We introduce the term “lump” here, but here we call that a 25-lump where an n-lump is n balls in which each of the balls touches each of the others. In 24D the 25-lump is the biggest possible lump. The ball centers of an n-lump form a regular (n−1)-simplex. An n-lump is congruent to every other n-lump in n! ways.
By connotation a lump is a region of high density. In n dimensions some fraction of the volume within the regular simplex defined by the n+1 centers of the balls of a (n+1)-lump is within one of the n+1 balls. That fraction represents an upper bound of sphere packing density in n dimensions. This upper bound is achieved only in 2D.
A central Leech lump or clump is one whose balls all belong to the Leech lattice and in particular includes the origin ball which is itself the unique 1-clump. There are K1 = 196560 2-clumps (one for each neighbor of origin) and each 2-clump is included in K2 = 4600 3-clumps (one for each pair of origin neighbors that touch each other). This double counts the 3-clumps and thus there are 196560∙4600/2 = 452088000 3-clumps. For n>2 it is not immediately clear that any two n-lumps in LL are situated in the lattice like every other n-lump in the lattice; thus Kn may not be constant. Indeed the program described below suggests that after 4600 comes 891, 336, 170 and 100 except that occasionally the last is 102 instead of 100. The Leech lattice is remarkably symmetrical but there are limits.
Here we use the standard clump to locate the lattice that we defined here.
We have a program to generate random clumps. It relies on a table of the origin neighbors. It begins with the 1-clump and given an n-clump searches for another origin neighbor that touches each of the current n balls. This produces an (n+1)-clump. I need to know if n ever fails to reach 25. Such a program can be modified to guess at the Kn’s.
Choose random neighbor N; enumerate all 4600 neighbor balls that are √32 from N;
All n-clumps that contain N must be a subset of these 4600.
This is much more efficient.
Now would an even more efficient scheme be simpler code?
This code performs the above and yields these results. If my program is correct there are maximal 15-lumps and 17-lumps that cannot be extended.