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_3F_2[n,-x,-y; x+n+1,y+n+1] =Gamma(x+n+1)Gamma(y+n+1)Gamma(1/2n+1)Gamma(x+y+1/2n+1) ×Gamma(n+1)Gamma(x+y+n+1)Gamma(x+1/2n+1)Gamma(y+1/2n+1), (1) where _3F_2(a,b,c;d,e;z) is a ...

Except for triangles, every simple polygon has at least two nonoverlapping ears.

If a plane cuts the sides AB, BC, CD, and DA of a skew quadrilateral ABCD in points P, Q, R, and S, then (AP)/(PB)·(BQ)/(QC)·(CR)/(RD)·(DS)/(SA)=1 both in magnitude and sign ...

The lines joining the vertices A, B, and C of a given triangle DeltaABC with the circumcenters of the triangles DeltaBCO, DeltaCAO, and DeltaABO (where O is the circumcenter ...

De Grey (2018) found the first examples of unit-distance graphs with chromatic number 5, thus demonstrating that the solution to the Hadwiger-Nelson problem (i.e., the ...

Letting Lk be the linking number of the two components of a ribbon, Tw be the twist, and Wr be the writhe, then Lk(K)=Tw(K)+Wr(K). (Adams 1994, p. 187).

For an algebraic curve, the total number of groups of a g_N^r consisting in a point of multiplicity k_1, one of multiplicity k_2, ..., one of multiplicity k_rho, where sumk_i ...

A partial solution to the Erdős squarefree conjecture which states that the binomial coefficient (2n; n) is never squarefree for all sufficiently large n>=n_0. Sárkőzy (1985) ...

Let G be a group, then there exists a piecewise linear knot K^(n-2) in S^n for n>=5 with G=pi_1(S^n-K) iff G satisfies 1. G is finitely presentable, 2. The Abelianization of ...

The six planes through the midpoints of the edges of a tetrahedron and perpendicular to the opposite edges concur in a point known as the Monge point.