1 (* code snippet "4. Kosovo" *)
2 NumQ[j_,k_,c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n),3n^2+(3+4k)n+j(2-j+2k),{n,1,15000}];
3 vD:=Function[{ab,j,k,c},If[c<2,ab[[1]]+ab[[2]]c,-2c(2c-j)(2c-2k+j-2)(2c-2k-1)vD[ab,j,k,c-2]+(8c^2+(2-8k)c+(j-2)(2k-j))vD[ab,j,k,c-1]]];
4 f[j_,k_,c_]:=Product[(2c-2k+2i-1)(k-i+1),{i,1,(j-1)/2}]CatalanNumber[(j-3)/2](j-2)(2k-1)*(2c-1)!!^2*CatalanNumber[k-1];
5 g[j_,k_,c_]:=Product[(2c-2i+1)(2k-2i+1),{i,1,(j-1)/2}]{2c)!2^(2k+(j-7)/2);
6 h[j_,k_,c_]:=Product[2c-2i-1,{i,0,(j-3)/2}]Product[2c-2i-1,{i,0,k-1}];
7 Q[j_,k_,c_,ab_]:=Simplify[g[j,k,c]/(vD[ab,j,k,c-1]*h[j,k,c]+f[j,k,c]Catalan)];
8 l[n_,j_,k_]:=(-1)^(k+1)(2k)!^2/k!^2(3*k-2)/Product[(k-i)(2k-2i-1)^2,{i,0,(j-3)/2}]/(2k-j)/(2k-1)/(n((2k-j-2)(3-2k)-1)+1);
9 kD:=Function[{n,ab,j,k},If[k<2,ab[[1]]+ab[[2]]k,(2k+2j-9-2n)(2k+j-8-2n)(-2k+5-j)(2k+j-6)*kD[n,ab,j,k-2]+(8k^2+k(10j-48-8n)+(3j^2-(28+4n)j+68+18n))kD[n,ab,j,k-1]]];
10 Descend:=Function[{j,k},abh[0]={{-1,4},{19,234},{5818/3,254456/3},{667115,60003486},{467946090,71121907440},
11 {554143204110,127451285438100},{994115449382940,322092692148962160},{2516347061651130075,1092094185270706446150},
12 {8546069024090201027250,4785798287838257081935200},{37508692924557081882027450,26331102038134635548392485900},
13 {206659254109760483703789089700,177726957997323983116663150902000},
14 {1396637676485497608584841260027550,1444123356588023432320434243315206700},
15 {11361110319787394788017568214856502500,13905999029609441333101619589964946580000},
16 {109509742351999832489255793094925601037500,156598931559029451368898717824937174831465000},
17 {1234320809247763942235545044494798498436195000,2039097976865181167119056627863149102390546140000},
18 {16085205915675471439195309128783843538512283666875,30401039180587356456007967587920548312623820610393750},
19 {239989379884263177615577263747245812249369757283461250,514537230471714428505965482811829861362838523445500920000}};
20 abh[1]={{-1/3,-14/3},{-17,-8},{-758,-27820},{-302117,-23010044},{-1091480994/5,-146282046156/5},
21 {-262476468810,-54596049230880},{-475443072646380,-141682352738003640},{-1211573031414907725,-489475664504671450500},
22 {-4135193781750207709650,-2175112041708995560914300},{-18218507239728799899288030,-12097088912487772715204794320},
23 {-100680148628028059378172563700,-82356361704096372069838207986600},
24 {-682078864161239229949889893754850,-673893917980353010760236819146271800},
25 {-5559692282317104149119150499246482500,-6527078739785105011529098668023829975000},
26 {-53681246247288656939970174534335708392500,-73865394837022289182570623863734010339760000},
27 {-605939306349175124039948450466713432304279000,-965867254322525126192328035746702493817188406000}
1 (* code snippet "4. Kosovo" *)
2 NumQ[j_,k_,c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n),3n^2+(3+4k)n+j(2-j+2k),{n,1,15000}];
3 vD:=Function[{ab,j,k,c},If[c<2,ab[[1]]+ab[[2]]c,-2c(2c-j)(2c-2k+j-2)(2c-2k-1)vD[ab,j,k,c-2]+(8c^2+(2-8k)c+(j-2)(2k-j))vD[ab,j,k,c-1]]];
4 f[j_,k_,c_]:=Product[(2c-2k+2i-1)(k-i+1),{i,1,(j-1)/2}]CatalanNumber[(j-3)/2](j-2)(2k-1)*(2c-1)!!^2*CatalanNumber[k-1];
5 g[j_,k_,c_]:=Product[(2c-2i+1)(2k-2i+1),{i,1,(j-1)/2}]{2c)!2^(2k+(j-7)/2);
6 h[j_,k_,c_]:=Product[2c-2i-1,{i,0,(j-3)/2}]Product[2c-2i-1,{i,0,k-1}];
7 Q[j_,k_,c_,ab_]:=Simplify[g[j,k,c]/(vD[ab,j,k,c-1]*h[j,k,c]+f[j,k,c]Catalan)];
8 l[n_,j_,k_]:=(-1)^(k+1)(2k)!^2/k!^2(3*k-2)/Product[(k-i)(2k-2i-1)^2,{i,0,(j-3)/2}]/(2k-j)/(2k-1)/(n((2k-j-2)(3-2k)-1)+1);
9 kD:=Function[{n,ab,j,k},If[k<2,ab[[1]]+ab[[2]]k,(2k+2j-9-2n)(2k+j-8-2n)(-2k+5-j)(2k+j-6)*kD[n,ab,j,k-2]+(8k^2+k(10j-48-8n)+(3j^2-(28+4n)j+68+18n))kD[n,ab,j,k-1]]];
10 Descend:=Function[{j,k},abh[0]={{-1,4},{19,234},{5818/3,254456/3},{667115,60003486},{467946090,71121907440},
11 {554143204110,127451285438100},{994115449382940,322092692148962160},{2516347061651130075,1092094185270706446150},
12 {8546069024090201027250,4785798287838257081935200},{37508692924557081882027450,26331102038134635548392485900},
13 {206659254109760483703789089700,177726957997323983116663150902000},
14 {1396637676485497608584841260027550,1444123356588023432320434243315206700},
15 {11361110319787394788017568214856502500,13905999029609441333101619589964946580000},
16 {109509742351999832489255793094925601037500,156598931559029451368898717824937174831465000},
17 {1234320809247763942235545044494798498436195000,2039097976865181167119056627863149102390546140000},
18 {16085205915675471439195309128783843538512283666875,30401039180587356456007967587920548312623820610393750},
19 {239989379884263177615577263747245812249369757283461250,514537230471714428505965482811829861362838523445500920000}};
20 abh[1]={{-1/3,-14/3},{-17,-8},{-758,-27820},{-302117,-23010044},{-1091480994/5,-146282046156/5},
21 {-262476468810,-54596049230880},{-475443072646380,-141682352738003640},{-1211573031414907725,-489475664504671450500},
22 {-4135193781750207709650,-2175112041708995560914300},{-18218507239728799899288030,-12097088912487772715204794320},
23 {-100680148628028059378172563700,-82356361704096372069838207986600},
24 {-682078864161239229949889893754850,-673893917980353010760236819146271800},
25 {-5559692282317104149119150499246482500,-6527078739785105011529098668023829975000},
26 {-53681246247288656939970174534335708392500,-73865394837022289182570623863734010339760000},
27 {-605939306349175124039948450466713432304279000,-965867254322525126192328035746702493817188406000}
28 {-7906287653442943862409767973858652552097126548125,-14452693830499521315903006473321900713406050369972500},
29 {-118090012323922699712409299094969252935070156336941250,-245391045609131483699190960852072336578359955374403917500}};
30 ab=Table[kD[u,abh[u][[(j-1)/2]],j,k-j+2]/l[u,j,k],{u,0,1}]{ab[[1]],ab[[2]]-ab[[1]]}];
31 Q[j_,k_,c_,ab_]:=Simplify[g[j,k,c]/(vD[ab,j,k,c-1]*h[j,k,c]+f[j,k,c]Catalan)];
32 Print[Union[Flatten[Table[N[NumQ[j,k,c]=Q[j,k,c,Descend[j,k]],200],{j,3,11,2},{k,j-2,10},{c,1,7}]]]];ClearAll["Global
'*"]];

<figure><img src="image_17.png" /><figcaption>Fig. 18. The points on which Kosovo is tested for $1 \le c \le 7$ by the snippets "4. Kosovo" and "18. Altogether" are shown in red.</figcaption></figure>

1 (* code snippet "5. resolution" *)
2 NumQ[j_,k_,c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n),3n^2+(3+4k)n+j(2-j+2k),{n,1,10000}];
3 f[j_,k_,c_]:=Product[(2c-2k+2i-1)(k-i+1),{i,1,(j-1)/2}]CatalanNumber[(j-3)/2](j-2)(2k-1)*(2c-1)!!^2*CatalanNumber[k-1];
4 g[j_,k_,c_]:=Product[(2c-2i+1)(2k-2i+1),{i,1,(j-1)/2}](2c)!!^2*(2k+(j-7)/2);
5 h[j_,k_,c_]:=Product[2c-2i-1,{i,0,(j-3)/2}]Product[2c-2i-1,{i,0,k-1}];
28 {-7906287653442943862409767973858652552097126548125, -14452693830499521315903006473321900713406050369972500},
29 {-118090012323922699712409299094969252935070156336941250, -245391045609131483699190960852072336578359955374403917500}};
30 ab=Table[kD[u,abh[u][[(j-1)/2]],j,k-j+2]/l[u,j,k],{u,0,1}];{ab[[1]],ab[[2]]-ab[[1]]}];
31 Q[j_,k_,c_,ab_]:=Simplify[g[j,k,c]/(vD[ab,j,k,c-1]*h[j,k,c]+f[j,k,c]Catalan)};
32 Print[Union[Flatten[Table[N[NumQ[j,k,c]=Q[j,k,c,Descend[j,k]],200],{j,3,11,2},{k,j-2,10},{c,1,7}]]]]];ClearAll["Global
'*"];

<figure><img src="image_19.png" /><figcaption>Fig. 18. The points on which Kosovo is tested for $1 \le c \le 7$ by the snippets "4. Kosovo" and "18. Altogether" are shown in red.</figcaption></figure>

1 (* code snippet "5. resolution" *)
2 NumQ[j_,k_,c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n),3n^2+(3+4k)n+j(2-j+2k),{n,1,10000}];
3 f[j_,k_,c_]:=Product[(2c-2k+2i-1)(k-i+1),{i,1,(j-1)/2}]CatalanNumber[(j-3)/2](j-2)(2k-1)*(2c-1)!!^2*CatalanNumber[k-1];
4 g[j_,k_,c_]:=Product[(2c-2i+1)(2k-2i+1),{i,1,(j-1)/2}](2c)!!^2*(2k+(j-7)/2);
5 h[j_,k_,c_]:=Product[2c-2i-1,{i,0,(j-3)/2}]Product[2c-2i-1,{i,0,k-1}];
6 l:=Function[{n,j,k},(-1)^(k+1)(2k)!^2/k!^2(3k-2)/Product[(k-i)(2k-2i-1)^2,{i,0,(j-3)/2}]/(2k-j)/(2k-1)/(n((2k-j-2)(3-2k)-1)+1)];

7 AB:=Function[{j,k},For[c=1,c<=2,r=N[NumQ[j,k,c],2000];v=FindIntegerNullVector[{1,r,N[Catalan*r,2000]}];d[c]=(-g[j,k,c]([v[[3]]Catalan+v[[2]])]/v[[1]]-f[j,k,c]Catalan)/h[j,k,c];c++];{d[1],d[2]-d[1]}];ABlists:=Function[lim,z[1]=z[0]={};

8 For[j=3,j<=lim,For[w=0,w<=1,e[w]=Table[AB[j,k].{1,w}l[w,j,k},{k,j-2,j-1}];

9 ab[w]={{a,b}}/.Solve[Table[a+b(u-1)==e[w][u],{u,1,2}],{a,b}];AppendTo[z[w],ab[w][[1]]];w++;j+=2];{z[1],z[0]}];

10 Print[ABlists[15]];ClearAll["Global*"];

1 (* code snippet "6. symmetry" *)

2 NumQ[j_,k_,c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n),3n^2+(3+4k)n+j(2-j+2k),{n,1,15000}];

3 g[j_,k_,c_]:=Product[(2c-2i+1)(2k-2i+1),{i,1,(j-1)/2}](2c)!2^(2k+(j-7)/2);

4 h[j_,k_,c_]:=Product[2c-2i-1,{i,0,(j-3)/2}]Product[2c-2i-1,{i,0,k-1}];

5 zeta[j_,u_]:=Product[2+2i-j,{i,0,u-1}]/2^u;Ratio:=Function[{j,k,c},r=N[NumQ[j,k,c],2000];

6 v=FindIntegerNullVector[{1,r,N[Catalan*r,2000]}];-v[[1]]/v[[2]]];

7 GenAB[j_,k_]:=({a,b}/.Solve[Table[{a,b}.{1,c-1}==g[j,k,c]/Ratio[j,k,c]/h[j,k,c],{c,1,2}],{a,b})][[1]];

8 tau[j_,u_]:=If[u>(j-1)/2,Sign[zeta[j,u]](2j-2u-3)!!(2u-j)!!(-2)^u,zeta[j,u](-4)^u];

9 For[j=3,j<=13,Print[MatrixForm[Table[{j,j-u-1},{j-2u,j-u-1},(GenAB[j,j-u-1]/GenAB[j-2u,j-u-1])/tau[j,u]={{1,1}},{u,1,j+3}]]];j+=2];ClearAll["Global*"];

1 (* code snippet "7. Croatia" *)

2 NumQ[j_,k_,c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n),3n^2+(3+4k)n+j(2-j+2k),{n,1,2000}];

3 g[j_,k_,c_]:=Product[(2c-2i+1)(2k-2i+1),{i,1,(j-1)/2}](2c)!2^(2k+(j-7)/2);

4 h[j_,k_,c_]:=Product[2c-2i-1,{i,0,(j-3)/2}]Product[2c-2i-1,{i,0,k-1}];

5 Ratio:=Function[{j,k,c},r=N[NumQ[j,k,c],4000];v=FindIntegerNullVector[{1,r,N[Catalan*r,4000]}];-v[[1]]/v[[2]]];

6 GenAB[j_,k_]:=({a,b}/.Solve[Table[{a,b}.{1,c-1}==g[j,k,c]/Ratio[j,k,c]/h[j,k,c],{c,1,2}],{a,b})][[1]];

7 psi1[i_,j_]:=ymphj-1,-464+58j-3j^2,27936-4692j+432j^2-15j^3,-2659968+542256j-67836j^2+4260j^3-105j
^4,367568640-86278560j+13203480j^2-1139700j^3+51450j^4-945j^5][[i]];

8 psi2[i_,j_]:=m4j-15,306-95j+4j^2,-13360+464j-357j^2+12j^3,999648-379692j+40368j^2-2457j^3+60j^4,-113885568+

9 46449360j-6124164j^2+513228j^3-22935j^4+420j^5,18333538560-7933530720j+1224286440j^2-126833100j^3+7864950j^4-266175j
^5+3780j^6][[i]];

10 mu[i_,j_]:=-Product[j-2q-2,{q,1,i}]/(-2)^((3j-11-4i)/2);
6 l:=Function[{n,j,k},(-1)^(k+1)(2k)!^2/k!^2(3k-2)/Product[(k-i)(2k-2i-1)^2,{i,0,(j-3)/2}]/(2k-j)/(2k-1)/(n((2k-j-2)(3-2k)-1)+1)];

7 AB:=Function[{j,k},For[c=1,c<=2,r=N[NumQ[j,k,c],2000];v=FindIntegerNullVector[{1,r,N[Catalan*r,2000]}];d[c]=(-g[j,k,c]([v[[3]]Catalan+v[[2]])]/v[[1]]-f[j,k,c]Catalan)/h[j,k,c];c++];{d[1],d[2]-d[1]}];ABlists:=Function[lim,z[1]=z[0]={};

8 For[j=3,j<=lim,For[w=0,w<=1,e[w]=Table[AB[j,k].{1,w}l[w,j,k},{k,j-2,j-1}];

9 ab[w]={{a,b}}/.Solve[Table[a+b(u-1)==e[w][u],{u,1,2}],{a,b}];AppendTo[z[w],ab[w][[1]]];w++;j+=2];{z[1],z[0]}];

10 Print[ABlists[15]];ClearAll["Global*"];

1 (* code snippet "6. symmetry" *)

2 NumQ[j_,k_,c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n),3n^2+(3+4k)n+j(2-j+2k),{n,1,15000}];

3 g[j_,k_,c_]:=Product[(2c-2i+1)(2k-2i+1),{i,1,(j-1)/2}](2c)!2^(2k+(j-7)/2);

4 h[j_,k_,c_]:=Product[2c-2i-1,{i,0,(j-3)/2}]Product[2c-2i-1,{i,0,k-1}];

5 zeta[j_,u_]:=Product[2+2i-j,{i,0,u-1}]/2^u;Ratio:=Function[{j,k,c},r=N[NumQ[j,k,c],2000];

6 v=FindIntegerNullVector[{1,r,N[Catalan*r,2000]}];-v[[1]]/v[[2]]];

7 GenAB[j_,k_]:={{a,b}}/.Solve[Table[{a,b}.{1,c-1}==g[j,k,c]/Ratio[j,k,c]/h[j,k,c],{c,1,2}],{a,b}))[[1]];

8 tau[j_,u_]:=If[u>(j-1)/2,Sign[zeta[j,u]](2j-2u-3)!!(2u-j)!!(-2)^u,zeta[j,u](-4)^u];

9 For[j=3,j<=13,Print[MatrixForm[Table[{j,j-u-1},{j-2u,j-u-1},(GenAB[j,j-u-1]/GenAB[j-2u,j-u-1])/tau[j,u]={{1,1}},{u,1,j+3}]]];j+=2];ClearAll["Global*"];

1 (* code snippet "7. Croatia" *)

2 NumQ[j_,k_,c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n),3n^2+(3+4k)n+j(2-j+2k),{n,1,2000}];

3 g[j_,k_,c_]:=Product[(2c-2i+1)(2k-2i+1),{i,1,(j-1)/2}](2c)!2^(2k+(j-7)/2);

4 h[j_,k_,c_]:=Product[2c-2i-1,{i,0,(j-3)/2}]Product[2c-2i-1,{i,0,k-1}];

5 Ratio:=Function[{j,k,c},r=N[NumQ[j,k,c],4000];v=FindIntegerNullVector[{1,r,N[Catalan*r,4000}];-v[[1]]/v[[2]]];

6 GenAB[j_,k_]:={{a,b}}/.Solve[Table[{a,b}.{1,c-1}==g[j,k,c]/Ratio[j,k,c]/h[j,k,c],{c,1,2}],{a,b})][[1]];

7 psi1[i_,j_]:=ymphj-1,j-1,-464+58j-3j^2,27936-4692j+432j^2-15j^3,-2659968+542256j-67836j^2+4260j^3-105j
^4,367568640-86278560j+13203480j^2-1139700j^3+51450j^4-945j^5][[i]];

8 psi2[i_,j_]:=m4j-15,306-95j+4j^2,-13360+464j-357j^2+12j^3,999648-379692j+40368j^2-2457j^3+60j^4,-113885568+

9 46449360j-6124164j^2+513228j^3-22935j^4+420j^5,18333538560-7933530720j+1224286440j^2-126833100j^3+7864950j^4-266175j
^5+3780j^6][[i]];

10 mu[i_,j_]:=-Product[j-2q-2,{q,1,i}]/(-2)^((3j-11-4i)/2);
<figure><img src="image_18.png" /><figcaption>Fig. 19. The points on which the Kosovo-Serbia symmetry is tested at the $(\alpha_{j,\kappa}, \beta_{j,\kappa})$-level by the snippet "6. symmetry". Tested points are shown in red and their symmetrical correspondents in pink.</figcaption></figure>

```python
Print[Union[Table[Union[Table[GenAB[j, (j-3)/2-i]]={psi1[1+i, j], psi2[1+i, j]}/mu[i, j], {j, 5+2i, 37, 2}]}, {i, 0, 5}]]]];ClearAll
["Global'*"]
```
<figure><img src="image_20.png" /><figcaption>Fig. 19. The points on which the Kosovo-Serbia symmetry is tested at the $(\alpha_{j,\kappa}, \beta_{j,\kappa})$-level by the snippet "6. symmetry". Tested points are shown in red and their symmetrical correspondents in pink.</figcaption></figure>

```python
Print[Union[Table[Union[Table[GenAB[j, (j-3)/2-i]]=[psi1[1+i,j], psi2[1+i,j]]/mu[i,j], {j,5+2i,37,2}]], {i,0,5}]];ClearAll
["Global'*"]
```
<figure><img src="image_19.png" /><figcaption>Fig. 20. The points on which Croatia is tested. Note that the test is performed directly at the $(\alpha_{j,\kappa}, \beta_{j,\kappa})$-level. Points tested by the snippet "7. Croatia" are shown in red. Points tested beyond the quadrant are not shown.</figcaption></figure>

1. (* code snippet "8. coefficients" *)

2. coef=CoefficientList[{3198013886925+(145296572850+(5207427225+(4353102000+(877052475+(78210090+(3023055+41580(X-29))(X-27))(X-25))(X-23))(X-21))(X-19))(X-17), -14487726825-(104826150+(452605725+(121200300+(13697775+(640710+10395(X-27))(X-25))(X-23))(X-21))(X-19))(X-17)}, X]; Print[Table[GCD[coef[[1, i]], coef[[2, i]]], {i, 1, 7}]]; ClearAll["Global'*"];

1. (* code snippet "9. ratio" *)

2. NumQ[j_, k_, c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n), 3n^2+(3+4k)n+j(2-j+2k), {n, 1, 50000}];

3. For[j=1, j<=7, For[k=Abs[j-2], k<=7, For[c=1, c<=7, r=N[NumQ[j, k, c], 3000]]; v=FindIntegerNullVector[{1, r, N[Catalan*r, 3000]}];

4. q=(-v[[1]])/v[[3]]=((2c)!2^(2k+(j-7)/2+Floor[1/j])/CatalanNumber[k-1]/(2k-1)/(2c-1))!!^2/(j-2)/Product[(2c-2k+2i-1)

5. 1)(k-i+1)/(2c-2i+1)/(2k-2i+1), {i, 1, (j-1)/2}/CatalanNumber[(j-3)/2]); Print[{j, k, c, q}]; c++; k++; j+=2]; ClearAll["Global'*");
<figure><img src="image_21.png" /><figcaption>**Fig. 20.** The points on which Croatia is tested. Note that the test is performed directly at the $(\alpha_{j,\kappa}, \beta_{j,\kappa})$-level. Points tested by the snippet "7. Croatia" are shown in red. Points tested beyond the quadrant are not shown.</figcaption></figure>

1. (* code snippet "8. coefficients" *)

2. `coef=CoefficientList[{3198013886925+(145296572850+(5207427225+(4353102000+(877052475+(78210090+(3023055+41580(X-29))(X-27))(X-25))(X-23))(X-21))(X-19))(X-17), -14487726825-(104826150+(452605725+(121200300+(13697775+(640710+10395(X-27))(X-25))(X-23))(X-21))(X-19))(X-17)}, X]; Print[Table[GCD[coef[[1, i]], coef[[2, i]]], {i, 1, 7}]]; ClearAll["Global'*"];

1. (* code snippet "9. ratio" *)

2. `NumQ[j_, k_, c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n), 3n^2+(3+4k)n+j(2-j+2k), {n, 1, 50000}];`

3. `For[j=1, j<=7, For[k=Abs[j-2], k<=7, For[c=1, c<=7, r=N[NumQ[j, k, c], 3000]]; v=FindIntegerNullVector[{1, r, N[Catalan*r, 3000]}];`

4. `q=(-v[[1]])/v[[3]]=((2c)!2^(2k+(j-7)/2+Floor[1/j])/CatalanNumber[k-1]/(2k-1)/(2c-1))!!^2/(j-2)/Product[(2c-2k+2i-1)`

5. `1)(k-i+1)/(2c-2i+1)/(2k-2i+1), {i, 1, (j-1)/2}/CatalanNumber[(j-3)/2]); Print[{j, k, c, q}]; c++; k++; j+=2]; ClearAll["Global'*"]`
<figure><img src="image_20.png" /><figcaption>Fig. 21. The points on which the ratio conjecture is tested for $1 \le c \le 7$ by the snippet "9. ratio" are shown in red.</figcaption></figure>

1. (* code snippet "10. log2-a" *)
2. Q[c_]:=c+ContinuedFractionK[-2n^2, 3n+c, {n, 1, 2000}];
3. For[c=2, c<=25, r=N[Q[c], 80];
4. v=FindIntegerNullVector[{1, r, N[Log[2]*r, 80]}];
5. Print[{c, Simplify[2/LerchPhi[1/2, 1, -1+c]] == v[[1]] / (v[[2]] + Log[2]*v[[3]]) == 1 / (Log[2]*2^(c-2) - Sum[2^(c-j-2)/j, {j, c-2}])}];
   c++]; ClearAll["Global*"];

1. (* code snippet "11. log2-b"*)
2. R[c_]:=c+ContinuedFractionK[-2n^2-2n, 3n+c, {n, 1, 2000}];
3. For[c=3, c<=20, r=N[R[c], 200]; v=FindIntegerNullVector[{1, r, N[Log[2]*r, 200]}];
4. Print[{c, 2^(c-4)*(c-3)*v[[1]]} == v[[3]], -v[[1]] / (v[[2]] + Log[2]*v[[3]])]; c++]; ClearAll["Global*"];

1. (* code snippet "12. Inostranstvol" *)
<figure><img src="image_22.png" /><figcaption>Fig. 21. The points on which the ratio conjecture is tested for $1 \le c \le 7$ by the snippet "9. ratio" are shown in red.</figcaption></figure>

1. (* code snippet "10. log2-a" *)
2. Q[c_]:=c+ContinuedFractionK[-2n^2, 3n+c, {n, 1, 2000}];
3. For[c=2, c<=25, r=N[Q[c], 80];
4. v=FindIntegerNullVector[{1, r, N[Log[2]*r, 80]}];
5. Print[{c, Simplify[2/LerchPhi[1/2, 1, -1+c]]==v[[1]]/(v[[2]]+Log[2]v[[3]])==1/(Log[2]2^(c-2)-Sum[2^(c-j-2)/j, {j, c-2}])}];
   c++];ClearAll["Global'*"];

1. (* code snippet "11. log2-b"*)
2. R[c_]:=c+ContinuedFractionK[-2n^2-2n, 3n+c, {n, 1, 2000}];
3. For[c=3, c<=20, r=N[R[c], 200]; v=FindIntegerNullVector[{1, r, N[Log[2]*r, 200]}];
4. Print[{c, 2^(c-4)*(c-3)*v[[1]]==v[[3]], -v[[1]]/(v[[2]]+Log[2]v[[3]])}]; c++];ClearAll["Global'*"];

1. (* code snippet "12. Inostranstvol" *)
2 `NumQ[j_]:=7+6j+ContinuedFractionK[-2n(n+1)^2(n+2j+1), 3n^2+(9+4j)n+(7+6j), {n,1,10000}];`

3 `D1:=Function[i, If[i<2,2+15i, 2(2i-1)^3(1-i)*D1[i-2]+(8i^2-2i+3)D1[i-1]]]; For[j=0, j<=14, r=N[NumQ[j], 80];`

4 `v=FindIntegerNullVector[{1, r, N[Catalan*r, 80]}]; Q=-v[[1]]/(v[[2]]+Catalan v[[3]]); Print[{j, (2*j+1)!(D1[j]-2 Catalan (2*j+1))!!^2==Q}]; j++]; ClearAll["Global*"];`

1 (* code snippet "13. Inostranstvo2" *)

2 `NumQ[j_]:=23+10j+ContinuedFractionK[-2n(n+1)(n+3)(n+2j+3), 3n^2+(17+4j)n+(23+10j), {n,1,30000}];`

3 `For[j=0, j<=30, r=N[NumQ[j], 180]; v=FindIntegerNullVector[{1, r, N[Catalan*r, 180]}]; Print[{j, (2j+5)!(2j+4)/(2j+5))!!^2==v[[1]]/v[[3]]}]; j++]; ClearAll["Global*"];`

1 (* code snippet "14. Inostranstvo" *)

2 `NumQ:=Function[{i, a, b, c}, z=a+b+1; x=a*b+(c+1)z; x+2z*i+ContinuedFractionK[-2n(n+a)(n+b)(n+2i+c),`

3 `3n^2+(2z+2c+1+4i)n+(x+2z*i), {n, 1, 5000}]]; lim=6; ToV:=Function[{i, a, b, c}, r=N[NumQ[i, a, b, c], 80]; v=FindIntegerNullVector[{1, r, N[Catalan*r, 80]}];`

4 `Q=-v[[1]]/(v[[2]]+Catalan v[[3]])]; Table[ToV[i, a+p, b+p, c+p], {p, 0, 1}, {a, 0, lim, 2}, {i, 0, lim}, {c, 0, lim, 2}, {b, a, lim, 2}]`

1 (* code snippet "15. Series" *)

2 `all={{Pi, {0,0,2,2,2,2,0}, {57153600, -33075, 103904}, 0}, {Pi, {0,1,1,2,2,1,1}, {69854400, -24255, 76192}, 0}, {Pi, {0,1,2,1,1,2,1}, {62868960, -14553, 45712}, 0}, {Pi, {0,2,2,1,2,2,0}, {-65318400, -4725, 14848}, 0}, {Pi, {0,2,2,2,2,0,0}, {-6350400, -3675, 11552}, 0}, {Pi, {1,1,2,2,1,1,0}, {-907200, -315, 992}, 0}, {Pi, {1,2,1,1,2,1,0}, {-635040, -147, 464}, 0}, {Pi, {1,2,2,2,2,1,0}, {16934400, -245, 768}, 0}, {Pi, {2,1,1,1,1,2,0}, {-59535000, -6615, 21292}, 0}, {Pi, {2,2,0,0,2,1,1}, {-93139200, -2695, 9344}, 0}, {Pi, {2,2,0,0,2,2,0}, {-52920000, -2695, 9056}, 0}, {Pi, {2,2,1,2,2,0,0}, {-50803200, 3675, -11264}, 0}, {Pi, {2,2,2,2,0,0,0}, {129600, -75, 224}, 0}, {Catalan, {0,1,2,2,2,1,0}, {-50803200, 66150, -60577}, 0}, {Catalan, {0,3,2,3,0,0,0}, {-3456000, -6750, 6197}, 0}, {Catalan, {1,2,2,2,1,0,0}, {-2419200, -3150, 2909}, 0}, {Catalan, {1,3,0,3,1,0,0}, {8064000, 8750, -8109}, 0}, {Catalan, {2,0,2,0,2,0,0}, {-33868800, -22050, 21131}, 0}, {Catalan, {3,2,3,0,0,0,0}, {-27648, 54, -25}, 0}, {Log[2], {0,0,2,2,2,2,2}, {-33177600, -38400, 26617}, 1}, {Log[2], {0,0,2,2,2,2,0}, {1382400, -1600, 1109}, 1}, {Log[2], {0,1,1,2,2,1,1}, {11059200, -7680, 5323}, 1}, {Log[2], {0,1,2,1,1,2,1}, {-22118400, 10240, -7097}, 1}, {Log[2], {0,2,2,0,0,2,2}, {17280000, -1760, 1219}, 1}, {Log[2], {0,2,2,2,2,0,0}, {442368, 512, -355}, 1}, {Log[2], {1,1,2,0,0,2,2}, {-88473600, 5120, -3539}, 1}, {Log
[2], {1,1,2,2,1,1,0}, {-230400, -160, 111}, 1}, {Log[2], {1,2,1,1,2,1,0}, {-22118400, -10240, 7109}, 1}, {Log
[2], {2,2,0,0,2,1,1}, {-88473600, -5120, 3627}, 1}, {Log[2], {2,2,0,0,2,2,0}, {69120000, 7040, -4951}, 1}, {Log
[2], {2,2,1,2,2,0,0}, {7077888, -1024, 707}, 1}, {Log[2], {2,2,2,2,0,0,0}, {-13824, 16, -11}, 1}};
3 G:=Function[{e,ep},(-1)^(n+1)/Product[(2n+2i-3+ep)^e[[i]},{i,1,Length[e]}]];F:=Function[{e,ep},Sum[G[e,ep]/.{n->j},{j
,1,100000}]];Union[Table[{F[all[[i,2]],all[[i,4]]],all[[i,1]],1}.all[[i,3]]<10^30,{i,1,Length[all]}]]]

1 (* code snippet "16. files" *)

2 NumQ[j_,k_,c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n),3n^2+(3+4k)n+j(2-j+2k),{n,1,17000}];

3 f[j_,k_,c_]:=Product[(2c-2k+2i-1)(k-i+1),{i,1,(j-1)/2}]CatalanNumber[(j-3)/2](j-2)(2k-1)*(2c-1)!!^2*CatalanNumber[k-1];

4 g[j_,k_,c_]:=Product[(2c-2i+1)(2k-2i+1),{i,1,(j-1)/2}](2c)!2^(2k+(j-7)/2);

5 h[j_,k_,c_]:=Product[2c-2i-1,{i,0,(j-3)/2}]*Product[2c-2i-1,{i,0,k-1}];

6 l:=Function[{n,j,k},(-1)^(k+1)(2k)!^2/k!^2^(3k-2)/Product[(k-i)(2k-2i-1)^2,{i,0,(j-3)/2}]/(2k-j)/(2k-1)/(n((2k-j-2)(3-2
k)-1)+1)];

7 AB:=Function[{j,k},

8 For[c=1,c<=2,r=N[NumQ[j,k,c],5000];

9 v=FindIntegerNullVector[{1,r,N[Catalan*r,5000]}];

10 d[c]=(-g[j,k,c](v[[3]])Catalan+v[[2]])/v[[1]]-f[j,k,c]Catalan)/h[j,k,c];c++;{d[1],d[2]-d[1]}];

11 z[1]=z[0]={};

12 For[j=3,j<=501,For[w=0,w<=1,e[w]=Table[AB[j,k].{1,w}*l[w,j,k],{k,j-2,j-1}]];

13 ab[w]={a,b}/.Solve[Table[a+b(u-1)==e[w][u]],{u,1,2}],{a,b}];

14 AppendTo[z[w],ab[w][[1]]];w++;j+=2;

15 z[1]>>file1.txt;

16 z[0]>>file2.txt;

17 ClearAll["Global*"];

1 (*code snippet "17. j-level"*)

2 l[1]=<<file1.txt;l[2]=<<file2.txt;

3 data[j_,x_,y_]:=l[x][[(j-1)/2,y]];

4 rho[j_]:=2^j CatalanNumber[1/2(j-3)]((1/2(j-1))!)^2;

5 th1[j_]:=(j-6)(j-4)(j-2)(j-1)j(2j-7)(2j-5)(-1-3j+2j^2);

6 th2[j_]:=4(-390+312j+653j^2-942j^3+442j^4-87j^5+6j^6);
[2], {1,1,2,2,1,1,0}, {-230400, -160, 111}, 1}, {Log[2], {1,2,1,1,2,1,0}, {-22118400, -10240, 7109}, 1}, {Log
[2], {2,2,0,0,2,1,1}, {-88473600, -5120, 3627}, 1}, {Log[2], {2,2,0,0,2,2,0}, {69120000, 7040, -4951}, 1}, {Log
[2], {2,2,1,2,2,0,0}, {7077888, -1024, 707}, 1}, {Log[2], {2,2,2,2,0,0,0}, {-13824, 16, -11}, 1}};
3 G:=Function[{e,ep},(-1)^(n+1)/Product[(2n+2i-3+ep)^e[[i]},{i,1,Length[e]}]];F:=Function[{e,ep},Sum[G[e,ep]/.{n->j},{j
,1,100000}]];Union[Table[{F[all[[i,2]],all[[i,4]]],all[[i,1]],1}.all[[i,3]]<10^30,{i,1,Length[all]}]]]

(* code snippet "16. files" *)

NumQ[j_,k_,c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n),3n^2+(3+4k)n+j(2-j+2k),{n,1,17000}];

f[j_,k_,c_]:=Product[(2c-2k+2i-1)(k-i+1),{i,1,(j-1)/2}]CatalanNumber[(j-3)/2](j-2)(2k-1)*(2c-1)!!^2*CatalanNumber[k-1];

g[j_,k_,c_]:=Product[(2c-2i+1)(2k-2i+1),{i,1,(j-1)/2}](2c)!2^(2k+(j-7)/2);

h[j_,k_,c_]:=Product[2c-2i-1,{i,0,(j-3)/2}]*Product[2c-2i-1,{i,0,k-1}];

l:=Function[{n,j,k},(-1)^(k+1)(2k)!^2/k!^2^(3k-2)/Product[(k-i)(2k-2i-1)^2,{i,0,(j-3)/2}]/(2k-j)/(2k-1)/(n((2k-j-2)(3-2
k)-1)+1)];

AB:=Function[{j,k},

For[c=1,c<=2,r=N[NumQ[j,k,c],5000];

v=FindIntegerNullVector[{1,r,N[Catalan*r,5000]}];

d[c]=(-g[j,k,c](v[[3]])Catalan+v[[2]])/v[[1]]-f[j,k,c]Catalan)/h[j,k,c];c++];{d[1],d[2]-d[1]}];

z[1]=z[0]={};

For[j=3,j<=501,For[w=0,w<=1,e[w]=Table[AB[j,k].{1,w}*l[w,j,k],{k,j-2,j-1}]];

ab[w]>{a,b}/.Solve[Table[a+b(u-1)==e[w][u]],{u,1,2}],{a,b}];

AppendTo[z[w],ab[w][[1]]];w++;j+=2;

z[1]>>file1.txt;

z[0]>>file2.txt;

ClearAll["Global'*"];

(*code snippet "17. j-level"*)

l[1]=<<file1.txt;l[2]=<<file2.txt;

data[j_,x_,y_]:=l[x][[(j-1)/2,y]];

rho[j_]:=2^j CatalanNumber[1/2(j-3)]((1/2(j-1))!)^2;

th1[j_]:=(j-6)(j-4)(j-2)(j-1)j(2j-7)(2j-5)(-1-3j+2j^2);

th2[j_]:=4(-390+312j+653j^2-942j^3+442j^4-87j^5+6j^6);
```pascal
th3[j_]:=(j-6)(j-4)^2(j-1)(1+j)(13-11j+2j^2);

test1=Union[Table(data[j+2,1,2]==4(data[j,1,2]*th1[j]-th2[j]*rho[j])/th3[j], {j,3,499,2})];

test2=Union[Table(data[j+2,1,1] == (4(data[j,1,1]j(j-6)(j-4)(j-2)(j-1)(2j-7)(2j-5)-12(2-4j+j^2)rho[j]))/((j-6)(j-4)^2(j^2-1)), {j,3,499,2})];

test3=Union[Table(data[j+2,2,2] == (4(data[j,2,2]j(2j+1)(j-1)(2j-5)(j-4)-(18+11j-17j^2+3j^3)rho[j]))/((j-4)(j-3)(1+j)), {j,5,499,2})];

test4=Union[Table(data[j+2,2,1] == (4(data[j,2,1](j-4)(j-1)j(2j-5)(2j-3)-(3j-2)rho[j]))/((j-4)(j-1)(1+j)), {j,3,499,2})];

Print[Union[Join(test1, test2, test3, test4)]];

ClearAll["Global'*"];
```

```pascal
(*code snippet "18. Altogether"*)

NumQ[j_,k_,c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n),3n^2+(3+4k)n+j(2-j+2k),{n,1,5000}];

rho[j_]:=2^(j-3)(j-5)!(j-3)/(j-6);

p[j_,r_,i_]:={(j-3)(j-2)(2j-9)(2j-r), (j-1)(j-r), (j-8)(j-r), (2j-7)(j-r)}[[i]];

aa[j_]:=If[j==3,-1,(4(aa[j-2]p[j,7,1]-(3j-8)rho[j]))/p[j,3,2]];

ab[j_]:=If[j<=5,115j-341,4(ab[j-2]p[j,3,1]-(3j^3-35j^2+115j-96)rho[j])/p[j,5,2]];

ba[j_]:=If[j==3,-1/3,(4(ba[j-2]p[j,11,1]p[j,4,3]-12(p[j,7,2]+7)rho[j]))/(p[j,6,3]p[j,3,2])];

bb[j_]:=If[j==3,-14/3,4(bb[j-2]p[j,11,1]p[j,4,3](p[j,2,4]-1)-4(6x^6-15x^5-68x^4+74x^3+89x^2-44x-18/x->j-4)rho[j])/p[j,3,2]p[j,6,3](p[j,6,4]+1))];

vD[z_,j_,k_,c_]:=If[c<2,z.{1,c},2c(j-2c)(2c-2k+j-2)(2c-2k-1)vD[z,j,k,c-2]+(8c^2+(2-8k)c+(j-2)(2k-j))vD[z,j,k,c-1]];

f[j_,k_,c_]:=Product[(2c-2k+2i-1)(k-i+1),{i,1,(j-1)/2}]CatalanNumber[(j-3)/2](j-2)(2k-1)(2c-1)!!^2*CatalanNumber[k-1];

g[j_,k_,c_]:=Product[2(2c-2i+1)(2k-2i+1),{i,1,(j-1)/2}](2c)!2^(2k-3);

h[j_,k_,c_]:=Product[2c-2i-1,{i,0,(j-3)/2}]Product[2c-2i-1,{i,0,k-1}];

l[n_,j_,k_]:=(2k)!2/(k!(-2)^(3k-2)Product[(k-i)(2k-2i-1)^2,{i,0,(j-3)/2}](j-2k)(2k-1)(n((2k-j-2)(3-2k)-1)+1));

kD[n_,z_,j_,k_]:=If[k<2,z.{1,k},(2k+2j-9-2n)(2k+j-8-2n)(5-2k-j)(2k+j-6)kD[n,z,j,k-2]+(8k^2+k(10j-48-8n)+(3j^2-(28+4n)j+68+18n))kD[n,z,j,k-1]];

M:=Function[{j,k},t=Table[kD[u,{aa[j],ab[j]},{ba[j],bb[j}]][[u+1]],j,k-j+2]/l[u,j,k],{u,0,1}];{t[[1]],t.{-1, 1}}];

Q[j_,k_,c_,z_]:=Together[g[j,k,c]/(vD[z,j,k,c-1]h[j,k,c]+f[j,k,c]Catalan)];

Print[Union[Flatten[Table[N[{NumQ[j,k,c]}==Q[j,k,c,M[j,k]}],200],{j,3,11,2},{k,j-2,10},{c,1,7}]]]];
```

```pascal
(*code snippet "19. j-level bis"*)
```
7 `th3[j_]:=(j-6)(j-4)^2(j-1)(1+j)(13-11j+2j^2);`
8 `test1=Union[Table[data[j+2,1,2]]==4(data[j,1,2]*th1[j]-th2[j]*rho[j])/th3[j],{j,3,499,2}]]];`
9 `test2=Union[Table[data[j+2,1,1]]=(4(data[j,1,1]j(j-6)(j-4)(j-2)(j-1)(2j-7)(2j-5)-12(2-4j+j^2)rho[j]))/((j-6)(j-4)^2(j^2-1)),{j,3,499,2}]]];`
10 `test3=Union[Table[data[j+2,2,2]]=(4(data[j,2,2]j(2j+1)(j-1)(2j-5)(j-4)-(18+11j-17j^2+3j^3)rho[j]))/((j-4)(j-3)(1+j)),{j,5,499,2}]]];`
11 `test4=Union[Table[data[j+2,2,1]]=(4(data[j,2,1](j-4)(j-1)j(2j-5)(2j-3)-(3j-2)rho[j]))/((j-4)(j-1)(1+j)),{j,3,499,2}]]];`
12 `Print[Union[Join(test1,test2,test3,test4]]];`
13 `ClearAll["Global'*");`

1 (*code snippet "18. Altogether"*)
2 `NumQ[j_,k_,c_]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n),3n^2+(3+4k)n+j(2-j+2k),{n,1,5000}];`
3 `rho[j_]:=2^(j-3)(j-5)!(j-3)/(j-6);`
4 `p[j_,r_,i_]:={(j-3)(j-2)(2j-9)(2j-r),(j-1)(j-r),(j-8)(j-r),(2j-7)(j-r)}[[i]];`
5 `aa[j_]:=If[j==3,-1,(4(aa[j-2]p[j,7,1]-(3j-8)rho[j]))/p[j,3,2]];`
6 `ab[j_]:=If[j<=5,115j-341,4(ab[j-2]p[j,3,1]-(3j^3-35j^2+115j-96)rho[j])/p[j,5,2]];`
7 `ba[j_]:=If[j==3,-1/3,(4(ba[j-2]p[j,11,1]p[j,4,3]-12(p[j,7,2]+7)rho[j]))/(p[j,6,3]p[j,3,2]);`
8 `bb[j_]:=If[j==3,-14/3,4(bb[j-2]p[j,11,1]p[j,4,3](p[j,2,4]-1)-4(6x^6-15x^5-68x^4+74x^3+89x^2-44x-18/x->j-4)rho[j])/p[j,3,2]p[j,6,3](p[j,6,4]+1))];`
9 `vD[z_,j_,k_,c_]:=If[c<2,z.{1,c},2c(j-2c)(2c-2k+j-2)(2c-2k-1)vD[z,j,k,c-2]+(8c^2+(2-8k)c+(j-2)(2k-j))vD[z,j,k,c-1]];`
10 `f[j_,k_,c_]:=Product[(2c-2k+2i-1)(k-i+1),{i,1,(j-1)/2}]CatalanNumber[(j-3)/2](j-2)(2k-1)(2c-1)!!^2*CatalanNumber[k-1];`
11 `g[j_,k_,c_]:=Product[2(2c-2i+1)(2k-2i+1),{i,1,(j-1)/2}](2c)!2^(2k-3);`
12 `h[j_,k_,c_]:=Product[2c-2i-1,{i,0,(j-3)/2}]Product[2c-2i-1,{i,0,k-1}];`
13 `l[n_,j_,k_]:=(2k)!^2/(k!(-2)^(3k-2)Product[(k-i)(2k-2i-1)^2,{i,0,(j-3)/2}](j-2k)(2k-1)(n((2k-j-2)(3-2k)-1)+1));`
14 `kD[n_,z_,j_,k_]:=If[k<2,z.{1,k},(2k+2j-9-2n)(2k+j-8-2n)(5-2k-j)(2k+j-6)kD[n,z,j,k-2]+(8k^2+k(10j-48-8n)+(3j^2-(28+4n)j+68+18n))kD[n,z,j,k-1]];`
15 `M:=Function[{j,k},t=Table[kD[u,{aa[j],ab[j]},{ba[j],bb[j]}][[u+1]],j,k-j+2]/l[u,j,k],{u,0,1}];{t[[1]],t.{-1,1}}];`
16 `Q[j_,k_,c_,z_]:=Together[g[j,k,c]/(vD[z,j,k,c-1]h[j,k,c]+f[j,k,c]Catalan)];`
17 `Print[Union[Flatten[Table[N[{NumQ[j,k,c]}==Q[j,k,c,M[j,k]}],200],{j,3,11,2},{k,j-2,10},{c,1,7}]]]];`

1 (*code snippet "19. j-level bis"*)
2 `{aa[3],ab[3],ba[3],bb[3],aa[5],ab[5],ba[5],bb[5],rho[j_],b[j_],a[j_],p[j_],d[j_],e[j_]}]:={-1,4,-1/3,-14/3,19,234,-17,-8,2^{j+1}(j-1)!/((j-2)(j-4)),(j-6)(j-2)(j-1)j(2j-7)(2j-5),4(j-1)j(2j-5),(j-6)(j-4)(j-1)(j+1)/4,6j^6-15j^5-68j^4+74j^3+89j^2-44j-18,(3j+1)(j^2-7)+3}`;

3 `For[j=5,j<=17,Print[{aa[j+2],ab[j+2],ba[j+2],bb[j+2]}={(aa[j]]a[j](2j-3)-(3j-2)rho[j])/((j-1)(j+1)),(ab[j]]a[j](2j+1)-e[j-2]rho[j])/((j-3)(j+1)),(ba[j]]b[j]-3((j-3)(j-1)-1)rho[j])/p[j],(bb[j]]b[j](j(2j-3)-1)-d[j-2]rho[j])/(p[j]((j-2)(2j-7)-1))]];j+=2]`
2 `{aa[3],ab[3],ba[3],bb[3],aa[5],ab[5],ba[5],bb[5],rho[j_],b[j_],a[j_],p[j_],d[j_],e[j_]}':={-1,4,-1/3,-14/3,19,234,-17,-8,2^{j+1}(j-1)!/((j-2)(j-4)),(j-6)(j-2)(j-1)j(2j-7)(2j-5),4(j-1)j(2j-5),(j-6)(j-4)(j-1)(j+1)/4,6j^6-15j^5-68j^4+74j^3+89j^2-44j-18,(3j+1)(j^2-7)+3};

3 `For[j=5,j<=17,Print[{aa[j+2],ab[j+2],ba[j+2],bb[j+2]}={(aa[j]]a[j](2j-3)-(3j-2)rho[j])/((j-1)(j+1)),(ab[j]]a[j](2j+1)-e[j-2]rho[j])/((j-3)(j+1)),(ba[j]]b[j]-3((j-3)(j-1)-1)rho[j])/p[j],(bb[j]]b[j](j(2j-3)-1)-d[j-2]rho[j])/p[j]((j-2)(2j-7)-1))]}];j+=2]`
## C Altogether

This appendix recaps the entire algorithm. It takes as input $j, \kappa, c$ and returns the exact expression of $Q_{j,\kappa,c}$ where:

$$Q_{j,\kappa,c} = j(2-j+2\kappa) + \sum_{n=1}^{\infty} \left( \frac{-2n(c+n)(j+n-1)(1-j+2\kappa+n)}{j(2-j+2\kappa)+(3+4\kappa)n+3n^2} \right)$$

### C.1 If $j \ge 2\kappa + 3$

In this case $(j, \kappa, c) \in \text{Bosnia & Herzegovina} \cup \text{Croatia}$, hence $Q_{j,\kappa,c}$ is computed by straightforward finite summation.

$$Q_{j,\kappa,c} = j(2-j+2\kappa) + \sum_{n=1}^{j-2\kappa-1} \left( \frac{-2n(c+n)(j+n-1)(1-j+2\kappa+n)}{j(2-j+2\kappa)+(3+4\kappa)n+3n^2} \right)$$

### C.2 If $3+\kappa \le j \le 2\kappa+1$

In this case we are in Serbia. We hence use the symmetry relation:

$$Q_{j,\kappa,c} = Q_{2(\kappa+1)-j,\kappa,c}$$

Indeed,

$$3+\kappa \le j \le 2\kappa+1 \Rightarrow 1 \le j' \le \kappa-1 \le \kappa+2 \Rightarrow (j', \kappa) \in \text{Montenegro} \cup \text{Kosovo}$$

Replace $j$ by $j' = 2(\kappa+1)-j$ and compute $Q_{j',\kappa,c}$ using the formula for Montenegro or for Kosovo given in the next subsections.

### C.3 If $j=1$

In this case we are in Montenegro. We hence define:

$$\Delta_{\kappa,c}(\alpha, \beta) = \begin{cases} \alpha + \beta c & \text{if } c < 2 \\ -2c(2c-1)(2(c-\kappa)-1)^2 \Delta_{\kappa,c-2}(\alpha, \beta) & \text{if } c \ge 2 \\ + (8c^2 + (2-8\kappa)c - 2\kappa + 1) \Delta_{\kappa,c-1}(\alpha, \beta) & \end{cases}$$

$$\Gamma_{\kappa,c}(\alpha, \beta) = (2c-1)!!^2 G + \Delta_{\kappa,c-1}(\alpha, \beta) \cdot \prod_{i=0}^{\kappa-1} (2(c-i)-1)$$

$$\delta_{\kappa} = \frac{4^{\kappa-1}}{(2\kappa-1)C_{\kappa-1}} \quad \text{and} \quad \rho_{\kappa} = \frac{\delta_{\kappa}(-1)^{\kappa}(1-2\kappa)}{(2\kappa)!(2\kappa-3)!!}$$

$$\alpha_{\kappa} = \rho_{\kappa} \Delta_{1,\kappa-1}(1,-2) \quad \text{and} \quad \beta_{\kappa} = -\rho_{\kappa} (2\kappa-3)^2 \Delta_{2,\kappa-1}(1,12) - \alpha_{\kappa}$$

And output $Q_{1,\kappa,c} = \frac{\delta_{\kappa}(2c)!}{\Gamma_{\kappa,c}(\alpha_{\kappa}, \beta_{\kappa})}$
## C Altogether

This appendix recaps the entire algorithm. It takes as input $j, \kappa, c$ and returns the exact expression of $Q_{j,\kappa,c}$ where:

$$Q_{j,\kappa,c} = j(2-j+2\kappa) + \sum_{n=1}^{\infty} \left( \frac{-2n(c+n)(j+n-1)(1-j+2\kappa+n)}{j(2-j+2\kappa)+(3+4\kappa)n+3n^2} \right)$$

### C.1 If $j \ge 2\kappa + 3$

In this case $(j, \kappa, c) \in \text{Bosnia & Herzegovina} \cup \text{Croatia}$, hence $Q_{j,\kappa,c}$ is computed by straightforward finite summation.

$$Q_{j,\kappa,c} = j(2-j+2\kappa) + \sum_{n=1}^{j-2\kappa-1} \left( \frac{-2n(c+n)(j+n-1)(1-j+2\kappa+n)}{j(2-j+2\kappa)+(3+4\kappa)n+3n^2} \right)$$

### C.2 If $3+\kappa \le j \le 2\kappa+1$

In this case we are in Serbia. We hence use the symmetry relation:

$$Q_{j,\kappa,c} = Q_{2(\kappa+1)-j,\kappa,c}$$

Indeed,

$$3+\kappa \le j \le 2\kappa+1 \Rightarrow 1 \le j' \le \kappa-1 \le \kappa+2 \Rightarrow (j', \kappa) \in \text{Montenegro} \cup \text{Kosovo}$$

Replace $j$ by $j' = 2(\kappa+1)-j$ and compute $Q_{j',\kappa,c}$ using the formula for Montenegro or for Kosovo given in the next subsections.

### C.3 If $j=1$

In this case we are in Montenegro. We hence define:

$$\Delta_{\kappa,c}(\alpha, \beta) = \begin{cases} \alpha + \beta c & \text{if } c < 2 \\ -2c(2c-1)(2(c-\kappa)-1)^2 \Delta_{\kappa,c-2}(\alpha, \beta) & \text{if } c \ge 2 \\ + (8c^2 + (2-8\kappa)c - 2\kappa + 1) \Delta_{\kappa,c-1}(\alpha, \beta) & \end{cases}$$

$$\Gamma_{\kappa,c}(\alpha, \beta) = (2c-1)!!^2 G + \Delta_{\kappa,c-1}(\alpha, \beta) \cdot \prod_{i=0}^{\kappa-1} (2(c-i)-1)$$

$$\delta_{\kappa} = \frac{4^{\kappa-1}}{(2\kappa-1)C_{\kappa-1}} \quad \text{and} \quad \rho_{\kappa} = \frac{\delta_{\kappa}(-1)^{\kappa}(1-2\kappa)}{(2\kappa)!(2\kappa-3)!!}$$

$$\alpha_{\kappa} = \rho_{\kappa} \Delta_{1,\kappa-1}(1,-2) \quad \text{and} \quad \beta_{\kappa} = -\rho_{\kappa} (2\kappa-3)^2 \Delta_{2,\kappa-1}(1,12) - \alpha_{\kappa}$$

And output $Q_{1,\kappa,c} = \frac{\delta_{\kappa}(2c)!}{\Gamma_{\kappa,c}(\alpha_{\kappa}, \beta_{\kappa})}$
### C.4 If $3 \le j \le \kappa + 2$

$3 \le j \le \kappa + 2$ is in Kosovo. In which case we proceed in three steps:

**Step 1:**

If $j = 3$ or $j = 5$ let:

$$ \{\alpha_3, \beta_3, \alpha_3, \beta_3\} = \{-1, 4, -\frac{1}{3}, -\frac{14}{3}\} \text{ and } \{\alpha_5, \beta_5, \alpha_5, \beta_5\} = \{19, 234, -17, -8\} $$

If $j \ge 7$ define:

$$ \rho_j = \frac{2^{j+1}(j-1)!}{(j-2)(j-4)} \quad \text{and} \quad b_j = (j-6)(j-2)(j-1)j(2j-7)(2j-5) $$

$$ a_j = 4(j-1)j(2j-5) \quad \text{and} \quad p_j = \frac{(j-6)(j-4)(j-1)(j+1)}{4} $$

$$ d_j = 6j^6 - 15j^5 - 68j^4 + 74j^3 + 89j^2 - 44j - 18 \quad \text{and} \quad e_j = (3j+1)(j^2-7)+3 $$

and iterate using the following formulae to compute $\{\alpha_{j+2}, \beta_{j+2}, \alpha_{j+3}, \beta_{j+3}\}$:

$$ \alpha_{j+2} = \frac{\alpha_j a_j (2j-3) - (3j-2)\rho_j}{(j-1)(j+1)} \quad \text{and} \quad \beta_{j+2} = \frac{\beta_j a_j (2j+1) - e_{j-2}\rho_j}{(j-3)(j+1)} $$

$$ \beta_{j+2} = \frac{\beta_j b_j - 3((j-3)(j-1)-1)\rho_j}{p_j} \quad \text{and} \quad \beta_{j+2} = \frac{\beta_j b_j (j(2j-3)-1) - d_{j-2}\rho_j}{p_j((j-2)(2j-7)-1)} $$

**Step 2:** Define:

$$ s(j, \kappa) = \prod_{i=0}^{\frac{j-3}{2}} (\kappa - i)(2\kappa - 2i - 1)^2 $$

$$ \ell(n, j, \kappa) = \frac{(-1)^{\kappa+1} (2\kappa)!^2}{\kappa! 2^{3\kappa-2} (2\kappa-j)(2\kappa-1)(n((2\kappa-j-2)(3-2\kappa)-1)+1) \cdot s(j, \kappa)} $$

$$ \eta(n, j, \kappa) = (2\kappa + 2j - 9 - 2n)(2\kappa + j - 8 - 2n)(-2\kappa + 5 - j)(2\kappa + j - 6) $$

$$ \phi(n, j, \kappa) = 8\kappa^2 + \kappa(10j - 48 - 8n) + 3j^2 - (28 + 4n)j + 68 + 18n $$
### C.4 If $3 \le j \le \kappa + 2$

$3 \le j \le \kappa + 2$ is in Kosovo. In which case we proceed in three steps:

#### Step 1:

If $j = 3$ or $j = 5$ let:

$$ \{\alpha_3, \beta_3, \alpha_3, \beta_3\} = \{-1, 4, -\frac{1}{3}, -\frac{14}{3}\} \text{ and } \{\alpha_5, \beta_5, \alpha_5, \beta_5\} = \{19, 234, -17, -8\} $$

If $j \ge 7$ define:

$$ \rho_j = \frac{2^{j+1}(j-1)!}{(j-2)(j-4)} \quad \text{and} \quad b_j = (j-6)(j-2)(j-1)j(2j-7)(2j-5) $$

$$ a_j = 4(j-1)j(2j-5) \quad \text{and} \quad p_j = \frac{(j-6)(j-4)(j-1)(j+1)}{4} $$

$$ d_j = 6j^6 - 15j^5 - 68j^4 + 74j^3 + 89j^2 - 44j - 18 \quad \text{and} \quad e_j = (3j+1)(j^2-7)+3 $$

and iterate using the following formulae to compute $\{\alpha_{j+2}, \beta_{j+2}, \alpha_{j+3}, \beta_{j+3}\}$:

$$ \alpha_{j+2} = \frac{\alpha_j a_j (2j-3) - (3j-2)\rho_j}{(j-1)(j+1)} \quad \text{and} \quad \beta_{j+2} = \frac{\beta_j a_j (2j+1) - e_{j-2}\rho_j}{(j-3)(j+1)} $$

$$ \beta_{j+2} = \frac{\beta_j b_j - 3((j-3)(j-1)-1)\rho_j}{p_j} \quad \text{and} \quad \beta_{j+2} = \frac{\beta_j b_j (j(2j-3)-1) - d_{j-2}\rho_j}{p_j((j-2)(2j-7)-1)} $$

#### Step 2: Define:

$$ s(j, \kappa) = \prod_{i=0}^{\frac{j-3}{2}} (\kappa - i)(2\kappa - 2i - 1)^2 $$

$$ \ell(n, j, \kappa) = \frac{(-1)^{\kappa+1} (2\kappa)!^2}{\kappa! 2^{3\kappa-2} (2\kappa-j)(2\kappa-1)(n((2\kappa-j-2)(3-2\kappa)-1)+1) \cdot s(j, \kappa)} $$

$$ \eta(n, j, \kappa) = (2\kappa + 2j - 9 - 2n)(2\kappa + j - 8 - 2n)(-2\kappa + 5 - j)(2\kappa + j - 6) $$

$$ \phi(n, j, \kappa) = 8\kappa^2 + \kappa(10j - 48 - 8n) + 3j^2 - (28 + 4n)j + 68 + 18n $$
$$ \bar{\Delta}_{n,j,\kappa}(\alpha, \beta) = \begin{cases} \alpha + \beta\kappa & \text{if } \kappa < 2 \\ \eta(n,j,\kappa) \cdot \bar{\Delta}_{n,j,\kappa-2}(\alpha, \beta) + \phi(n,j,\kappa) \cdot \bar{\Delta}_{n,j,\kappa-1}(\alpha, \beta) & \text{if } \kappa \ge 2 \end{cases} $$

Using the values $\alpha_j, \beta_j, \alpha_{j-1}, \beta_{j-1}$ compute:

$$ \alpha_{j,\kappa} = \frac{\bar{\Delta}_{0,j,\kappa-j+2}(\alpha_j^\alpha, \beta_j^\beta)}{\ell(0, j, \kappa)} \quad \text{and} \quad \beta_{j,\kappa} = \frac{\bar{\Delta}_{1,j,\kappa-j+2}(\beta_j^\alpha, \beta_j^\beta)}{\ell(1, j, \kappa)} - \alpha_{j,\kappa} $$

**Step 3: Define:**

$$ \Delta_{j,\kappa,c} = \begin{cases} \alpha_{j,\kappa} + \beta_{j,\kappa}c & \text{if } c < 2 \\ -2c(2c-j)(2c-2\kappa+j-2)(2c-2\kappa-1)\Delta_{j,\kappa,c-2} & \text{if } c \ge 2 \\ + (8c^2 + (2-8\kappa)c + (j-2)(2\kappa-j))\Delta_{j,\kappa,c-1} & \text{if } c < 2 \end{cases} $$

$$ f_{j,\kappa,c} = C_{\frac{j-3}{2}} C_{\kappa-1} (j-2) (2\kappa-1) (2c-1)!!^2 \prod_{i=1}^{\frac{j-1}{2}} (2c-2\kappa+2i-1)(\kappa-i+1) $$

$$ g_{j,\kappa,c} = (2c)! 2^{\frac{j+4\kappa-7}{2}} \prod_{i=1}^{\frac{j-1}{2}} (2c - 2i + 1)(2\kappa - 2i + 1) $$

$$ h_{j,\kappa,c} = \prod_{i=0}^{\frac{j-3}{2}} (2c - 2i - 1) \prod_{i=0}^{\kappa-1} (2c - 2i - 1) $$

Output:

$$ Q_{j,\kappa,c} = \frac{g(j, \kappa, c)}{\Delta_{j,\kappa,c-1} \cdot h(j, \kappa, c) + f(j, \kappa, c) \cdot G} $$

Because the nontrivial part of the algorithm is Kosovo we only provide a code snippet ("18. Altogether") testing the final formulae over Kosovo. It seems amazing that such a short yet complex code perfectly captures exactly the behavior of $Q_{j,\kappa,c}$.

## D Lazy evaluation

A very useful trick allowing to speed-up the computation of $Q_{j,\kappa,c}$ is lazy evaluation. We illustrate it in the case of Kosovo.
$$ \bar{\Delta}_{n,j,\kappa}(\alpha, \beta) = \begin{cases} \alpha + \beta\kappa & \text{if } \kappa < 2 \\ \eta(n,j,\kappa) \cdot \bar{\Delta}_{n,j,\kappa-2}(\alpha, \beta) + \phi(n,j,\kappa) \cdot \bar{\Delta}_{n,j,\kappa-1}(\alpha, \beta) & \text{if } \kappa \ge 2 \end{cases} $$

Using the values $\alpha_j, \beta_j, \alpha_{j-1}, \beta_{j-1}$ compute:

$$ \alpha_{j,\kappa} = \frac{\bar{\Delta}_{0,j,\kappa-j+2}(\alpha_j^\alpha, \beta_j^\beta)}{\ell(0, j, \kappa)} \quad \text{and} \quad \beta_{j,\kappa} = \frac{\bar{\Delta}_{1,j,\kappa-j+2}(\beta_j^\alpha, \beta_j^\beta)}{\ell(1, j, \kappa)} - \alpha_{j,\kappa} $$

**Step 3: Define:**

$$ \Delta_{j,\kappa,c} = \begin{cases} \alpha_{j,\kappa} + \beta_{j,\kappa}c & \text{if } c < 2 \\ -2c(2c-j)(2c-2\kappa+j-2)(2c-2\kappa-1)\Delta_{j,\kappa,c-2} & \text{if } c \ge 2 \\ + (8c^2 + (2-8\kappa)c + (j-2)(2\kappa-j))\Delta_{j,\kappa,c-1} & \text{if } c < 2 \end{cases} $$

$$ f_{j,\kappa,c} = C_{\frac{j-3}{2}} C_{\kappa-1} (j-2) (2\kappa-1) (2c-1)!!^2 \prod_{i=1}^{\frac{j-1}{2}} (2c-2\kappa+2i-1)(\kappa-i+1) $$

$$ g_{j,\kappa,c} = (2c)! 2^{\frac{j+4\kappa-7}{2}} \prod_{i=1}^{\frac{j-1}{2}} (2c - 2i + 1)(2\kappa - 2i + 1) $$

$$ h_{j,\kappa,c} = \prod_{i=0}^{\frac{j-3}{2}} (2c - 2i - 1) \prod_{i=0}^{\kappa-1} (2c - 2i - 1) $$

Output:

$$ Q_{j,\kappa,c} = \frac{g(j, \kappa, c)}{\Delta_{j,\kappa,c-1} \cdot h(j, \kappa, c) + f(j, \kappa, c) \cdot G} $$

Because the nontrivial part of the algorithm is Kosovo we only provide a code snippet ("18. Altogether") testing the final formulae over Kosovo. It seems amazing that such as short yet complex code perfectly captures exactly the behavior of $Q_{j,\kappa,c}$.

## D Lazy evaluation

A very useful trick allowing to speed-up the computation of $Q_{j,\kappa,c}$ is lazy evaluation. We illustrate it in the case of Kosovo.
g, h, f are all smooth. Hence it is inefficient to compute them and then simplify the resulting fraction. Instead, it is best to keep for each of those functions a vector whose coordinates count the occurrence of small prime factors met during calculation – without actually multiplying integers.

When the three vectors $V_g, V_h, V_f$ are ready we can identify at each coordinate $i$ the value:

$$t_i = \min(V_g[i], V_h[i], V_f[i])$$

and update: $V_{\bullet}[i] = V_{\bullet}[i] - t_i$ before performing the actual multiplication.

Estimating the largest primes potentially appearing in each of the functions is easy. $f$'s largest prime is $\max(2\kappa - 1, 2c + j - 2)$, $g$'s is $\max(2\kappa - 1, 2c)$ and $h$'s is $2c - 1$. Hence it suffices to consider primes smaller than $\max(2\kappa - 1, 2c + j - 2)$.

Given that the coefficients of $Q_{j,\kappa,c}$ would explode for $j, \kappa, c$ values exceeding 1000 it suffices to tabulate the factorization of integers between 2 and 2998 and use this table to distribute factors in the vectors.

## References

1. D. H. Bailey and H. R. P. Ferguson. Numerical Results on Relations Between Numerical Constants Using a New Algorithm. Mathematics of Computation, 53(188):649–656, 1989.

2. G. Bana, W. Jamroga, D. Naccache, and P. Y. A. Ryan. Convergence Voting: From Pairwise Comparisons to Consensus. CoRR, abs/2102.01995, 2021.

3. H. Barral, Éric Brier, R. Géraud-Stewart, A. Léonard, D. Naccache, Q. Vermande, and S. Vivien. Discovering New L-Function Relations Using Algebraic Sieving. Cryptology ePrint Archive, Paper 2021/1060, 2021. https://eprint.iacr.org/2021/1060.

4. B. Cloître. A101269. oeis.org/A101269, 2004.

5. H. Cohen. Elementary Continued Fractions for Linear Combinations of $\zeta$ and $L$ Values, 2022.

6. O. David. The Conservative Matrix Field, 2023.

7. H. R. P. Ferguson. PSOS: A New Integral Relation Finding Algorithm Involving Partial Sums of Squares and No Square Roots. Abstracts of Papers Presented to the American Mathematical Society, 9(56):88T–11–75, 214, March 1988.

8. H. R. P. Ferguson, D. H. Bailey, and S. Arno. Analysis of PSLQ, an Integer Relation Finding Algorithm. Mathematics of Computation, 68(227):351–369, 1999.

9. R. Flórez, R. A. Higuita, and A. Mukherjee. Characterization of the Strong Divisibility Property for Generalized Fibonacci Polynomials, 2019.

10. B. Gosper. Continued Fraction Arithmetic, 1970s (exact date unknown). Unpublished manuscript.

11. J. Håstad, B. Helfrich, J. Lagarias, and C. P. Schnorr. Polynomial Time Algorithms for Finding Integer Relations Among Real Numbers. In B. Monien and G. Vidal-Naquet, editors, STACS 86, pages 105–118, Berlin, Heidelberg, 1986. Springer Berlin Heidelberg.

12. A. K. Lenstra, H. W. Lenstra Jr., and L. Lovász. Factoring Polynomials with Rational Coefficients. Mathematische Annalen, 261(4):515–534, 1982.
g, h, f are all smooth. Hence it is inefficient to compute them and then simplify the resulting fraction. Instead, it is best to keep for each of those functions a vector whose coordinates count the occurrence of small prime factors met during calculation – without actually multiplying integers.

When the three vectors $V_g, V_h, V_f$ are ready we can identify at each coordinate $i$ the value:

$$t_i = \min(V_g[i], V_h[i], V_f[i])$$

and update: $V_{\bullet}[i] = V_{\bullet}[i] - t_i$ before performing the actual multiplication.

Estimating the largest primes potentially appearing in each of the functions is easy. $f$'s largest prime is $\max(2\kappa - 1, 2c + j - 2)$, $g$'s is $\max(2\kappa - 1, 2c)$ and $h$'s is $2c - 1$. Hence it suffices to consider primes smaller than $\max(2\kappa - 1, 2c + j - 2)$.

Given that the coefficients of $Q_{j,\kappa,c}$ would explode for $j, \kappa, c$ values exceeding 1000 it suffices to tabulate the factorization of integers between 2 and 2998 and use this table to distribute factors in the vectors.

## References

1. D. H. Bailey and H. R. P. Ferguson. Numerical Results on Relations Between Numerical Constants Using a New Algorithm. Mathematics of Computation, 53(188):649–656, 1989.

2. G. Bana, W. Jamroga, D. Naccache, and P. Y. A. Ryan. Convergence Voting: From Pairwise Comparisons to Consensus. CoRR, abs/2102.01995, 2021.

3. H. Barral, Éric Brier, R. Géraud-Stewart, A. Léonard, D. Naccache, Q. Vermande, and S. Vivien. Discovering New L-Function Relations Using Algebraic Sieving. Cryptology ePrint Archive, Paper 2021/1060, 2021. https://eprint.iacr.org/2021/1060.

4. B. Cloître. A101269. oeis.org/A101269, 2004.

5. H. Cohen. Elementary Continued Fractions for Linear Combinations of $\zeta$ and $L$ Values, 2022.

6. O. David. The Conservative Matrix Field, 2023.

7. H. R. P. Ferguson. PSOS: A New Integral Relation Finding Algorithm Involving Partial Sums of Squares and No Square Roots. Abstracts of Papers Presented to the American Mathematical Society, 9(56):88T–11–75, 214, March 1988.

8. H. R. P. Ferguson, D. H. Bailey, and S. Arno. Analysis of PSLQ, an Integer Relation Finding Algorithm. Mathematics of Computation, 68(227):351–369, 1999.

9. R. Flórez, R. A. Higuita, and A. Mukherjee. Characterization of the Strong Divisibility Property for Generalized Fibonacci Polynomials, 2019.

10. B. Gosper. Continued Fraction Arithmetic, 1970s (exact date unknown). Unpublished manuscript.

11. J. Håstad, B. Helfrich, J. Lagarias, and C. P. Schnorr. Polynomial Time Algorithms for Finding Integer Relations Among Real Numbers. In B. Monien and G. Vidal-Naquet, editors, STACS 86, pages 105–118, Berlin, Heidelberg, 1986. Springer Berlin Heidelberg.

12. A. K. Lenstra, H. W. Lenstra Jr., and L. Lovász. Factoring Polynomials with Rational Coefficients. Mathematische Annalen, 261(4):515–534, 1982.
13. D. Naccache and O. Yifrach-Stav. On Catalan Constant Continued Fractions. In S. El Hajji, S. Mesnager, and E. M. Souidi, editors, *Codes, Cryptology and Information Security*, pages 43–54, Cham, 2023. Springer Nature Switzerland.

14. A. S. Nimbran and P. Levrie. Some Continued Fractions for $\pi$ and G, 2018.

15. G. Raayoni, S. Gottlieb, Y. Manor, G. Pisha, Y. Harris, U. Mendlovic, D. Haviv, Y. Hadad, and I. Kaminer. Generating Conjectures on Fundamental Constants with the Ramanujan Machine. *Nature*, 590(7844):67–73, Feb 2021.

16. D. Sanderson. אמא ודני עם סאונד מוב. www.youtube.com/watch?v=OnrKa4H7Foc, 1979. Second 2:22.
13. D. Naccache and O. Yifrach-Stav. On Catalan Constant Continued Fractions. In S. El Hajji, S. Mesnager, and E. M. Souidi, editors, *Codes, Cryptology and Information Security*, pages 43–54, Cham, 2023. Springer Nature Switzerland.

14. A. S. Nimbran and P. Levrie. Some Continued Fractions for $\pi$ and G, 2018.

15. G. Raayoni, S. Gottlieb, Y. Manor, G. Pisha, Y. Harris, U. Mendlovic, D. Haviv, Y. Hadad, and I. Kaminer. Generating Conjectures on Fundamental Constants with the Ramanujan Machine. *Nature*, 590(7844):67–73, Feb 2021.

16. D. Sanderson. אמא ודני עם סאונד מוב. www.youtube.com/watch?v=OnrKa4H7Foc, 1979. Second 2:22.