A130702 Possible sides in the Euler V=E-F+2 as roots in a cubic polynomial of the form: P(x)=(x-V)*(x-F)*(x+E) =x^3+(E-V-F)*x^2+(V*F-E(V+F))*x=E*F*V Solves here for F ( Face, Edge, Vertex).

4, 8, 12, 16, 18, 20, 22, 24, 28, 30, 32, 36, 40, 42, 44, 48, 52, 54, 56, 60, 64, 66, 72

Offset

1,1

Comments

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}

Links

Table of n, a(n) for n=1..23.

Formula

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)

Example

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}}

Mathematica

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}]

Crossrefs

Cf. edge: A008458; vertex: A118081.

Sequence in context: A020647 A274919 A196032 * A355740 A371089 A370348

Adjacent sequences: A130699 A130700 A130701 * A130703 A130704 A130705

Keyword

nonn,uned

Author

Roger L. Bagula, Jul 06 2007

Status

approved