-2

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:

  1. 113
  2. 7253
  3. 464213
  4. 29709653
  5. 1901417813
  6. 121690740053
  7. 7788207363413
  8. 498445271258453
  9. 31900497360541013
  10. 2041631831074624853
  11. 130664437188775990613
  12. 8362523980081663399253
  13. 535201534725226457552213
  14. 34252898222414493283341653
  15. 2192185486234527570133865813
  16. 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.

Tsar Asterov XVII
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    $a_{n+1}=a_n*64+21$ ? – Collag3n Apr 08 '23 at 16:17
  • Can it be written as a summation perhaps? Just to kind of keep with consistency of the formulae I'm working on – Tsar Asterov XVII Apr 08 '23 at 16:22
  • $64^n\cdot 113 + \frac{64^n-1}{3}$ and the last part is $21\cdot\sum64^i$ if you want a sum – Collag3n Apr 08 '23 at 16:27
  • @Collag3n A bit confused, can you show it in the format of summation that was in the question? That would be greatly appreciated! – Tsar Asterov XVII Apr 08 '23 at 16:53
  • There are many ways to write this as a sum – Collag3n Apr 08 '23 at 16:57
  • Hmm, late question: did you look for a method, how to develop the recursion-formula from the given values? (Even for the general/similar case?) For seuqences of this type it is not difficult to find the actual recursion-formula. – Gottfried Helms Apr 12 '23 at 18:57
  • @GottfriedHelms I had actually done it for Branch 1 of all the patterns I had, but due to lack of, well, experience with summations, I couldn't really figure out the formula for Branch-2 (the one I was asking about here), even after 2 weeks, so I decided to ask here. – Tsar Asterov XVII Apr 14 '23 at 02:56
  • Ah, well. Didn't want to keep you busy for nothing. I'd just done one old exercise in getting the parameters for a recursive sequence and thought you might be interested in such a method. Happy you've got your sought answer! – Gottfried Helms Apr 14 '23 at 04:53
  • @GottfriedHelms Would you be so kind as to share me the exercise? I would like a good challenge in pattern finding. As for the answer, Well, I might come back later for a new larger pattern, but I think I might have gotten what to do to find the answer, might, but we'll see. Again, thanks for yours and Collag3n's help. – Tsar Asterov XVII Apr 14 '23 at 09:58

1 Answers1

3

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 $z=3n, n\geq 0$

Collag3n
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  • My apologies, as I have given a wrong formula for the summation in the question. I have corrected it now. Could you perhaps give me the right formula now that I have corrected it? – Tsar Asterov XVII Apr 08 '23 at 17:07
  • @Aster17, what is $\mathbb W$ and what are your branch-1 values (first few are enough) ? – Collag3n Apr 08 '23 at 17:14
  • W just signifies that n can be any whole number, and some of the branch-1 values are 53, 3413, 218453, 894784843. I've checked, the formula in the question does work out for Branch-1. – Tsar Asterov XVII Apr 08 '23 at 17:26
  • @Aster17, see my update – Collag3n Apr 08 '23 at 18:19
  • thanks a lot! The formula you gave works out perfectly! I also slightly altered the formula to get the right values for other variations I had in this project of mine, and it works perfect! Thank you, I highly appreciate your help! – Tsar Asterov XVII Apr 09 '23 at 03:48
  • @Aster17 - you might click on the "accept" mark at the answers head to close the case. – Gottfried Helms Apr 09 '23 at 12:43
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    @GottfriedHelms I know, thank you, I just got busy with testing the formulae and forgot. Will do so now. – Tsar Asterov XVII Apr 09 '23 at 14:58