Difficult Sequence
- Thread starter artyom_p
- Start date
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,962
Posting Guidelines (Summary)
Welcome to our tutoring boards! :) This page summarizes main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to help...
www.freemathhelp.com
Yeah, I of course tried interpolation polynomials, but this sequence is from combinations of cards so I think it grows with factorial speed. This is why I can't calculate more terms (program worked on 3-rd term about 44 minutes). May be we can approximate the required function with this polynomial, but I need an accurate formula.
Imagine that you have \(\displaystyle 4\) equations that have \(\displaystyle 4\) unknowns.
\(\displaystyle an^3 + bn^2 + cn + d = 1\)
\(\displaystyle an^3 + bn^2 + cn + d = 36\)
\(\displaystyle an^3 + bn^2 + cn + d = 11472\)
\(\displaystyle an^3 + bn^2 + cn + d = 15349848\)
Say, when \(\displaystyle n = 0\), you want to get \(\displaystyle 1\), so the first equation will be
\(\displaystyle a0^3 + b0^2 + c0 + d = d = 1\)
Repeat for other equations when \(\displaystyle n = 1,2,3\) and solve for \(\displaystyle a, b, c\). You have already found \(\displaystyle d\).
\(\displaystyle an^3 + bn^2 + cn + d = 1\)
\(\displaystyle an^3 + bn^2 + cn + d = 36\)
\(\displaystyle an^3 + bn^2 + cn + d = 11472\)
\(\displaystyle an^3 + bn^2 + cn + d = 15349848\)
Say, when \(\displaystyle n = 0\), you want to get \(\displaystyle 1\), so the first equation will be
\(\displaystyle a0^3 + b0^2 + c0 + d = d = 1\)
Repeat for other equations when \(\displaystyle n = 1,2,3\) and solve for \(\displaystyle a, b, c\). You have already found \(\displaystyle d\).
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,619
As the others have indicated, there are many formulas that will give any four terms you choose for a sequence. You need to provide more information to narrow it down; where does this come from, and what connects it to combinatorics?
But it doesn't look like it will be anything known:
This is my own research about one card game, I need number of combinations of cards that works according special rules. I can't get the formula of such combinations so I tried to simulate the situation with a program. But it doesn't give me even a 5th term, it needs so much time for it. About polynomials: I know that there are some given 5th member, but it's not a simple polynomial sequence, it grows much faster (I can check the result of 5th and more terms, polynomials give much less than necessary, for example 5th number is about 61 million).
BigBeachBanana
Senior Member
- Joined
- Nov 19, 2021
- Messages
- 2,274
How did you come up with the first 4 numbers?
Why do you think the 5th number is around 61M?
What are the special rules you are modelling for the card game?
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,619
If even one of the four numbers is off by one, the sequence will not be recognizable. You'd need a very accurate simulation to be able to reverse-engineer the formula.
So I'd have a lot more hope for actually working out the numbers (including letting a computer count all the possibilities, which is not what I would call a simulation). Alternatively, depending on your goal, merely approximating the counts by a formula might be reasonable.
But you're right,polynomials are not relevant here; that was just an illustration of the problem. The same could probably be done using formulas with factorials, though not as easily.
Program gave me the first 4 terms.
I'm sorry, the 5th number is not about 61M, it's just a value of polynomial above for x = 4. A true value should be much more.
Okay, I'll try to explain the rules. It's kinda games named "Patience" in my country, when you don't need any actions except laying out the cards. The rules: when you lay out a current card you check if the value of this card is equal to any of cards on the table. If it is, you should erase these cards and cards between them from the table. For example: 6 7 8 6. When I put the second 6 I remove 6 7 8 and 6 from the table and go on. In the end of deck If there is no one card left on the table, it's a win. I'm studying in the universityon the math specialty so I wondered what is the probability of winning this game. Numbers that I wrote in post are the numbers of winning combinations of n*4 cards. For example 2 means all winning combinations of 2 fours of cards: 6 6 6 6 7 7 7 7. It equals to 36.
Thank you, I made sure this is non-recognizable sequence, so I'll work on it for myself, finding some patterns or coincidences may be)