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From the identity $\ln\zeta(s) = \frac{q(1)}{1^{s}}+\frac{q(2)}{2^{s}}+\frac{q(3)}{3^{s}}+\frac{q(4)}{4^{s}}+\ldots$, where $q(n)= \begin{cases} \frac{1}{k} & \text {$n={p}^{k}, k \in \mathbb{N} $} \\ \text{0} & \text{otherwise} \\ \end{cases} $: $$ \begin{pmatrix} \frac{1}{1^2} & \frac{1}{2^2} & \frac{1}{3^2} & \frac{1}{4^2} & \cdots\\ \frac{1}{1^4} & \frac{1}{2^4} & \frac{1}{3^4} & \frac{1}{4^4}& \cdots\\ \frac{1}{1^6} & \frac{1}{2^6} & \frac{1}{3^6} & \frac{1}{4^6}& \cdots\\ \frac{1}{1^8} & \frac{1}{2^8} & \frac{1}{3^8} & \frac{1}{4^8}& \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix} \cdot \begin{pmatrix} q(1) \\ q(2) \\ q(3) \\ q(4) \\ q(5) \\ \vdots \end{pmatrix} = \begin{pmatrix} \ln\zeta(2) \\ \ln\zeta(4) \\ \ln\zeta(6) \\ \ln\zeta(8) \\ \vdots \end{pmatrix} $$ let $A= \begin{pmatrix} \frac{1}{1^2} & \frac{1}{2^2} & \frac{1}{3^2} & \frac{1}{4^2} & \cdots\\ \frac{1}{1^4} & \frac{1}{2^4} & \frac{1}{3^4} & \frac{1}{4^4}& \cdots\\ \frac{1}{1^6} & \frac{1}{2^6} & \frac{1}{3^6} & \frac{1}{4^6}& \cdots\\ \frac{1}{1^8} & \frac{1}{2^8} & \frac{1}{3^8} & \frac{1}{4^8}& \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}, C=\begin{pmatrix} \ln\zeta(2) \\ \ln\zeta(4) \\ \ln\zeta(6) \\ \ln\zeta(8) \\ \vdots \end{pmatrix}$, then $$\begin{pmatrix} q(1) \\ q(2) \\ q(3) \\ q(4) \\ q(5) \\ \vdots \end{pmatrix} = A^{-1} \cdot C$$ since $A^{-1}$ is just an inverse Vandermonde, and C is expressible by using Bernoulli numbers ($\ln\zeta(2n)=\ln\left|B_{2n}\right|+2n\ln 2\pi-\ln 2-\ln (2n)!$), we could theoretically get q(n) with only Bernoulli numbers as something more complex. Maybe we have even skipped the use of Riemann zeta zeros.
My question is, is there already something similar to this idea proposed?

P.S. the above is similar to what I posted about a year ago: Get prime number identifying function?

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  • $\begingroup$ Have you already tried to get an impression how an inverted $A$ wood look like? Perhaps as an extrapolation from the occurence with finite small sizes to finite greater sizes? It might be that the inversion, perhaps with the help of $LR$-decomposition, leads to divergent series? $\endgroup$ – Gottfried Helms Jul 14 '18 at 6:45
  • $\begingroup$ I know that lagrange interpolating can help give the formula of inverse Vandermonde matrices (see proofWiki), which is the method I use to try to solve this particular $A^{-1}$, though I haven't made it yet. I don't feel that it will diverge; though A has infinite amounts of rows and columns, I believe taking a limit will make it fine. $\endgroup$ – HankY Jul 14 '18 at 11:46
  • $\begingroup$ Hmm, in the attempt to invert the Carleman-matrix(see wikipedia) for the function $\exp(x)$ using the LR-decomposition the dot-products of rows- and columns of the inverted LR-components(Pascal-matrix and matrix of Stirlingnumbers 1'st kind) occur the (infinite) harmonic series , or equivalently the formal series for $\log(0)$ so this vandermonde matrix cannot be inverted - just one example to back my warning (or "reservations" or "doubts") here. $\endgroup$ – Gottfried Helms Jul 14 '18 at 12:32
  • $\begingroup$ What if let's not make A a matrix with infinite rows/cols - let's make it an n*n matrix, i.e. $A=[a_{ij}]_{n \times n }=[\frac{1}{j^{2i}}]_{n \times n }$. Then $A^{-1}$ should exist because I've checked that det(A) does not equal 0. $\endgroup$ – HankY Jul 14 '18 at 12:52
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I'm extending on my comments.

To define an inverse of a vandermonde-matrix of infinite size it needs some special care; I don't think, that observation of the behave of inverses with increasing sizes is reliable - it might, however, give a hint for the perspective.
A better parh is in my view to find an analytical expression for the entries in the inverse (depending on r(ow) and c(olumn) index) if this is possible, and use that analytical description for the dot-product with the infinite vector of logarithm of zetas.

One possibility to do this is to look at the triangular and diagonal components of the LDU-decomposition of the Vandermondematrix; that components have exactly determinable inverses, except that in the inversion of the diagonal-matrix expressions like $1/0$ might occur.

Let us use your matrix $A$ and find the three matrices $L$,$D$,$U$ by ldu-decomposition for some finite size. We find the following top-left segments of the matrices $L$,$D$,$U$ one below the other:

  L:
  1          0                  0                    0                        0                    0  |
  1          1                  0                    0                        0                    0  |
  1        5/4                  1                    0                        0                    0  |
  1      21/16              49/36                    1                        0                    0  |
  1      85/64          1897/1296              205/144                        1                    0  |
  1    341/256        69553/46656          32197/20736                5269/3600                    1  |
  1  1365/1024    2515513/1679616      4741165/2985984        20881861/12960000            5369/3600  |
  1  5461/4096  90663937/60466176  686641813/429981696  77087691109/46656000000      7139587/4320000  |
  -          -                  -                    -                        -                    -  +
  D:            
  1          0                  0                    0                        0                    0  |
  0      -3/16                  0                    0                        0                    0  |
  0          0             10/729                    0                        0                    0  |
  0          0                  0            -35/65536                        0                    0  |
  0          0                  0                    0              126/9765625                    0  |
  0          0                  0                    0                        0        -77/362797056  |
  0          0                  0                    0                        0                    0  |
  0          0                  0                    0                        0                    0  |
  -          -                  -                    -                        -                    -  +
  U:                
  1        1/4                1/9                 1/16                     1/25                 1/36  |
  0          1            128/243                 5/16                  128/625               35/243  |
  0          0                  1            6561/8192              45927/78125                 7/16  |
  0          0                  0                    1          2097152/1953125            2048/2187  |
  0          0                  0                    0                        1  244140625/181398528  |
  0          0                  0                    0                        0                    1  |
  0          0                  0                    0                        0                    0  |
  0          0                  0                    0                        0                    0  |
  -          -                  -                    -                        -                    -  +

There are some patterns visible here, but I did not really try to detect a set of formulae depending on r(ow) and c(olumn)-indexes so far.
Only we have by definition $$ A = L \cdot D \cdot U$$ for any finite dimension. And because the entries do not change when we change the size, we can assume, that this holds also for the infinite size.

Now we can easily find inverses for that triangular matrices, let's write $ K = L^{-1}, T=U^{-1}$. For the inverse of $D$ we need that no zero occurs on the diagonal - and the pattern of the entry suggests that no zero would occur, so we assume existence and compute also $C=D^{-1}$ with finite size.

The top-left segments of that matrices ($K$,$C$,$T$ below each other) look like this:

   K:
            1           0             0              0                 0                     0  |
           -1           1             0              0                 0                     0  |
          1/4        -5/4             1              0                 0                     0  |
        -1/36        7/18        -49/36              1                 0                     0  |
        1/576       -5/96        91/192       -205/144                 1                     0  |
     -1/14400     11/2880     -341/4800      1529/2880        -5269/3600                     1  |
     1/518400  -91/518400   1001/172800  -44473/518400       37037/64800            -5369/3600  |
  -1/25401600    1/181440  -533/1814400   9581/1270080   -353639/3628800           54613/90720  |
            -           -             -              -                 -                     -  +
   C: 
            1           0             0              0                 0                     0  |
            0       -16/3             0              0                 0                     0  |
            0           0        729/10              0                 0                     0  |
            0           0             0      -65536/35                 0                     0  |
            0           0             0              0       9765625/126                     0  |
            0           0             0              0                 0         -362797056/77  |
            0           0             0              0                 0                     0  |
            0           0             0              0                 0                     0  |
            -           -             -              -                 -                     -  +
   T:
            1        -1/4         5/243        -7/8192        42/1953125          -11/30233088  |
            0           1      -128/243           7/64    -24576/1953125              55/59049  |
            0           0             1     -6561/8192    531441/1953125              -55/1024  |
            0           0             0              1  -2097152/1953125          90112/177147  |
            0           0             0              0                 1  -244140625/181398528  |
            0           0             0              0                 0                     1  |
            0           0             0              0                 0                     0  |
            0           0             0              0                 0                     0  |
            -           -             -              -                 -                     -  +

Now, to proceed, it is crucial to find analytical expressions for the entries in the matrices, so that we can explicate series-formulae for the dot-products of rows in $T$ with columns in $K$ multiplied by the diagonal-elements in $C$.

I've not yet really tried to find some meaningful pattern here. But doing the dot-products with increasing sizes of the matrices by numerical evaluation by Pari/GP-software seems to give zeros for all entries in $T \cdot C \cdot K$ ! So we seem to get a result that $$\lim_{\text{size} \to \infty} A^{-1} = \mathbb 0$$

Of course, using such a matrix as left-multiplicator for the column-vector $\Lambda$ containing the logarithms of the zetas would again result in a zero-vector and would thus be useless.

What we could try, is, to separate the dotproducts by changing the order of computation. Instead as suggested so far $$ (T \cdot C \cdot K) \cdot \Lambda = \mathbb 0 \cdot \Lambda = \mathbb 0\\ \tag {dismissed} $$ we do with different order of computation $$ T \cdot (C \cdot K \cdot \Lambda) = T \cdot Y = X \tag {proposed} $$

This gives for determination of $Y$ convergence with increasing size of the matrices involved to

 0.497700
  2.23248
  3.11517
  4.48353
  7.32370
  11.9564
  20.6917
  35.4022
  ...     (furtherly increasing by similar growth rate)

and $ T \cdot Y = X$ seems to have as well convergent, but not zero, dot-products giving in the limit the expected "indicator" for prime-powers in $X$:

   i   X[i]   approx by size=128x128  //  i is the row-index 
  ---+------+-------------------------------------------------
   1    0    -3.13140E-128
   2    1          1.00000
   3    1          1.00000
   4  1/2         0.500000
   5    1          1.00000
   6    0     2.38367E-104
   7    1          1.00000
   8  1/3         0.333333
   9  1/2         0.500000
  10    0      1.98922E-91
  11    1          1.00000
  12    0      1.74025E-86
  13    1          1.00000
  14    0      4.24518E-81
  15    0     -1.15516E-78
  16  1/4         0.250000
  17    1          1.00000
  18    0      9.14500E-72
  19    1          1.00000
  20    0      1.05168E-67
  21    0     -1.06579E-65
  22    0      8.29498E-64
  23    1          1.00000
  24    0      3.13720E-60
  25  1/2         0.500000
  26    0      1.13847E-56
  27  1/3         0.333333
  28    0      1.33656E-53
  29    1          1.00000
  30    0      2.17444E-50
  31    1          1.00000
  32  1/5         0.200000
  33    0     -1.85178E-45
  34    0      4.41824E-44
  35    0     -7.29484E-43
  36    0      8.34838E-42
  37    1          1.00000
  38    0      3.17639E-39
  39    0     -9.40728E-38
  40    0      1.92052E-36
  41    1          1.00000
  42    0      4.07160E-34
  43    1          1.00000
  44    0      1.11334E-31
  45    0     -1.92810E-30
  46    0      3.19415E-29
  47    1          1.00000
  48    0      4.50101E-27
  49  1/2         0.500000
  50    0      1.13855E-25
  51    0     -1.12392E-26
  52    0     -5.45288E-25
  53    1          1.00000
  54    0      1.16300E-21
  55    0     -9.83767E-21
  56    0      2.71287E-20
  57    0     -3.88663E-19
  58    0     -3.27127E-18
  59    1          1.00000
  60    0     -4.67250E-15
  61    1          1.00000
  62    0      8.51704E-13
  63    0      6.72168E-11
  64  1/6         0.166667
  ...       approximation quality too low from here with matrixsize 128

That values for the entries in $X$ are estimated by the results of multiplication of the matrices with increasing sizes.


I have written a short treatize with perhaps some better explanations on a very similar problem some years ago, where I also succeeded to get a proof for my version of the Vandermonde-matrix: that the "inverse" taken by this method indeed is zero because I could decode the systematic formulae for the entries in $K,C,T$ in that versions and state explicitely series-definitions for the involved dot-products. The proof for the zero-result in all entries of $A^{-1}$ was given in an MO-answer a couple of years ago.

See my own text here which does not yet include the MO-answer. A bit introduction is already on the index-page for my math-papers.


Couldn't resist; at least a partial decomposition of the entries in $T$, the numbers beside the systematic fractions are integer:

  T:
  1*1^1/1^1  -2*1^1/2^3    5*1^1/3^5  -14*1^1/4^7     42*1^1/5^9    -132*1^1/6^11      429*1^1/7^13      -1430*1^1/8^15
          .   1*2^3/2^3  -16*2^3/3^5  224*2^3/4^7  -3072*2^3/5^9   42240*2^3/6^11  -585728*2^3/7^13    8200192*2^3/8^15
          .           .    1*3^5/3^5  -54*3^5/4^7   2187*3^5/5^9  -80190*3^5/6^11  2814669*3^5/7^13  -96722262*3^5/8^15
          .           .            .    1*4^7/4^7   -128*4^7/5^9   11264*4^7/6^11  -851968*4^7/7^13   59637760*4^7/8^15
          .           .            .            .      1*5^9/5^9    -250*5^9/6^11    40625*5^9/7^13   -5468750*5^9/8^15
          .           .            .            .              .      1*6^11/6^11    -432*6^11/7^13    116640*6^11/8^15
          .           .            .            .              .                .       1*7^13/7^13      -686*7^13/8^15
          .           .            .            .              .                .                 .         1*8^15/8^15
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