12.07.2015 Views

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

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

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Set(15)Φ = Φ(a, b, n) =A <strong>new</strong> <strong>lower</strong> <strong>bound</strong> <strong>for</strong> ‖(3/2) k ‖ 7(⌊ ⌋ ⌊ ⌋ ⌊ ⌋a + n + µ a + n µ= min− −µ∈Z pp p⌊ ⌋ ⌊ ⌋ ⌊ ⌋)a + b + n a + b + µ n − µ+−−ppp( )( )a + n + µ a + b + n≤ min ord p.0≤µ≤n µ n − µ∏p> √ a+b+nFrom (8), (10) and (13), (15) we deducep e pand Φ ′ = Φ ′ (a, b, n) =∏p> √ a+b+nLemma 3. The following inclusions are valid:( )( )a + n − 1 + µ a + b + nΦ −1 ·∈ Zµ n − µ<strong>for</strong> µ = 0, 1, . . . , n,henceΦ −1 Q n (x) ∈ Z[x] and Φ −1 P n (x) ∈ Z[x].Supplementary arithmetic in<strong>for</strong>mation <strong>for</strong> the case n replaced by n + 1is given inLemma 4. The following inclusions are valid:(16) ( )( )(n+1)Φ ′ −1 a + n + µ a + b + n + 1·∈ Z <strong>for</strong> µ = 0, 1, . . . , n+1,µ n + 1 − µhence(n + 1)Φ ′ −1 Qn+1 (x) ∈ Z[x] and (n + 1)Φ ′ −1 Pn+1 (x) ∈ Z[x].Proof. Write( )( ) ( )( )a + n + µ a + b + n + 1 a + n + µ a + b + n=µ n + 1 − µµ n − µThere<strong>for</strong>e, if p ∤ n + 1 − µ then(17) ( )( )a + n + µ a + b + n + 1ord pµ n + 1 − µ≥ ord p( a + n + µµp e′ p.· a + b + n + 1n + 1 − µ .)( ) a + b + n≥ e ′n − µp;

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