A new lower bound for (3/2)k - Wadim Zudilin

10 Wadim Zudilinand{−(α + γ)x − y} + {y} − {−(α + γ)x} ={0 if 0 ≤ y ≤ {−(α + γ)x},1 if {−(α + γ)x} < y < 1.The above conditions (23) on ̂ϕ(x, y) imply 1 − {(α + β)x} = {−(α + γ)x}or, equivalently, (β − γ)x ∈ Z.Finally, the sets {x ∈ R : (α + γ)x ∈ Z} and {x ∈ R : (β − γ)x ∈ Z} havemeasure 0, thus proving the required assertion.□Our final aim is estimating the absolute value of ε k from below, where( ) 3 k= M k + ε k , M k ∈ Z, 0 < |ε k | < 1 22 .Write k ≥ 3 in the form k = 3(βm + 1) + j with non-negative integersm and j < 3β. Multiply both sides of (9) by ˜Φ −1 3 b−2a+j+1 , where ˜Φ =Φ(αm, βm, γm) if n = γm and ˜Φ = Φ ′ (αm, βm, γm)/(γm+1) if n = γm+1,and substitute z = 1/9:( ) 3 j (Q n (9)˜Φ −1 2 j · 3 b−2a+1 F a, b; 1 )(24)29( ) 1= P n (9)˜Φ −1 3 b−2a+j+1 + R n˜Φ −1 3 b−2a+j+1 .9From (6), (7) we see that( ) 3 j (3 b−2a+1 F a, b; 1 ) ( ) 3 3(b+1)+j ( ) 3 k≡(mod Z) = ,29 22hence the left-hand side equals Mk ′ + ε k for some Mk ′ ∈ Z and we may writeequality (24) in the form( 1(25) Q n (9)˜Φ −1 2 j · ε k = M k ′′ + R n˜Φ9)−1 3 b−2a+j+1 ,whereM k ′′ = P n(9)˜Φ −1 3 b−2a+j+1 − Q n (9)˜Φ −1 2 j M k ′ ∈ Zby Lemmas 3 and 4. Lemma 2 guarantees that, for at least one of n = γm orγm+1, we have Mk ′′ ≠ 0; we make the corresponding choice of n. Assumingfurthermore that( ) 1(26) C 0 − C 2 + (β − 2α) log 3 < 0,9from (19) and (21) we obtain)∣ 1 ∣∣∣∣ n( R ˜Φ −1 3 b−2a+j+1 < 1 92for all m ≥ N 1 ,