0185132人目の素数さん垢版 | 大砲2019/08/12(月) 16:47:05.09ID:uLwjs1DH [ (n-1)^2 /4 ] + [ n/2 ] = [ nn/4 ] (short proof) δ = mod(n, 2) δ = 0 (n:even) δ = 1 (n:odd) then [ (n-1)^2 /4 ] = ((n-1)^2 -1+δ)/4, [ n/2 ] = (n-δ)/2, [ nn/4 ] = (nn-δ)/4,