1,1

Polynomial cubic of Euler's V,F,E: V=E-F+2 P(x)=(x-V)*(x-F)*(x+E) =x^3+(E-V-F)*x^2+(V*F-E(V+F))*x=E*F*V letting (E-V-F)=-2 V+F=E+2 and product: p=V*F I got P(x)=x3-2*x2+(p-E*(E+2))*x+E*p Setting that polynomial equal to zero gives roots that agree with Euler's equation. In the exceptional groups: ( down to two integer variables) p=16*m ; m-> {1,3,15} E=6*n ; n->{1,2,5} The program works to produce the right roots for {-E,V,F}

F roots such that:x^3+(E-V-F)*x^2+(V*F-E(V+F))*x=E*F*V and that are exceptional like ( tetrahedron, cube, octahedron, dodecahedron, icosahedron)

Program to get roots for tetrahedron, (cube, octahedron),

(dodecahedron,

icosahedron):

a = {1, 2, 5}

b = {1, 3, 15}

g[n_, m_] := x /. Solve[e [a[[m]]]*p[b[[m]]] - e [a[[m]]]*(e[a[[

m]]] + 2)*x + p[b[[m]]]* x - 2* x^2 + x^3 == 0, x][[n]]

Table[g[n, m], {n, 1, 3}, {m, 1, 3}]

{{-6, -12, -30}, {4, 6, 12}, {4, 8, 20}}

ExpandAll[(x - v)*(x - f)*(x + e)]; e[n_] := 6*n; p[m_] := 16*m; a0 = Table[If[IntegerQ[x /. Solve[e [m]*p[p0] - e [m]*(e[m] + 2)*x +p[p0]* x - 2* x^2 + x^3 == 0, x][[1]]] && IntegerQ[x /. Solve[e [m]*p[p0] - e [m]*(e[m] + 2)*x + p[p0]* x - 2* x^2 + x^3 == 0, x][[2]]] && IntegerQ[x /. Solve[e [m]*p[p0] - e [m]*(e[m] + 2)*x +p[p0]* x - 2* x^2 + x^3 == 0, x][[3]]], {Abs[x] /. Solve[e [m]*p[p0] - e [m]*(e[m] + 2)*x + p[p0]* x - 2* x^2 + x^3 == 0, x][[3]]}, {}], {m, 1, 12}, {p0, 1, 33}]

Cf. edge: A008458; vertex: A118081.

Sequence in context: A133803 A081747 A020647 this_sequence A053806 A068306 A113645

Adjacent sequences: A130699 A130700 A130701 this_sequence A130703 A130704 A130705

nonn,uned

Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jul 06 2007