Almost-all results on the p^\lambda problem

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Obviously, we have S(∆) := � n, r ∼ Y, |t| ≤ ∆ Combining (7) and (8), we obtain the desired estimate. ✷ To optimize the estimate in Lemma 4, we choose 1. S(∆) ≪ ∆Y 2−λ + Y. (8) Qh := 1 + � h 1/3 X (λ−2)/3 Y (λ+2)/3� . Then, by a short calculation using J ≤ H = N t and XY = N, we obtain when Lemma 5: We have � h≤H This implies |Lh| 2 ≪ � N 1+λ/2+3t/2 + N 2−λ + N 2+t Y −1 + N 4/3+λ/3+4t/3 Y 1/3 + N 2/3+2λ/3+5t/3 Y 2/3 + N 4/3−2λ/3+t/3 Y 4/3� (log N). Lemma 6: For every sufficiently sm<strong>all</strong> fixed η > 0 we have provided the conditions are satisfied. 4 Treatment of Kh � |Lh| h≤H 2 ≪ N 2−2η N t+100η ≤ Y ≤ N 2−λ−4t−100η , t < (2 − λ)/3, t > 2(1 − λ)/5 (9) Heath-Brown established the following two estimates for Kh (see [11, pages 261-262]). Lemma 7: (i) We have 6
� 2 Kh ≪ (log N) N 1−λ h −1 + Xh 1/2 N λ/2� . (ii) Let (p, q) be any exponent pair for which 0 the exponent pair (p, q) = (2/7, 4/7), and choosing P := � N (2−(1+2p)(λ+t))/(2p+3) Y (4p+1−2q)/(2p+3)� = � N (14−11(λ+t))/25 Y 7/25� , from Lemma 7, we obtain Lemma 8: (i) We have (ii) We have � h≤H � h≤H |Kh| 2 � 4 ≪ (log N) N 2(1−λ) + X 2 N λ+2t� . |Kh| 2 � 8 ≪ (log N) N (36+11λ+36t)/25 Y −7/25 + N (43−7λ+18t)/25 Y 9/25 + N 1+t Y + provided the condition is satisfied. when N 2+t Y −1 + N (57−18λ+7t)/25 Y −9/25� , From the preceding lemma, we derive t < 14/11 − λ (10) Lemma 9: (i) For every sufficiently sm<strong>all</strong> fixed η > 0 we have � |Kh| h≤H 2 ≪ N 2−2η N (λ+2t)/2+100η ≤ Y. (ii) For every sufficiently sm<strong>all</strong> fixed η > 0 we have (11) when N max{(11λ+36t)/7−2,t}+100η ≤ Y ≤ N min{1−t,7(1+λ)/9−2t}−100η , 7 (11)
Page 1 and 2: Almost-all
Page 3 and 4: 2 Reduction to exponential sums We
Page 5: We shall prove the
Page 9 and 10: Then the conditions in (9), (10), (
Page 11: Using this inequality, (22) and Tay

Almost-all results on the p^\lambda problem

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