Periodica Mathematica Hungarica

, Volume 10, Issue 4, pp 261–271 | Cite as

On the order of the error function of the square-full integers

  • D. Suryanarayana


LetL(x) denote the number of square full integers ≤x. By a square-full integer, we mean a positive integer all of whose prime factors have multiplicity at least two. It is well known that
$$\left. {L(x)} \right| \sim \frac{{\zeta ({3 \mathord{\left/ {\vphantom {3 2}} \right. \kern-\nulldelimiterspace} 2})}}{{\zeta (3)}}x^{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-\nulldelimiterspace} 2}} + \frac{{\zeta ({2 \mathord{\left/ {\vphantom {2 3}} \right. \kern-\nulldelimiterspace} 3})}}{{\zeta (2)}}x^{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern-\nulldelimiterspace} 3}} ,$$
where ζ(s) denotes the Riemann Zeta function. Let Δ(x) denote the error function in the asymptotic formula forL(x). On the basis of the Riemann hypothesis (R.H.), it is known that\(\Delta (x) = O(x^{\tfrac{{13}}{{81}} + \varepsilon } )\) for every ε>0. In this paper, we prove the following results on the assumption of R.H.:
$$\frac{1}{x}\int\limits_1^x {\Delta (t)dt} = O(x^{\tfrac{1}{{12}} + \varepsilon } ),$$
$$\int\limits_1^x {\frac{{\Delta (t)}}{t}\log } ^{v - 1} \left( {\frac{x}{t}} \right) = O(x^{\tfrac{1}{{12}} + \varepsilon } )$$
for any integer ν≥1.

In fact, we prove some general results and deduce the above from them.

On the basis of (1) and (2) above, we conjecture that\(\Delta (x) = O(x^{{1 \mathord{\left/ {\vphantom {1 {12}}} \right. \kern-\nulldelimiterspace} {12}} + \varepsilon } )\) under the assumption of R.H.

AMS (MOS) subject classifications (1970)

Primary 10H15 Secondary 10H25 

Key words and phrases

Square-full integers Riemann Zeta function Riemann hypothesis 


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Copyright information

© Akadémiai Kiadó 1979

Authors and Affiliations

  • D. Suryanarayana
    • 1
  1. 1.Department of MathematicsAndhra UniversityWaltairIndia

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