-3|,1))+14*(1-min(|n-4|,1))+42*(1-min(|n-5|,1))+132*(1-min(|n-6|,1))$$

.. which would generate: $1,2,5,14,42,132,0,0,0,0,...$ (all $0$ for $n \ge 7$)