Good day.
I am looking for a closed-form formula for this kind of integer triangle:
$$ 1 \\ 2 \quad 2 \\ 2 \quad 6 \quad 4 \\ 2 \quad 10 \quad 16 \quad 8 \\ 2 \quad 14 \quad 36 \quad 40 \quad 16 \\ 2 \quad 18 \quad 64 \quad 112 \quad 96 \quad 32 \\ \dots $$
The triangle is filled from top to bottom. The initial state is "1" at the top and "2" at the begening of the second row (exception to the rule which follows). Further the rule applies:
$$ a(r,c)=2*a(r-1,c-1)+a(r-1,c) $$
or by words: two times upper left element plus (one time) upper right element.
Thank you.
Solved: As suggested by Gerry Myerson (with small modification), one divides first element of each row by 1, second by 2, third by 4, fourth by 8, fifth by 16 and $i$-th by $2^{(i-1)}$.
Then one arrives to the so-called "(2,1)-Pascal triangle" and the information can be found on internet: