20 Questions Type Puzzel

[TheRealJoeFish]

How so?

Yes, how so?

The intervals between the numbers are ascending primes times alternating ascending powers of 2 and 3.

6-2=4=2x2
15-6=9=3x3
35-15=20=5x2x2
77-35=42=7x3x3
165-77=88=11x2x2x2
516-165=351=13x3x3x3

1 is the prime before 2 and any number to the power of zero is one. So the difference between 2 and the number before it would be one. Thus
2-1=1=1* (3 to the zero power). So the number in sequence before 2 would be 1.

7x3x3 = 63, not 42, so your 4th step doesn't hold. But if we just fix that arithmetic error, we get:
2
2+(2*2*1) = 6
6+(3*3*1) = 15
15+(5*2*2) = 35
35+(7*3*2) = 77
77+(11*2*3) = 143
143+(13*3*3)= 260 (which breaks from the p_(n-1)*p_n pattern).

And then you can go

260+(17*2*4) = 396
396+(19*3*4) = 624.

This sequence would be defined as n_0 = 2, n_i = n_(i-1) + (p_i)*(2.5+(.5*(-1)^i))*(ceiling(i/2))

The point, I think, is that if you give a finite sequence portion of a sequence, there are an infinite number of sequences that can follow, if you're creative!

To bookend, the "other" solution, other than 143, was 165.

2
2*2+2 = 6
6*6+3 = 15
15*2+5 = 35
35*2+7 = 77
77*2+11 = 165

and so on!