This list contains the Williamson matrices of order m, i.e. four circulant symmetric (1,-1) matrices A,B,C,D of order m which satisfy the Williamson matrix equation

A^2 + B^2 + C^2 + D^2 = 4m I_m

Then using A,B,C,D in the following Williamson array, Yamada constructed Hadamard matrices of order 4m.

A B C D
-B A D -C
-C -D A B
-D C -B A

In the following list - stands for -1.

We define Williamson matrices of type 3 or Yamada type if: 4n=x^2+x^2+y^2+y^2 and the Williamson equation satisfied.

Find a j such that j^2=-1(mod n) and take A,B,jA,jB a partition of omega={1,2,...,(n-1)/2}. There exists such a j if and only if all prime divisors of n are =1(mod 4). If A,B,jA,jB is a solution we call this a Yamada type Williamson matrix.

m=5, 4m=20=3^2+1^2+3^2+1^2, one solution: A, B, 2A, 2B
1----
11--1
1----
1-11-

m=13, 4m=52=5^2+1^2+5^2+1^2, one solution: A, B, 5A, 5B
111-11--11-11
11--1-11-1--1
11-1-1111-1-1
1---111111---

m=17, 4m=68=5^2+3^2+5^2+3^2, one solution: A, B, 4A, 4B
1--1-11111111-1--
11-1---1--1---1-1
1-11-111--111-11-
1---111----111---

m=25, 4m=100=7^2+1^2+7^2+1^2, two solutions: A, B, 7A, 7B
1-----1-11--11--11-1-----
1-11-1-111------111-1-11-
1-----1-11--11--11-1-----
11--1-1---111111---1-1--1

1---1-111--------111-1---
1--11-1--111--111--1-11--
11-1--1-1--------1-1--1-1
1-1111--1---11---1--1111-

m=25, 4m=100=5^2+5^2+5^2+5^2, one solutions: A, B, 7A, 7B
1-1--111--111111--111--1-
11--1--1-11111111-1--1--1
111--1--1111--1111--1--11
1111-1-1-1--11--1-1-1-111

m=29, 4m=116=7^2+3^2+7^2+3^2, Yamada type. No solution by exhaustive search

m=37, 4m=148=7^2+7^2+5^2+5^2, one solution: A, B, 6A, 6B
11---1-----1-1-11-11-11-1-1-----1---1
1-1111-1-11--1-1--11--1-1--11-1-1111-
1--11-111------11----11------111-11--
1--111-1----111-111111-111----1-111--

m=41, 4m=164=9^2+1^2+9^2+1^2, Yamada type. No solution by exhaustive search

m=45, 4m=180; Since 45=3x3x5, i.e. contains a prime divisor 3=3(mod 4) doesn't belongs in this class of solutions.

m=53, 4m=212= 9^2+9^2+5^2+5^, Yamada type. No solution for j=23 : A,B, 23A, 23B

m=61, 4m=244=11^2+11^2+1^2+1^2, one solution: A, B, 11A, 11B
11--1--11--1-1-1111--1-----1------1-----1--1111-1-1--11--1--1
1---1-1-1111---11--1-11-1---111111---1-11-1--11---1111-1-1---
11--1--11--1-1-1111--1-----1------1-----1--1111-1-1--11--1--1
1111-1-1----111--11-1--1-111------111-1--1-11--111----1-1-111

m=73, 4m=292=11^2+11^2+5^2+5^2, solution possible: A, B, 27A, 27B

m=85, 4m=340=11^2+11^2+7^2+7^2, one solution possible: A, B, 13A, 13B

m=101, 4m=404=11^2+11^2+9^2+9^2, one solution possible: A, B, 10A, 10B

Notes: The decomposition m=31, 4m=124=7^2+5^2+5^2+5^2, was found by Turyn (1972)
The exhaustive search for the orders 29 and 31 is confirmed in: D.Z.Djokovic, Williamson matrices of orders 4.29 and 4.31, J.Combin. Theory, Ser. A, 59 (1992), 309-311.
M.Yamada, On the Williamson type j matrices of orders 4x29, 4x41, and 4x37, J. Combin. Theory Ser. A, 27 (1979), 378-381.