Harmonic sets and the harmonic prime number theorem

Harmonic sets and the harmonic prime number theorem Version: 9th September 2004 Kevin A. Broughan and Rory J. Casey University of Waikato, Hamilton, New Zealand E-mail: kab@waikato.ac.nz We restrict primes and prime powers to sets H() = ( 2n, 2n ]. Let θ H () = p H() log p. Then the error in θ H() has, unconditionally, the epected order of magnitude θ H () = log 2 + O ( ). However, if ψ H () = p m H() log p then ψ H() = log 2 + O(log ). Some reasons for and consequences of these sharp results are eplored. A proof is given of the harmonic prime number theorem, π H ()/π() log 2.. INTRODUCTION ( 2n ) a good estimate for the square full part of the binomial coefficient nusing derived by [4] it was shown in [] that, if an is the square-free part of n!, then log a n = n log 2+O( n). Of course log a n is the sum of the logarithms of primes which appear to odd powers in the prime factorization of n!, and includes, according to the given result, more than half of the primes up to n. If n < p n then these primes are easy to describe: p divides the square-free part of n! if and only if [n/p] is odd, or in other words for some whole number m n 2m < p n 2m. This shows that, at least for primes sufficiently large, dividing the square free part of n! has an alternative description as membership in the union of a finite set of half open intervals. As the primes get smaller the description of the intervals becomes more comple. For eample if n /3 < p n /2, p divides the square-free part of n! if and only if for some j N n j + < p n j and [n ] has opposite parity to j. p

2 BROUGHAN,CASEY (To see this consider p α n! such that α = [n/p] + [n/p 2 ] with j = [n/p 2 ].) The net level would be n /4 < p n /3 with an even more comple description in terms of intervals. So, instead of proceeding in this manner a restricted sum of logarithms of primes is defined assuming all of the intervals have the first form leading to the set H() = ( 2n, 2n ] and restricted sums θ H () = p H() log p, ψ H() = m,p m H() log p, and π H () = p H(). By bridging across to log a n we derive θ H () = log 2 + O( ). After getting a better epression for log a n we are able to obtain a lower bound π H () log 2/ log + O( / log ). One suspects this lower bound is also an upper bound as evidenced by the first approimation π H ()/π() log 2 which is proved. Quite independent of log a n, by rewriting ψ H () = and using Chebychev s result log([]!) = ( ) n+ ψ( n ) ψ( n ), N the natural error for these sums, ψ H () = log 2 + O(log ), is derived. An eplanation for the unepected regularity of these sums comes from an integral epression for ψ H () in terms of the Riemann zeta function: ψ H () does not depend on the non-trivial zeros of ζ(s), whereas ψ() ψ H () depends on all of them. A study of arithmetic progressions in the contet of the sets H() shows that no restricted analogue of Dirichlet s theorem (for primes in an arithmetic progression) can be true. 2. RESTRICTED FORMS OF CHEBYCHEV S FUNCTIONS Definition 2.. Let > 0 be a real number. Then the harmonic set with parameter is defined to be H() = ( 2n, 2n ].

HARMONIC SETS 3 Definition 2.2. For R, > 0 let θ H () = log p = log p, where p is any prime. p H() p, /p odd Theorem 2.. As, θ H () = log 2 + O ( ). Proof. Let 4 and N = By [, Theorem ], if a N is the square free part of N! ( ) () log a N = N log 2 + O N, where the implied constant is absolute. Let n N, let k = 2n be odd and 2N p N a rational prime with p k N!. Then k = N/p and k(k + ) N. Therefore, by [, Lemma ], N 2n < p N 2n Conversely, if p satisfies this inequality for some n N and 2N p N, k = 2n, then 2N p N/k which implies 2k N so k(k + ) N. Hence k is odd and p k N!. Now split the sum which defines θ H (): θ H () = n 2N 2n <p 2n = + 2 log p + By what has been shown, ( ) 2 = a N + O 2N ( ) = N log 2 + O N = log 2 + O ( ) 2N<n N 2n <p 2n log p Also = O 2 p O( )(log p) = O ( )

4 BROUGHAN,CASEY so the result follows. Definition 2.3. ψ H () = log p 2n <pm 2n Since 0 ψ H () θ H () log 2 2 log 2 we can write ψ H () = log 2 log + O( log 2 ). However it is easy to derive a more accurate error estimate. Theorem 2.2. As,ψ H () = log 2 log + O(log ) Proof. Let T () := log([]!) when N and T (n) = (T (n+) + T (n ))/2 for n N. Then [2, Page 282] From the definition T () = ψ H () = ψ( n ). ( ) n+ ψ( n ). Therefore T () ψ H () = 2T (/2). Using Stirling s approimation T () = log + O(log ), the given epression for ψ H () follows directly. Definition 2.4. π H () = 2n <p 2n Since when p, log p/ log we see π H () θ H() log = log 2 log + O( log ). However we have not been able to verify the conjecture that this lower bound is also an upper bound. This would deliver the epected error bound in what might be called the harmonic prime number theorem. Below the asymptotic form is derived. Some numerical evidence seems to indicate that Li() log 2 is not as good an approimation to π H () as Li() is to π(). See Figure.

HARMONIC SETS 5 400 300 200 00 000 2000 3000 4000 5000 FIG.. Comparison of Li() log 2 (top), π H () (middle), log 2/ log (bottom). Theorem 2.3. lim inf π H () π() log 2. Proof. First using Theorem 2. write θ H () θ() Then use the integral formula log 2 + O( ) log 2 + o() = = log 2. + o() + o() π() = θ() log + θ(t) 2 t log 2 t dt and the lower bound for π H () given above to write π H () π() = θ H ()/ log θ() log + θ(t) 2 t log 2 t dt θ H ()/θ() + log θ() 2 θ(t) t log 2 t dt. and the second term in the denominator tends to zero.

6 BROUGHAN,CASEY To derive the asymptotic form of the upper bound for the ratio π H ()/π() we use the theorem of Heath-Brown [3], namely uniformly for 7/2 y. Theorem 2.4. π() π( y) = y log + O( y log 45/44 ) π H () lim = log 2. π() Proof. First we find an upper bound for the ratio in terms of integrals and error terms, then show that four of the terms in the ratio tend to zero. Finally we show that lim sup π H ()/π() log 2 so the theorem follows from this inequality and the previous theorem. We use the notation I(a, b) := b (/ log t)dt. a () π H () π() = 7/2 I( 2n, 2n ) + O( 7/2 I(2, ) + O( 2n(2n ) log 45/44 ( 2n log 2 ) )) + O( 7/2 log ) 7/2 I( 2n, 2n ) O( 7/2 2n(2n ) log I(2,) + 45/44 ( 2n ) ) I(2,) + O(7/2 / log ) I(2,) + O(/ log2 ) I(2,). (2) The last term in the denominator: O( log 2 ) I(2, ) (3) The last term in the numerator: = O( log ) + O( log ) 0. 7/2 log I(2, ) = O( 5/2 ) + O( log ) 0. (4) The second term in the numerator:

HARMONIC SETS 7 7/2 log + O( 2n(2n ) log 45/44 ( 2n ) log 2 ) log 7/2 log /44 2n(2n ) (log log ) 45/44 + O( log ) 7/2 2n(2n ) + O( log ) 0. (5) Finally we show that lim sup 7/2 I( 2n, 2n ) log 2 : I(2, ) Firstly, by L Hôpital s rule, for each n N, I( lim 2n, 2n ) I(2, ) Let > 4 2/5 and N < /4. Then = 2n(2n ). so 7/2 I( 2n, I(2, ) 2n ) /4 I( 2n, I(2, ) 2n ) I(2, ) I( 3, 2 ) I( I(2, ) 2N+, 2N ) lim sup 7/2 I( 2n, I(2, ) 2n ) 2.3 2N.(2N + ). But this is true for all N so therefore lim sup 7/2 I( 2n, I(2, ) 2n ) log 2. 3. ARITHMETIC PROGRESSIONS Definition 3.. If > 0 let K() := [(2n ), 2n).

8 BROUGHAN,CASEY Lemma 3.. If, y R are positive, then y H() K(y). Lemma 3.2. Let {q,..., q m } N be odd, q = q... q m, and r = min{q,..., q m }. Then if [(2n )q, (2n )q + r) we have {q,..., q m } H(). Proof. Let n N be such that [(2n )q, (2n )q + r). Then for j m, (2n )q... q m (2n )q < + r (2n )q + q j q j q j q j q j Hence q j = (2n )q... ˆq j... q m is odd so q j H(). Note: This shows, for eample, that each finite subset of odd primes {p,..., p m } H( j ) for a sequence j. Lemma 3.3. If α R then as #(N + α) H() = log 2 + O ( ) where the implied constant is absolute. ( Proof. For n N and > 0 let I n = H() = 2n, I n ] 2n so and for n, (N + α) I n has at most one point. For n >, #(N + α) I n = and hence 2n(2n ) #(N + α) H() = 2n(2n ) = log 2 + O ( ). + O ( )

HARMONIC SETS 9 Lemma 3.4. Let A(h, k) = {mk + h : m = 0,, 2,...} be an arithmetic progression with h, k Z and k > 0. Then, as #{A(h, k) H()} = log 2 k where the implied constant is absolute. ( ) + O /k Proof. The term km + h A(h, k) H() if and only if for some n 2n < km + h 2n if and only if /k 2n < m + h k /k 2n the result follows directly from Lemma 3.3 with α replaced by n/k and replaced by /k. Theorem 3.. Let h and k Z be such that (h, k) =. Then there eists a sequence m in R such that #A(h, k) P H ( m ). Proof. Number the infinite sequence of distinct primes in A(h,k) as {p, p 2,...}. Then, by Lemma 3.2, there is a number m m such that {p,..., p m } A(h, k) P H ( m ). It might be thought that with the same notation as in the theorem above, for all #{A(h, k) P H ()}. However if we let m = 2p... p m n i = p... ˇp i... p m then m [2n i p i, (2n i + ) p i ) But this last set has an empty intersection with K(p i ), so therefore p i / H ( m ) so A(h, k) P H ( m ) =. It follows that p h mod k,p H( m ) log p p = 0 log 2 φ(k) log m + O() so the natural version of Dirichlet s theorem is false. 4. MORE ACCURATE EXPRESSIONS

0 BROUGHAN,CASEY Lemma 4.. For all n N let a n be the square free part of n!. Let ɛ > 0 be given. Then for all n n ɛ, log a 2n = 2n log 2 c n + A n where A n ɛ n + 2 log(4n π ) + 2n + 5n 2. if Proof. By Stirling s formula there eists an n such that for all n n R n R n < := log n! (n log n n + log(2πn)), then 2 44n 2. If θ n is the square free part of n + then log θ n log(n + ) so for all n if T n := log θ 2n 2 log(a 2n, θ 2n ), then T n < log(2n + ). It follows from these bounds that, for n n ɛ, log a 2n = (2n + 2 ) log 2 c n 2 log(2πn) + R 2n + R n + S n + T n and R 2n + R n + S n + T n < ɛ n + log(2n + ) +, so therefore 5n2 log a 2n = 2n log 2 c n + A n, where A n ɛ n + 2 log(4n π ) + 2n + 5n 2. If n is odd then the value of a n is the same as that of a n ecept for n prime in which case an etra error of log n is incurred leading to: Lemma 4.2. Let ɛ > 0 be given. Then for all n n ɛ, log a 2n+ = 2n log 2 c n + B n

HARMONIC SETS where B n ɛ n + 2 log(8n3 π ) + n + 5n 2. Since these epressions are valid for all ɛ > 0, by first using, say, ɛ/2 and adjusting n ɛ, we can make the uniform statement Theorem 4.. Let ɛ > 0 be given. Then for all n n ɛ, where n log a n = n log 2 c 2 + C n C n ɛ n. 5. CONNECTION WITH THE RIEMANN ZETA FUNCTION Theorem 5.. Let Z be the non-trivial zeros of ζ(s). Then ρ Z ρ ρ 2 ( ) n+ n ρ ( ) = O 3/4 log 2. Proof. The von Mangoldt formula is ψ() = ρ Z ρ ρ ( 2 log ) 2 log 2π for > 0 and where the conditionally convergent sum is taken with increasing Imρ. Hence θ() = ρ Z ρ ρ + O ( log 2 ). Write θ H () = n θ( 2n ) θ( 2n ) + O ( )

2 BROUGHAN,CASEY so, by Theorem 2., 2 ( ) n+ log 2 = ρ [ n ρ ρ + O ( ) + O n log 2 ( n ( ( )) = log 2 O ρ ( ) +O 3/4 log 2 (2n ) ρ (2n) ρ ρ ρ ) [ (2n ) ρ ] (2n) ρ ] Therefore ρ ρ ρ 2 ( ) n+ n ρ ( ) = O 3/4 log 2. Theorem 5.2. Let a > and > 0. Then ψ H () = ( ζ (s))( 2 s ) s ds 2πi s. (a) Proof. Since [2, Section 3.2] it follows that ψ() = ( ζ (s)) ds s 2πi (a) ζ(s) s T () = ψ( n ) = 2πi = 2πi (a) (a) ( ζ (s)) ζ(s) s [ n s ]ds s ( ζ (s)) s ds s. (The interchange of integration and summation in the second line is justified by the finiteness of the sum and vanishing of the integral for all but a

HARMONIC SETS 3 finite number of terms.) Therefore ψ H () = T () 2T ( 2 ) = ( ζ (s))( s 2( 2πi 2 )s ) ds s = 2πi (a) (a) ( ζ (s))( 2 s ) s ds s. This representation shows that ψ H () depends only on the (double) pole of ζ (s) at s = and the value ζ (0). Since ( ) n+ η(s) := n s = ( 2 s )ζ(s), we can write ψ K () := p m H() log p = ψ() ψ H () = ( ζ (s) ds )(η(s) )s 2πi (a) ζ(s) s. This epression shows that the complimentary function ψ K () depends on (all of) the non-trivial zeros of ζ(s), and this dichotomy thus offers an eplanation for the unepected regularity of ψ H (). REFERENCES. Broughan, K. A. Asymptotic order of the squarefree part of n!. Integers: Electronic Journal of Combinatorial Number Theory, 2 A0, (2002), p-6. 2. Edwards, T.M. Riemann s zeta function. New York: Dover, 974. 3. Heath-Brown, D.R. Sieve identities and gaps between primes., Journèes Arithmètiques, Metz, (98), p6-65. 4. Sárközy, A. On divisors of binomial coefficients I, J. Number Theory 20 (985), p70-80.