E. Landau, Primzahlen, Chelsia reprint, New York, 1958.  Home/Search   Document Not in Database   Summary   Related Articles

....all square free integers z not divisible by the first k primes. We have N(z) xf(P k ) nz S P k n (n) nz S (n) P k n x nz S (n) f( n x mod P k , k) If we calculate the sum in order then any intermediate result is bounded by r k x (1 f(P k ) y (cf. [8], p. 582) When x = 4 10 16 and k = 5 this is bounded by 0.85 10 16 . The special leaves. The contribution of the special leaves is n S (n) f( n x , p(d(n) 1) where the sum is taken over all special n. Now we have f( n x , r) P r n x ] f(P r ) f( n x ....

....of this n is (x n) r k 2f(P k ) Hence any intermediate sum is bounded by n S (r k n x 2f(P k ) 6.1) where the sum is taken over all square free n [x 1 3 , x 2 3 ) not divisible by primes p k . If we denote by Q k (x) the number of such integers x, then, by Landau ([8], pp. 633 636) we have Q k (x) s k x R(x) and R(x) f(P k ) x . The sum may be written as a Stieltjes integral 42 x 1 3 x 2 3 (r k t x 2f(P k ) dQ k (t) Using the above expression for Q k and integration by parts we find that the sum in equation (1) is bounded by ....

E. Landau, Primzahlen, Chelsia reprint, New York, 1958.

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