# Product of a Laver table and a Hadamard matrix has mostly 0 rows

I recently noticed (while playing around) that the product of a Laver matrix with a Hadamard matrix gives a very sparse matrix. In particular, all but logarithmically few rows are all zero. The nonzero rows all seem to occur near powers of 2, or multiples of large powers of 2. For instance, multiplying the 256x256 Laver table by the 256x256 Hadamard matrix, the only nonzero rows are

{1, 2, 4, 8, 9, 16, 32, 33, 64, 65, 96, 97, 128, 129, 160, 161, 192, 193, 224, 225, 256}


Is this a known phenomenon? Is there an explanation for it, other than the high periodicity of the Laver table?

In particular since all current proofs of the unbounded growth of the period of the first row of a Laver table rely on strong cardinal axioms, I wonder if understanding this phenomenon could help us understand it.

• Can you provide a definition (or a link) of what you mean by Laver table and Hadamard matrix? Sep 11 '16 at 15:21
• Edited, please check if the links point where they should. Sep 11 '16 at 19:42
• Laver tables are broken by row and column permutations whereas Hadamard matrices are not. Are you asserting this for an arbitrary Hadamard matrix, or just those in a standard format? Gerhard "Permutations Affect Sparsity Of Products" Paseman, 2016.09.11. Sep 11 '16 at 20:07
• Just to add to what Gerhard said: You may also multiply columns (and rows, but this does not affect anything) by $\pm 1$; also, I'm not sure whether a Hadamard matrix of order $2^n$ is unique even up to these transformations. Sep 11 '16 at 21:32
• For small orders the equivalence classes are not many. Once you get past order 24, there are a lot of such classes of inequivalent Hadamard matrices. Gerhard "Check Out Hadamard Matrix Websites" Paseman, 2016.09.11. Sep 11 '16 at 22:12