A343607 Minimal number of colors required for an edge-coloring of the complete graph K_n with no monochromatic triangle.
0, 0, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4
Offset: 0
Examples
For n = 0 and n = 1, the empty graph K_0 and the singleton graph K_1 don't have any edge, so zero colors are needed. For n = 2 we have one edge, so one color is needed. For n = 3 we have a triangle, so we need a second color for the third edge. For n = 4 (square + diagonals) and n = 5 (pentagon + diagonals forming a pentagram) two colors are still enough: one can use one color for the "circumference", i.e., edges (i,i+1), and the other color for the diagonals. For n >= 6, a third color is needed.
Links
- James Cummings, Daniel Král', Florian Pfender, Konrad Sperfeld, Andrew Treglown and Michael Young, Monochromatic triangles in three-coloured graphs, Journal of Combinatorial Theory B 103 no. 4 (2013) 489-503 (also: arXiv:1206.1987).
- Wikipedia, Monochromatic triangle.
Crossrefs
Programs
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PARI
A343607(n)=if(n>1, vecmax(color(n))+1, 0) \\ using the helper function: {M343607=List([[]]); color(n, i=matrix(n,n,r,c,r+c--*c--/2), C, k)= C|| C = if(#M343607 >= n, M343607[n], n>2, color(n-1,i)); k|| k=if(n>3, vecmax(C)+1, n-1); C=Vec(C, n*(n-1)/2); my(bad(C)= for(a=1,n-2, my(c=C[i[a,n]]); for(b=a+1, n-1, C[i[a,b]] !=c || C[i[b,n]] !=c || return( i[b,n] ))), C0=C, j); while(j=bad(C), until(j-- < i[1,n], if(C[j]++ < k, while(j<#C, C[j++]=0); next(2))); while(C[j]++ >= k, C[j]=0; j--); C=Vec(color(n-1,i,C[1..-n]),#C); if(C[1..n] != C0[1..n], k++; C=C0)); #M343607
= 13. Changing 1..n to 1..2-2*n is much faster but yields suboptimal solution for n >= 12 (using 4 instead of 3 required colors).
Formula
a(n) = max { A343606(n,k)+1; k >= 1 } U { 0 }.
a(n) = min {k >= 0; A003323(k) > n}. - Pontus von Brömssen, Aug 01 2021
Extensions
More terms from Pontus von Brömssen, Jul 21 2021
Comments