I've heard (and believed even without proof) that given any finite sequence there is more than one formula for which the same first inputs give the same first outputs.

Given that:

f(1)=1 f(2)=3 f(3)=6

One possible function is f(x) = x(x+1)/2.

So, I tried to find another polynomial function for which f(1)=1, f(2)=3 and f(3)=6, but for which f(4) is not 10, but got no success. So I ask: Can it be done? (please provide an example :))

I find this question interesting because often we can make a conjecture by finite induction which is wrong (like the proposed "prime-producing polynomials" which always start to fail after some time :P)

  • $\begingroup$ The next number of the sequence is 42, obviously. $\endgroup$ – Did Feb 2 '14 at 16:47
  • $\begingroup$ @did do you mean becuase it is the meaning of life? $\endgroup$ – Jorge Fernández Hidalgo Feb 2 '14 at 17:15
  • $\begingroup$ @user4140 Afraid I do. $\endgroup$ – Did Feb 2 '14 at 17:22
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    $\begingroup$ There is also $f(x)=\begin{cases}1&x=1\\3&x=2\\6&x=3\\47.9027347&otherwise \end{cases}$ which is a fine formula $\endgroup$ – Ross Millikan Mar 5 '14 at 0:09

Let us take the simplest case after a quadratic. Then let $$f(x)=a+b x+c x^2+d x^3$$ So we can build three equations from which we can eliminate $b$,$c$ and $d$ as a function of $a$. As a result, we have
$$b=\frac{1}{6} (3-11 a)$$ $$c=a+\frac{1}{2}$$ $$d=-\frac{a}{6}$$

You can now select any number for $a$. The cubic will always go through the three data points and $f(4)=10-a$

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  • $\begingroup$ For the lazy, WolframAlpha can fit that cubic for you ;) $\endgroup$ – David H Feb 2 '14 at 16:42
  • $\begingroup$ Also, in case you're interested, the cubic corresponding to $f(4)=k$ is apparently $f(x)=\frac{x(x+1)}{2}+(10-k)(1-\frac{11}{6}x+x^2-\frac16x^3)$. $\endgroup$ – David H Feb 2 '14 at 16:45

Let $a$ be any real number.

Then, by Lagrange interpolation formula the function:


satisfies $f(1)=1, f(2)=3, f(3)=6, f(4)=a$.

So anything is the right answer...

Just for fun: setting $a=cat$ you get the pattern, $1,2,3,cat$.

P.S. This is not the only formula which satisfies these conditions. If $g$ is any function, then

$$f(x)+g(x)(x-1)(x-2)(x-3)(x-4)$$ also satisfies the pattern $1,3,6,a$.

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Yes, you are correct. There are indeed more than a sequence that starts with $1,3,6$. The function you input is the triangular number function. OEIS gives other types of sequences too!

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Theorem: For any finite sequence of numbers $(a_1,a_2\dots a_n)$ there is at least 1 polynomial such that $f(i)=a_i$ for all $i$ between $1$ and $n$.

Proof: First we prove for all i there is polynomial suc that p(i)=a_i and for other integers betwenn 1 and n it is 0.


$g_i(x)=(x+1)(x+2)\dots(x-(i-1))(x-(i+1)\dots(x-n)=k$ when $x=i$ and 0 for other integers between 1 and n. Therefore the function

$p_i(x)=\dfrac{(x+1)(x+2)\dots(x-(i-1))(x-(i+1)\dots(x-n)}{k}\cdot a_i=a_i$when $x=i$ and $0$ for other integers between 1 and n.

Now check the polynomial $p(x)=p_1(x)+p_2(x)\dots+p_n(x)$ fits the sequence perfectly.

Theorem: For any list $(a_1,a_2\dots a_n)$ there exist infinite polynomials such that $f(i)=a_i$ for all $i$ between $1$ and $n$.

Proof: by the above theorem there are polynomials satisfying $(a_1,a_2\dots a_n,k)$ for any real k. Clearly all those polynomials also satisfy $(a_1,a_2\dots,a_n,k)$ and they are uncountably many.

This also shows the next number in the sequence could be any number, and there would still be a "polynomial formula" to back you up

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  • $\begingroup$ There is a mistake in your proof of Lagrange Iterpolation. $k$ is not a constant, it depends on $i$, thus $k$ and $p_i$ are not polynomials. You need to divide by the right constant and multiply by $a_i$. $\endgroup$ – N. S. Feb 2 '14 at 16:39
  • $\begingroup$ I'm sorry, I have no Idea what lagrange interpolation is, I am building n different polynomials $p_i$ and then I'm adding them to get a new one p(x) that satisfies everything.@N.S. so why is $p_i$ not a polynomial? $\endgroup$ – Jorge Fernández Hidalgo Feb 2 '14 at 16:41
  • $\begingroup$ Oh, I got you, got it. $\endgroup$ – Jorge Fernández Hidalgo Feb 2 '14 at 16:42
  • $\begingroup$ You have the right idea, is just that $k$ is not constant. Set $p_i(x)=a_i\frac{g_i(x)}{g_i(i)}$ and it works.. And you rediscover Lagrange interpolation ;) $\endgroup$ – N. S. Feb 2 '14 at 16:43
  • $\begingroup$ Got it, thanks.Is it fine now? $\endgroup$ – Jorge Fernández Hidalgo Feb 2 '14 at 16:46

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