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I would like to ask if anyone could please suggest me accurate ( ! ) sequence number announcer tool. It can be web based application, working within website, or desktop software. It can be free of charge (preferable) or paid. Would be willing to pay for it but it has to announce the number correctly. There are no math formulas behind the sequence. Numbers are ordered from left to right - the first number is the oldest one and the last number is the most recent one. I would like to use it mainly for binary digits (0 or 1 only) but sometimes also for ''normal'' numbers. So far the only website I know is:

http://oeis.org

But everything is announced wrong. I am testing like 50+ situations and being aware of the most recent number (the one listed on the most right side so the last one). Obviously I am testing the said website without typing in the most recent number, even if I know it. But it is all the time giving me wrong announcement.

I am Not asking for assistance regarding accurately announcing first next number in the sequence but, to clarify again and to avoid misunderstanding, I am asking for suggestion about which other website/tool/software could I use for this. So it will tell me which number will occur next.

Your reply would be highly appreciated, thank you in advance.

Puppy00
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    This is not at all clear, at least not to me. It appears that you are asking for a program that, when given the first $n$ terms of a sequence, outputs the $(n+1)^{st}$. Clearly no such program exists in general. There are polynomials which interpolate every $n+1$ data points so we can justify any "next entry" in a polynomial manner (though of course the desired sequence might not even be polynomial). Or have I misunderstood the question? – lulu Dec 10 '17 at 21:30
  • Can you clarify your question? As it stands, it is not clear what you are asking. If you can't clarify it, I think it should be closed. – lulu Dec 10 '17 at 22:16
  • Voting to close the question, as it is unclear what you are asking. If you can, please edit for clarity. – lulu Dec 10 '17 at 22:35
  • To further clarify, I will give examples and for each ''X'', I am searching for the tool/website/software which will accurately announce what first next number, in this case number X is.

    first example: 1,2,2,2,1,2,1, X (result: X is 4) second example: 2,0,0,2,0,0,0, X (result: X is 0) third example: 1,0,1,1,2,1,4,4,3,4, X (result: X is 3)

    So i am looking for tool which accurately announces first next digit. There are no math formulas behind.

     – Puppy00 Dec 10 '17 at 23:05
  • How do you get X's in your examples? – kingW3 Dec 10 '17 at 23:50
  • Those examples look entirely random. – lulu Dec 11 '17 at 00:38
  • Yes they are random. Please kindly see my four replies to Tiwa whos answer I am still waiting. – Puppy00 Dec 11 '17 at 09:09

1 Answers1

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Claim. If you give me an arbitrary sequence, I can produce a random number and prove that it is, in fact, the next term of the sequence. Indeed, suppose your terms are $a_1, a_2, a_3, ..., a_n$. Then I choose a random number $X$. We can construct a polynomial of degree $n$ that satisfies the sequence $a_1, a_2, ..., a_n, X$; that is, a polynomial $p$ such that $p(k) = a_k$.

Proof sketch. First, we fit the points $(1, a_1)$ and $(2, a_2)$ with a line:

$$f_1(x) = (x-1)(a_2-a_1) + a_2$$

Now consider the quadratic function

$$f_2(x) = Q_1 \cdot (x-1)(x-2) + f_1(x)$$

Observe that regardless of the value of $Q_1$, $f(1) = a_1$ and $f(2) = a_2$, as needed. Consider $$f_2(3) = 2Q_1 + f_1(3)$$ which we need to equal $a_3$. Then let $$Q_1 = \frac{a_3 - f_1(3)}2$$ and so $f_2(x)$ satisfies the sequence for $a_1, a_2, a_3$.

Now consider the cubic function

$$f_3(x) = Q_2 \cdot (x-1)(x-2)(x-3) + f_2(x)$$

We can solve for $Q_2$ such that $f_3(4) = a_4$:

$$Q_2 = \frac{a_4 - f_2(4)}6$$

Notice a pattern? We can do this indefinitely for any sequence, which means regardless of the $X$ chosen, we can create a function that gives the next term in the sequence $a_1, a_2, a_3, ..., a_n, X$.

More succinctly: Suppose your terms are $a_1, a_2, a_3, ..., a_n$. Then I choose a random number $X$. Then the polynomial that satisfies the sequence $a_1, a_2, ..., a_n, X$ is $f_n(x)$, where $f$ satisfies the recurrence relation

$$f_n(x) = \frac{a_{n+1} - f_{n-1}(n+1)}{n!} \prod_{k=1}^n (x-k) + f_{n-1}(x)$$ with the basis $$f_1(x) = (x-1)(a_2-a_1) + a_2$$

Because the next term in an arbitrary sequence can be proven to be arbitrary, a more meaningful request might be to find the next entry in a well-known sequence of numbers (e.g. 1, 1, 2, 3, 5, 8, X), and OEIS is almost certainly the best tool for this.

actinidia
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  • Tiwa thank you very much for your time spent for me but I am afraid your idea won't work. Let me try to explain further what I work with: In MS Office Excel, I have particular number of columns with 10 lines. I am separating the columns. Each column is one digit number. I am separating 10 lines in three subgroups. First subgroup of lines consist of first four lines. Second subgroup of lines consist of second four lines. Third consist of remaining two lines. Each number (consisting of multiple digits, e.g. 0221002121) where i want to announce next digit, belong to 1 subgroup. (next reply) – Puppy00 Dec 11 '17 at 00:15
  • What i do is that i count in each of the columns how many values (note: only first four lines located in this same column!!!) are larger than 0. Then i write down what i count. E.g. lets say in L column i have in four different cells of this column (first subgroup consisting of first four cells) 0,2,0,0. I would write down 1. Then in next column i have 1,1,1,0. I would write down 3 and so on. Now in this very small example there are for my explanation only two columns. My produced number is 13. I want to know what will be next so what will be on the right side of 3. (next reply) – Puppy00 Dec 11 '17 at 00:15
  • I have over 30 columns but always one number (in my example the number is 13) consist of all the columns (in the same line) but for only one subgroup so three subgroups in total. One number (multiple digits: two digits in said example for two columns) will always belong to only one subgroup. First subgroup (first number) will always consist of first four lines. Second, second four lines. (next reply) – Puppy00 Dec 11 '17 at 00:16
  • Third, remaining two lines. I am adding new columns twice per week and values are random. This means 100% accuracy is impossible (i think so) but i would like to come as close as possible to it. My ULTIMATE GOAL is to announce how many numbers, larger than 0, will each of those three subgroups (three different numbers!) have. – Puppy00 Dec 11 '17 at 00:16