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Let’s see, we should be able to show this.
If N=2M,
N∏j=1Γ(j/2)=M∏j=1Γ(j)⋅M∏j=1Γ(j−12)=G(M+1)⋅G(M+1/2)/G(1/2)=G(N/2+1)⋅G(N/2+1/2)/G(1/2).If N=2M+1,
N∏j=1Γ(j/2)=M∏j=1Γ(j)⋅M+1∏j=1Γ(j−12)=G(M+1)⋅G(M+3/2)/G(1/2)=G(N/2+1/2)⋅G(N/2+1)/G(1/2).1 to 4 of 4