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

A new lower bound for ‖(3/2) k ‖ 11where N 1 > 0 is an effective absolute constant.| ≥ 1 we have|M ′′k|Q n (9)˜Φ −1 2 j | · |ε k | ≥ |M ′′k | − ∣ ∣∣∣R n( 19hence from (19), (20) we conclude that|ε k | >Therefore, by (25) and)˜Φ −1 3 b−2a+j+1 ∣ ∣∣∣> 1 2 ,˜Φ2 j+1 |Q n (9)| ≥ ˜Φ2 3β |Q n (9)| > e−m(C 1(1/9)−C 2 +δ)for any δ > 0 and m > N 2 (δ), provided that C 1 (1/9) − C 2 + δ > 0; hereN 2 (δ) depends effectively on δ. Finally, since k > 3βm, we obtain theestimate(27) |ε k | > e −k(C 1(1/9)−C 2 +δ)/(3β)valid for all k ≥ K 0 (δ), where the constant K 0 (δ) may be determined interms of max(N 1 , N 2 (δ)).Taking α = γ = 9 and β = 19 (which is the optimal choice of the integerparameters α, β, γ, at least under the restriction β ≤ 100) we find that( )( )1 1C 0 = 3.28973907 . . . , C 1 = 35.48665992 . . . ,99and⎧1 if {x} ∈ [ 237 , 1 ] [18 ∪ 337 , 1 [10)∪437 , 1 ) [9 ∪ 637 , 1 [6]∪737 , 1 )5⎪⎨∪ [ 837 , 2 ) [9 ∪ 1037 , 5 ] [18 ∪ 1137 , 3 ) [10 ∪ 1237 , 1 ) [3 ∪ 14ϕ(x) =∪ [ 1637 , 4 ) [9 ∪ 1837 , 1 ) [2 ∪ 2037 , 5 ) [9 ∪ 2237 , 3 ) [5 ∪ 2437 , 2 )3∪ [ 2837⎪⎩, 7 ) [9 ∪ 3237 , 8 ) [9 ∪ 3637 , 1) ,0 otherwise,henceC 2 = C ′ 2 = 4.46695926 . . . .Using these computations we verify (26),( ) 1C 0 − C 2 + (β − 2α) log 3 = −0.07860790 . . . ,9and find that with δ = 0.00027320432 . . .e −(C 1(1/9)−C 2 +δ)/(3β) = 0.5803.37 , 718]This result, in view of (27), completes the proof of Theorem 1.□