Hard combinatorics question The origin of the coordinates is a pixel. After 1 second, it splits into two particles, one
shifts to the left and the other to the right. This process is repeated every second, and the two particles at the same point are mutually destroyed (so that, for example,
two seconds later, two particles remain.) How many particles will there be in $n$ seconds?
I calculated first values using programm ($1,$
$2,$ $2,$ $4,$ $2,$ $4,$ $4,$ $8,$ $2,$ $4,$ $4,$ $8,$ $4,$ $8,$ $8,$ $16,$ $2,$ $4,$ $4,$ $8,$ $4,$ $8,$ $8,$ $16,$ $4,
$ $8,$ $8,$ $16,$ $8,$ $16,$ $16,$ $32,$ $2,$ $4,$ $4,$ $8,$ $4,$ $8,$ $8,$ $16,$ $4,$ $8,$ $8,$ $16,$ $8,$ $16,$ $16,$
$32,$ $4,$ $8,$ $8,$ $16,$ $8,$ $16,$ $16,$ $32,$ $8,$ $16,$ $16,$ $32,$ $16,$ $32,$ $32,$ $64,$ $2,$ $4,$ $4,$ $8,$ $4,
$ $8,$ $8,$ $16,$ $4,$ $8,$ $8,$ $16,$ $8,$ $16,$ $16,$ $32,$) but I still cant find any formula.

This is called Gould's sequence, or Dress' sequence, or Glaisher's sequence; it appears in the OEIS.  The $n$-th term is the number of odd entries entries in row $n$ of Pascal's triangle.  The specific operation in OP is Rule 90 for cellular automata, and we are counting how many $1$'s there are in every stage.  Much more is available in the OEIS link above.
