Below is a introduction that contains some background to my question. The question is found at the bottom.

By calculating the eigenvalues of the matrix defined by the recurrence:

$\displaystyle T(n,1)=1, T(1,k)=1, n \geq k: -\sum\limits_{i=1}^{k-1} T(n-i,k), n<k: -\sum\limits_{i=1}^{n-1} T(k-i,n)$

mentioned in: Do these series converge to the von Mangoldt function? we get eigenvalues whose signs appear to be equal to a rearrangement of the Möbius function but whose magnitudes differ:

Example: The $6 \times 6$ determinant equal to zero:

$$\begin{vmatrix} 1-\lambda&1&1&1&1&1 \\ 1&-1-\lambda&1&-1&1&-1 \\ 1&1&-2-\lambda&1&1&-2 \\ 1&-1&1&-1-\lambda&1&-1 \\ 1&1&1&1&-4-\lambda&1 \\ 1&-1&-2&-1&1&2-\lambda \end{vmatrix}=0$$

gives eigenvalues:

{-5.2439, -3.4641, 3.4641, 2.5169, -2.2730, 0}

Compare this with the eigenvalues from the $7$x$7$ determinant:

{-6.8444, -5.2217, -3.4641, 3.4641, 3.0150, -1.9489, 0}

From Todd Timberlake's paper "Random numbers and random matrices: Quantum chaos meets number theory" (2005-2006) I learned that one needs to unfold the eigenvalues in order to plot their distribution. I did not understand the unfolding process so I asked Todd Timberlake via email how to do it for a finite list of eigenvalues (all in all $11$ eigenvalues in my mail). In the answer he said that the unfolding approximation method should not be applied to only $11$ eigenvalues.

From his paper I also read the following sentence:

"Furthermore, it indicates that if the prime numbers are eigenvalues of some quantum system, then the classical counterpart of that system should be integrable."

Putting the lambdas "$\lambda$" along the diagonal rarely seems to give integer values. Instead we can try putting them at all elements:

$$\begin{vmatrix} 1-\lambda&1-\lambda \\ 1-\lambda&-1-\lambda \end{vmatrix}=0$$

This gives the equation:

$$-2 + 2 \lambda = 0$$

and we see that $\lambda$ is integer.

Now this works up to a $4$x$4$ matrix or so, after that the $\lambda$:s become zero. But this is not the right thing to do as this is like taking the determinant of the original matrix. And because $\lambda$ is not an eigenvalue anymore we will instead of $\lambda$ use $x$.

Another option is to insert the $-x$ into the recurrence instead. Now this gives the sequence of real numbers I am going to ask a question about, but with negative signs. Because I want the sequence to be positive I therefore insert "$x$" into the recurrence which then becomes:

$\displaystyle T(n,1)=1, T(1,k)=1, n \geq k: x -\sum\limits_{i=1}^{k-1} T(n-i,k), n<k: x -\sum\limits_{i=1}^{n-1} T(k-i,n)$

This then gives a matrix that starts as this $6$x$6$ determinant equal to zero:

$$\begin{vmatrix}1&1&1&1&1&1 \\ 1&-1+x&1&-1+x&1&-1+x \\ 1&1&-2+x&1&1&-2+x \\ 1&-1+x&1&-1&1&-1+x \\ 1&1&1&1&-4+x&1 \\ 1&-1+x&-2+x&-1+x&1&2-2x \end{vmatrix} = 0$$

which results in the equation:

$\displaystyle 180 x - 306 x^2 + 184 x^3 - 46 x^4 + 4 x^5 = 0$

for which the $5$ solutions are:

$\displaystyle x=0,\; x=3/2,\; x=2,\; x=3,\; x=5$

The polynomials for the $1$ x $1$ to $n$ x $n$ determinants start:

$\displaystyle \begin{align*} &1=0\\ &-2+x=0\\ &6-5 x+x^2=0\\ &-6 x+5 x^2-x^3=0\\ &30 x-31 x^2+10 x^3-x^4=0\\ &180 x-306 x^2+184 x^3-46 x^4+4 x^5=0\\ &-1260 x+2322 x^2-1594 x^3+506 x^4-74 x^5+4 x^6=0\\ &1260 x^2-2322 x^3+1594 x^4-506 x^5+74 x^6-4 x^7=0\\ &-2520 x^3+4644 x^4-3188 x^5+1012 x^6-148 x^7+8 x^8=0\\ &-25200 x^3+61560 x^4-59744 x^5+29248 x^6-7552 x^7+968 x^8-48 x^9=0 \end{align*}$

The $n$:th polynomial appears to have $n-1$ solutions, when counting solutions equal to zero, and $n$:th polynomial also appears to have those $n-1$ solutions in common with the $n+1$:th polynomial. This allows us to order the solutions which then become the following infinite sequence $a_n$ starting:

$\displaystyle a_n = (), 2,3,0,5,\frac{3}{2},7,0,0,\frac{5}{3},11,0,13,\frac{7}{4},\frac{15}{7},0,17,0,19,0,\frac{7}{3},\frac{11}{6},23,0,0,\frac{13}{7},0,0,29,\frac{15}{11},31...$

The number of distinct primes dividing the number $n$, $\omega(n)$, appears to describe this sequence and there are similarities with the Möbius function. But I find it difficult to understand $\dfrac{15}{7}$, for example.

The Mathematica program for printing the matrix for the determinant with its polynomial and solutions is:

Clear[nn, t, n, k, M, x];
nn = 12; (*size of matrix*)
t[n_, 1] = 1;
t[1, k_] = 1;
t[n_, k_] := 
  t[n, k] = 
   If[n < k, 
    If[And[n > 1, k > 1], x - Sum[t[k - i, n], {i, 1, n - 1}], 0], 
    If[And[n > 1, k > 1], x - Sum[t[n - i, k], {i, 1, k - 1}], 0]];
M = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}];
Det[M] (*polynomial*)
Solve[Det[M] == 0, x] (*solutions to the polynomial*)

My question is: Are all the primes found among the solutions to the polynomials above? And can you describe the sequence of solutions $a_n$ using $\omega(n)$?

Edit: 16.9.2011

Plot of the 32:nd polynomial:

Plot of the 32:nd polynomial

Edit 2: 16.9.2011

The Mathematica code for the plot is:

Clear[nn, t, n, k, M, x];
nn = 32;(*size of matrix*)t[n_, 1] = 1;
t[1, k_] = 1;
t[n_, k_] := 
  t[n, k] = 
       If[n < k, 
    If[And[n > 1, k > 1], x - Sum[t[k - i, n], {i, 1, n - 1}], 0], 
    If[And[n > 1, k > 1], x - Sum[t[n - i, k], {i, 1, k - 1}], 0]];
M = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}];
Plot[{Log[Det[M]], -Log[-Det[M]]}, {x, 0, nn + 2}, Filling -> Axis]

Edit 18.11.2012: It appears that by using the formula by Wolfgang Schramm, one gets the terms of the sequence $a_n$ as solutions to much simpler polynomials that have the form:

$$\sum\limits_{k=1}^{k=n} T(n,k) \cdot \cos(-2 \pi \frac{k}{n}) = 0$$

But this way there is only one solution per polynomial compared to the determinant polynomial that gives n-1 consecutive terms of the sequence $a_n$ as its solutions.

  • $\begingroup$ "But I find it difficult to understand \frac{15}{7}, for example." Why $\frac{15}{7}$? Is there an explanation for $\frac{3}{2}$ or $\frac{7}{4}$ preceding it? (I am still trying to understand the question, so this clarification may not make any sense :-)) $\endgroup$ – Srivatsan Sep 13 '11 at 14:34
  • $\begingroup$ @Srivatsan Narayanan: Here is how I reasoned: $\frac{3}{2}=\frac{6}{6/2+1}$ where $6/2$ in the denominator is $6$ divided by the number of distinct primes dividing $6$. Similarly: $\frac{5}{3}=\frac{10}{10/2+1}$ and $\frac{7}{4}=\frac{14}{14/2+1}$. $\endgroup$ – Mats Granvik Sep 13 '11 at 14:45
  • 2
    $\begingroup$ I just have an observation. If you build up the fractions in your sequence, you obtain fractions with numerators that fall in sequence between the primes. $2,3,0,5,\frac{6}{4},7,0,0,\frac{10}{6},11,0,13,\frac{14}{8},\frac{15}{7},0,17,0,19,0,\frac{21}{9},\frac{22}{12},23,0,0,\frac{26}{14},0,0,29,\frac{30}{22},31...$ And the zeros appear where non-square-free integers would lie. $\endgroup$ – alex.jordan Sep 13 '11 at 14:56
  • 3
    $\begingroup$ @Mats: Funny, I saw the top page of this post and thought, "Hmm, this must be by Mats Granvik" :-) $\endgroup$ – joriki Sep 13 '11 at 18:08
  • 2
    $\begingroup$ @Eric Naslund and Brian M. Scott: I have put the 250 first terms of the sequence here on my blog: sequence from question on math stackexchange It took me a while to sort the values. $\endgroup$ – Mats Granvik Sep 14 '11 at 11:49

Your sequence is


I will try thinking of a proof in a bit.

Note: I deleted my other answer with the extremely long and roundabout way I used to come up with this.

  • $\begingroup$ Related generalization of that formula in terms of matrix powers of matrix A: pastebin.com/4Rkg9Rcz $\endgroup$ – Mats Granvik Oct 30 '16 at 16:36

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