3
$\begingroup$

Let $g(n)$ be the Dirichlet inverse of the Euler totient function:

$$g(n) = \sum\limits_{d|n} d \cdot \mu(d)$$

and let $f(x,y)$ be the elliptic equation: $$f(x,y)=x^3 - x^2 - y^2 - y$$

Show that the OEIS sequence A002070 is equal to:

$$A002070(n) = \sum _{k=1}^{p_n} \left(\sum _{y=1}^{p_n} \left(\sum _{x=1}^{p_n} \frac{g(k) \left[\gcd (f(x,y),p_n)=k\right]}{p_n}\right)\right)$$ starting:

{-2, -1, 1, -2, 1, 4, -2, 0, -1, 0, 7, 3, -8, -6, 8, -6, 5, 12, -7, 
 -3, 4, -10, -6, 15, -7,...}

where: $$\left[\gcd (f(x,y),p_n)=k\right]$$ is the Iverson bracket.

Associated Mathematica 8.0.1 program to verify the conjecture for the 25 first terms:

(*start*)
nn = 25; (*set nn=12 for faster computation*) f = x^3 - x^2 - y^2 - y; 
g[n_] := DivisorSum[n, MoebiusMu[#] # &]; Monitor[
 Table[Sum[
   Sum[Sum[If[GCD[f, Prime[n]] == k, 1, 0]*g[k]/Prime[n], {x, 1,
       Prime[n]}], {y, 1, Prime[n]}], {k, 1, Prime[n]}], {n, 1, nn}],
  n]
 (*end*)

OEIS sequence A002070 has the name:

Coefficient of x^p (p = n-th prime) in x * Product_{k>=1} (1-x^k)^2*(1-x^11k)^2.

Edit 10.10.2023:

$$A366450(n) = \sum _{k=1}^{n} \left(\sum _{y=1}^{n} \left(\sum _{x=1}^{n} \frac{g(k) \left[\gcd (f(x,y),n)=k\right]}{n}\right)\right)$$

Question: Prove or disprove: $\text{A366450(A005117(n)) = A006571(A005117(n))}$

where $\text{A006571 = Expansion of q*Product_{k>=1} (1-q^k)^2*(1-q^(11*k))^2.}$ https://oeis.org/A006571

https://oeis.org/A005117
$\text{A005117 = Squarefree numbers: numbers that are not divisible by a square greater than 1.}$

A366450 = {1, -2, -1, -4, 1, 2, -2, -8, -3, -2, 1, 4, 4, 4, -1, -16, -2, 6, 0, \
-4, 2, -2, -1, 8, 5, -8, -9, 8, 0, 2, 7, -32, -1, 4, -2, 12, 3, 0, \
-4, -8, -8, -4, -6, -4, -3, 2, 8, 16, -14, -10, 2, -16, -6, 18, 1, \
16, 0, 0, 5, 4, 12, -14, 6, -64, 4, 2, -7, 8, 1, 4, -3, 24, 4, -6, \
-5, 0, -2, 8, -10, -16, -27, 16, -6, -8, -2, 12, 0, -8, 15, 6, -8, 4, \
-7, -16, 0, 32, -7, 28, -3, -20, 2, -4, -16, -32, 2, 12, 18, 36, 10, \
-2, -3, 32, 9, 0, -1, 0, -12, -10, 4, 8, 11, -24, 8, -28, 25, -12, 8, \
-128, 6, -8, -18, 4, 0, 14, -9, 16, -7, -2, 10, 8, -8, 6, 4, 48, 0, \
-8, 14, -12, -10, 10, 2, 0, 6, 4, 7, 16, -7, 20, 6, -32}

Compared to A006571:

A006571 = {1, -2, -1, 2, 1, 2, -2, 0, -2, -2, 1, -2, 4, 4, -1, -4, -2, 4, 0, 2, \
2, -2, -1, 0, -4, -8, 5, -4, 0, 2, 7, 8, -1, 4, -2, -4, 3, 0, -4, 0, \
-8, -4, -6, 2, -2, 2, 8, 4, -3, 8, 2, 8, -6, -10, 1, 0, 0, 0, 5, -2, \
12, -14, 4, -8, 4, 2, -7, -4, 1, 4, -3, 0, 4, -6, 4, 0, -2, 8, -10, \
-4, 1, 16, -6, 4, -2, 12, 0, 0, 15, 4, -8, -2, -7, -16, 0, -8, -7, 6, \
-2, -8, 2, -4, -16, 0, 2, 12, 18, 10, 10, -2, -3, 8, 9, 0, -1, 0, -8, \
-10, 4, 0, 1, -24, 8, 14, -9, -8, 8, 0, 6, -8, -18, -2, 0, 14, 5, 0, \
-7, -2, 10, -4, -8, 6, 4, 8, 0, -8, 3, 6, -10, -8, 2, 0, 4, 4, 7, -8, \
-7, 20, 6, 8}


(A366450 - A006571) =  

{0, 0, 0, -6, 0, 0, 0, -8, -1, 0, 0, 6, 0, 0, 0, -12, 0, 2, 0, -6, 0, \
0, 0, 8, 9, 0, -14, 12, 0, 0, 0, -40, 0, 0, 0, 16, 0, 0, 0, -8, 0, 0, \
0, -6, -1, 0, 0, 12, -11, -18, 0, -24, 0, 28, 0, 16, 0, 0, 0, 6, 0, \
0, 2, -56, 0, 0, 0, 12, 0, 0, 0, 24, 0, 0, -9, 0, 0, 0, 0, -12, -28, \
0, 0, -12, 0, 0, 0, -8, 0, 2, 0, 6, 0, 0, 0, 40, 0, 22, -1, -12, 0, \
0, 0, -32, 0, 0, 0, 26, 0, 0, 0, 24, 0, 0, 0, 0, -4, 0, 0, 8, 10, 0, \
0, -42, 34, -4, 0, -128, 0, 0, 0, 6, 0, 0, -14, 16, 0, 0, 0, 12, 0, \
0, 0, 40, 0, 0, 11, -18, 0, 18, 0, 0, 2, 0, 0, 24, 0, 0, 0, -40}

Position of zeros in (A366450 - A006571) =

{1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, \
30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 46, 47, 51, 53, 55, 57, \
58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 76, 77, 78, 79, 82, \
83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 101, 102, 103, 105, 106, 107, \
109, 110, 111, 113, 114, 115, 116, 118, 119, 122, 123, 127, 129, 130, \
131, 133, 134, 137, 138, 139, 141, 142, 143, 145, 146, 149, 151, 152, \
154, 155, 157, 158, 159}

Removing numbers 76,116 and 152 we get:

A005117 = {1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, \
30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 46, 47, 51, 53, 55, 57, \
58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 82, 83, \
85, 86, 87, 89, 91, 93, 94, 95, 97, 101, 102, 103, 105, 106, 107, \
109, 110, 111, 113, 114, 115, 118, 119, 122, 123, 127, 129, 130, 131, \
133, 134, 137, 138, 139, 141, 142, 143, 145, 146, 149, 151, 154, 155, \
157, 158, 159}

https://oeis.org/A005117
A005117 = Squarefree numbers: numbers that are not divisible by a square greater than 1.

Program to verify conjecture up to $n = 160$:

(*start*) (* takes a few hours to run *) nn = 160; f = x^3 - x^2 - y^2 - y;
g[n_] := DivisorSum[n, MoebiusMu[#] # &]; Monitor[
 b = Table[
   Sum[Sum[Sum[If[GCD[f, n] == k, 1, 0]*g[k]/n, {x, 1, n}], {y, 1, 
      n}], {k, 1, n}], {n, 1, nn}], n]
(*end*)
a[n_] := SeriesCoefficient[
   q (Product[(1 - q^k), {k, 11, n, 11}] Product[
        1 - q^k, {k, n}])^2, {q, 0, n}];
c = Table[a[n], {n, 1, nn}]
b - c
Flatten[Position[%, 0]]

Edit 16.10.2023:

jj = 128;
fraction = {z};
Monitor[Do[
  f = x^3 - x^2 - (y^2 + y);
  nn = Length[fraction];
  A023900[n_] := DivisorSum[n, MoebiusMu[#] # &]; 
  c = Table[
    Sum[If[Mod[m, k] == 0, fraction[[m/k]]*A023900[k], 0], {k, 1, 
      m}], {m, 1, nn}]; 
  AXXXXXXz = 
   Numerator[
     Table[Sum[
       Sum[Sum[If[GCD[f, n] == m, 1, 0]*c[[m]]/n, {x, 1, n}], {y, 1, 
         n}], {m, 1, n}], {n, nn, nn}]][[1]];
  aa[n_] := 
   SeriesCoefficient[
    q (Product[(1 - q^k), {k, 1, n, 1}] Product[
         1 - q^k, {k, 11, n, 11}])^2, {q, 0, n}];
  AXXXXXX = Table[aa[n], {n, nn, nn}][[1]];
  append = z /. Solve[AXXXXXXz == AXXXXXX, z][[1]];
  fraction = Flatten[{fraction[[Range[nn - 1]]], append, z}];, {j, 1, 
   jj}], j]
fraction
Flatten[Position[Abs[Sign[fraction]], 1]]

{1, 0, 0, 3, 0, 0, 0, 4, 3/4, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, \
0, 0, -(45/
  4), 0, 21/2, 0, 0, 0, 0, 20, 0, 0, 0, 9/4, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 77/9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 28, 0, 0, \
0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 21, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -(135/
  4), 0, 0, 0, 0, 0, 0, 0, 63/2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
-(110/9), 0, 0, 0, -(85/2), 0, 0, 64, z}

 Flatten[Position[Abs[Sign[fraction]], 1]]

{1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 72, 81, 100, 108, 121, 125, \
128} https://oeis.org/A001694

 fraction[[2^Range[7]]]

 {0, 3, 4, 6, 20, 28, 64}

https://oeis.org/A038520 A038520 Number of elements of GF(2^n) with trace 1 and subtrace 0.

Edit 18.10.2023:

If one asks what sequence bb should we convolve the L-Series with to get the GCD-Möbius function sequence then there are patterns in the sequence bb:

(*Mathematica start*)
Clear[z, nn, jj];
jj = 160 - 1;
A366450 = {1, -2, -1, -4, 1, 2, -2, -8, -3, -2, 1, 4, 4, 
   4, -1, -16, -2, 6, 0, -4, 2, -2, -1, 8, 5, -8, -9, 8, 0, 2, 
   7, -32, -1, 4, -2, 12, 3, 0, -4, -8, -8, -4, -6, -4, -3, 2, 8, 
   16, -14, -10, 2, -16, -6, 18, 1, 16, 0, 0, 5, 4, 12, -14, 6, -64, 
   4, 2, -7, 8, 1, 4, -3, 24, 4, -6, -5, 0, -2, 8, -10, -16, -27, 
   16, -6, -8, -2, 12, 0, -8, 15, 6, -8, 4, -7, -16, 0, 32, -7, 
   28, -3, -20, 2, -4, -16, -32, 2, 12, 18, 36, 10, -2, -3, 32, 9, 
   0, -1, 0, -12, -10, 4, 8, 11, -24, 8, -28, 25, -12, 8, -128, 
   6, -8, -18, 4, 0, 14, -9, 16, -7, -2, 10, 8, -8, 6, 4, 48, 0, -8, 
   14, -12, -10, 10, 2, 0, 6, 4, 7, 16, -7, 20, 6, -32};
aa = {1, -2, -1, 2, 1, 2, -2, 0, -2, -2, 1, -2, 4, 4, -1, -4, -2, 4, 
   0, 2, 2, -2, -1, 0, -4, -8, 5, -4, 0, 2, 7, 8};
A006571 = {1, -2, -1, 2, 1, 2, -2, 0, -2, -2, 1, -2, 4, 4, -1, -4, -2,
    4, 0, 2, 2, -2, -1, 0, -4, -8, 5, -4, 0, 2, 7, 8, -1, 4, -2, -4, 
   3, 0, -4, 0, -8, -4, -6, 2, -2, 2, 8, 4, -3, 8, 2, 8, -6, -10, 1, 
   0, 0, 0, 5, -2, 12, -14, 4, -8, 4, 2, -7, -4, 1, 4, -3, 0, 4, -6, 
   4, 0, -2, 8, -10, -4, 1, 16, -6, 4, -2, 12, 0, 0, 15, 
   4, -8, -2, -7, -16, 0, -8, -7, 6, -2, -8, 2, -4, -16, 0, 2, 12, 18,
    10, 10, -2, -3, 8, 9, 0, -1, 0, -8, -10, 4, 0, 1, -24, 8, 
   14, -9, -8, 8, 0, 6, -8, -18, -2, 0, 14, 5, 0, -7, -2, 10, -4, -8, 
   6, 4, 8, 0, -8, 3, 6, -10, -8, 2, 0, 4, 4, 7, -8, -7, 20, 6, 8};
bb = {z};
Monitor[Do[
  nn = Length[bb];
  Flatten[Position[Abs[Sign[bb]], 1]];
  cc = (Table[
        Table[If[Mod[n, k] == 0, A006571[[n/k]], 0], {k, 1, nn}], {n, 
         1, nn}].Table[
        Table[If[Mod[n, k] == 0, bb[[n/k]], 0], {k, 1, nn}], {n, 1, 
         nn}])[[All, 1]] - A366450[[Range[nn]]];
  TableForm[{bb, Range[Length[cc]]}];
  TableForm[{cc, Range[Length[cc]]}];
  append = z /. (Solve[cc[[nn]] == 0, z][[1]]);
  bb = Flatten[{bb[[Range[nn - 1]]], append, z}];
  , {j, 1, jj}], j]
bb
(*end*)

bb = {1, 0, 0, -6, 0, 0, 0, -20, -1, 0, 0, 0, 0, 0, 0, -40, 0, 0, 0, 0, 0, \
0, 0, 0, 9, 0, -15, 0, 0, 0, 0, -80, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, -11, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -160, \
0, 0, 0, 0, 0, 0, 0, 20, 0, 0, 0, 0, 0, 0, 0, 0, -45, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -54, 0, 0, 0, 0, 0, 0, 0, 90, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 25, 0, 0, -320, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 40, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, z}

Position of ones in the characteristic sequence of bb = https://oeis.org/A001694

At multiples of 8 there are multiples of 20. There are also multiples of 15 at other numbers.

Improve this question
$\endgroup$
8
  • $\begingroup$ If someone knows how to tag this better you are welcome. $\endgroup$
    – Mats Granvik
    Commented Oct 9, 2023 at 15:58
  • $\begingroup$ Should I post this on Mathematics stack exchange instead? $\endgroup$
    – Mats Granvik
    Commented Oct 9, 2023 at 16:11
  • 6
    $\begingroup$ Since $p_n$ is prime, $\gcd$ can take only 2 values: 1 or $p_n$. Then the conjecture is reduced to just saying that the order of the elliptic curve $y^2+y=x^3-x^2$ over the field $\mathbb{F}_{p_n}$ equals $$p_n + 1 - \texttt{A002070}(n).$$ $\endgroup$
    – Max Alekseyev
    Commented Oct 9, 2023 at 21:49
  • 6
    $\begingroup$ .. and that is true as the modular form associated to this elliptic curve of conductor 11 is an eta quotient. I believe this is the contents of the article by Shimura linked in the oeis. $\endgroup$
    – Chris Wuthrich
    Commented Oct 10, 2023 at 8:58
  • 1
    $\begingroup$ I am not sure why you keep updating this. Max and my comment solved the initial question. I doubt many people will want to follow your coding experiments without extra motivation or explanations. (But if there is, maybe a new question is better than adding to this). $\endgroup$
    – Chris Wuthrich
    Commented Oct 18, 2023 at 16:30

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