So, I was looking at the Collatz conjecture, and I thought of trying to reverse engineer the patterns in a certain sense, forming branches and trees. I figured it our for Branch-1, the formula, but Branch-2 is giving me trouble. It would be greatly helpful for me if one could help me find a formula to generate the values in the below table:
- 113
- 7253
- 464213
- 29709653
- 1901417813
- 121690740053
- 7788207363413
- 498445271258453
- 31900497360541013
- 2041631831074624853
- 130664437188775990613
- 8362523980081663399253
- 535201534725226457552213
- 34252898222414493283341653
- 2192185486234527570133865813
- 140299871119009764488567412053
I'll also share the formula for Branch-1 I got for this type of sequence.
$$3 + \sum_{n=0}^z 10 \cdot 4^k$$ $$z = (3n + 3) - 2$$ $$(n \in \mathbb W)$$
So something around these lines would be extremely helpful! I appreciate all the help, and I can try and calculate more values if needed.
something that fits your previous sum: $113+\sum 7140\cdot 64^k$
EDIT: Your first sum can be written $53+\sum 3360\cdot 64^k$ and is also defined by $a_{n+1}=a_n*64+21$ but the way you want to write the second sum cannot be done the way you want (the $4^k$ sum comes from the sequence $5,341,21845,...$ that you see in Branch-1, but the sequence $11, 725, 46421,...$ which you see in Branch-2 does not produce this nice form)
The best I can do is $3+100\cdot4^z+\sum\limits_{k=0}^z10\cdot 4^k$ with $z=3n, n\geq 0$