%I #10 Nov 13 2024 17:16:11

%S 1,-2,4,16,-32,-128,-512,4096,-8192,-32768,-131072,1048576,4194304,

%T -33554432,268435456,4294967296,-8589934592,-34359738368,

%U -137438953472,1099511627776,4398046511104,-35184372088832,281474976710656

%N Determinant of n X n partial Hadamard matrix with coefficient m(i,j) 1<=i,j<=n (see comment).

%C Let M(infinity) be the infinite matrix with coefficient m(i,j) i>=1, j>=1 defined as follows : M(0)=1 and M(k) is the 2^k X 2^k matrix following the recursion : +M(k-1)-M(k-1) M(k)= -M(k-1)-M(k-1)

%F It appears that abs(a(n))=2^A000788(n). What is the rule for signs? Does sum(k=1, n, a(k+1)/a(k))=0 iff n is in A073536 ?

%F Conjecture: a(n) = (-2)^A000788(n-1). - _Chai Wah Wu_, Nov 12 2024

%e M(2)=/1,-1/-1,-1/ then a(2)=detM(2)=-2

%o (Python)

%o from sympy import Matrix

%o def A094384(n):

%o m = Matrix([1])

%o for i in range((n-1).bit_length()):

%o m = Matrix([[m, -m],[-m, -m]])

%o return m[:n,:n].det() # _Chai Wah Wu_, Nov 12 2024

%Y Cf. A000788, A073536.

%K sign,changed

%O 1,2

%A _Benoit Cloitre_, Jun 03 2004