

A130702


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


0



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

1,1


COMMENTS

Polynomial cubic of Euler's V,F,E: V=EF+2 P(x)=(xV)*(xF)*(x+E) =x^3+(EVF)*x^2+(V*FE(V+F))*x=E*F*V letting (EVF)=2 V+F=E+2 and product: p=V*F I got P(x)=x32*x2+(pE*(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+(EVF)*x^2+(V*FE(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 * A053806 A068306 A311118
Adjacent sequences: A130699 A130700 A130701 * A130703 A130704 A130705


KEYWORD

nonn,uned


AUTHOR

Roger L. Bagula, Jul 06 2007


STATUS

approved



