On sums involving divisor function, Euler's totient function, and floor function

Every positive integer l N can be formed l = (m + n) d, provided gcd (m, n) =1. From this point of view, the next formulas n=₃|₋ (d) and n (n+1) 2=₊=₁^n (k) nk, and these equivalence had been proved. In this paper on an extension of these results, the next identity is proved: ₊=₁^n (₀+₁) ₂=₊ \\ ₆₂₃ (₀, ₁) =₁ f (a, b) g (c) = ₊=₁^n ₀+₁=₊ \\ ₆₂₃ (₀, ₁) =₁ f (a, b) ₈ ₍₊ g (i) = ₀+₁ ₍ f (agcd (a, b), bgcd (a, b) ) g (gcd (a, b) ). We also show the next formulas are corollaries of it: ₊=₁^n (k) =₊=₁^n nk = ₀+₁ ₍ 1 (a+b{gcd (a, b) ) }, ₃|₍ f (d) g (nd) = ₊ = ₁^n f (gcd (k, n) ) (n{gcd (k, n) ) } (n{gcd (k, n) ) }, (n) = ₊ = ₁^n n{gcd (k, n) } (n{gcd (k, n) ) } = ₊ = ₁^n gcd (k, n) (n{gcd (k, n) ) }, n = ₊ = ₁^n (gcd (k, n) ) (n{gcd (k, n) ) }, ₀+₁=₍ \\ ₆₂₃ (₀, ₁) =₁ gcd (a-1, b+1) = ₀+₁=₍ (n) (n{gcd (a, b) ) }, and so on. In addition to it, we evaluate a sequence ₊=₁^n (k) (k).