ROSE WONG. f(1) f(n) where L the average value of f(n). In this paper, we will examine averages of several different arithmetic functions.

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1 AVERAGE VALUES OF ARITHMETIC FUNCTIONS ROSE WONG Abstract. I this paper, we will preset problems ivolvig average values of arithmetic fuctios. The arithmetic fuctios we discuss are: (1)the umber of represetatios of atural umbers as a sum of two squares ad ()as a sum of three squares, ()the umber of decompositios of atural umbers ito sums of cosecutive primes, ()the umber of primitive pythagorea triagles with a hypoteuse of a give legth, ad (5)the umber of divisors of atural umbers. 1. Itroductio A arithmetic fuctio is defied to be a fuctio f(), defied for N, which maps to a complex umber such that f: N C. Examples of arithmetic fuctios iclude: the umber of primes less tha a give umber, the umber of divisors of, ad the umber of ways ca be represeted as a sum of two squares. While the behavior of values of arithmetic fuctios are hard to predict, it is easier to aalyze the behavior of the averages of arithmetic fuctios which we defie as: f(1) f() lim = L where L the average value of f(). I this paper, we will examie averages of several differet arithmetic fuctios.. Average umber of represetatios of a atural umber as a sum of two squares Let r() be the umber of represetatios of as a sum of two squares = x + y. Theorem.1. The average umber of represetatios of a atural umber as a sum of two squares is π. That is, Date: November 0,

2 ROSE WONG r(1) + r() r() lim = π. Proof. For a fixed, let R() = r(1) r() We iterpret r() to be the umber of lattice poits (poits with iteger coordiates) o the circle x + y =. R() is the umber of lattice poits i the disc x + y except for the origi. If we place a uit square (cetered o the lattice poit with sides parallel to the coordiate axes) o each lattice poit covered by the dis, the total area of these squares is 1+R(), because R() does ot iclude the origi. This is ot exactly the area of the dis, as some squares go beyod the dis s boudary ad some areas of the circle are ot covered by ay square. However, we ca fid a circle that circumscribes all the squares ad a circle that is iscribed by all the squares. These will be the boudig outer ad ier circles, respectively. 1 The radius of the outer circle is +, because the greatest distace from the origi (0,0) to a outermost lattice poit is ad the greatest distace from that lattice poit to the edge of its correspodig uit square is 1. Hece, the area of the outter circle is larger tha the area of the squares. 1 + R() < π( + 1 ). By a similar argumet, the area of the ier circle is less tha the area of the squares. 1 + R() > π( 1 ) We ca simplify the iequalities usig the fact that π < 5 ad 0 < π 1 < ( = 1,,,...). Therefore, ad R() < π( + 1 ) 1 = π + π + ) R() > π( 1 It follows that 1 = π π + π 6 < R() < π + 6, π 1 < π + 6 π 1 > π 6.

3 AVERAGE VALUES OF ARITHMETIC FUNCTIONS which is the same as R() π < 6. If we divide by, we get R() 6 π <. Taig the limit as, we have R()/ complete. π ad the proof is. Average Number of Represetatios of a Natural Number as a sum of three squares We will ow exted the previous problem, that is, we will fid the average umber of represetatios of a atural umber as a sum of three squares. Agai, we will use geometry with iteger lattice poits to approach this problem. Let f() deote the umber of itegral solutios of x + y + z =. Theorem.1. The average umber of represetatios of a atural umber as a sum of three squares is π. Proof. For a fixed, let f(1) + f() f() lim = π / F() = f(1) f(). We iterpret f() to be the umber of lattice poits o the sphere x + y + z =, so F() to be the umber of lattice poits i the sphere x + y + z except for the origi. If we place a uit cube o each lattice poit covered i the sphere, the total volume of these cubes is 1 + F(). We wat to fid a sphere that ecloses all of the cubes ad a sphere that is iscribed iside the cubes. The radius of the outer sphere is +. Sice the greatest distace betwee outermost lattice poit ad the edge of its correspodig cube is ( 1 ) + ( 1 ) + ( 1 ) =. Similarly, the radius of the ier sphere is. Comparig the volumes, we have 1 + R() < π( + )

4 ROSE WONG ad 1 + R() > π( ). Expadig the iequalities out, we have ad R() < ( π / + π 1) + ( π + π) R() > ( π / + π 1) ( π + π). Usig the fact that π < 11 ad π <, we have Dividig by / gives R() π / π + 1) < 11. R() π 1 11 π + / ) < /. / Taig the limit as, we have complete. R() / π ad the proof is. Average value of umber of decompositios of a atural umber ito a sum of cosecutive primes Let f() be the umber of decompositios of a atural umber ito a sum of oe or more cosecutive prime umbers. For example, f(95) = because 95 = = Theorem.1. The average umber of decompositios of a atural umber ito a sum of oe or more cosecutive prime umbers is log. [1] f(1) f() lim = log Proof. Let P be a sequece of cosecutive primes: p 1 < p < p <... Every set of cosecutive primes whose sum is less tha or equal to x cotributes 1 to the sum f(1)+f()+... +f(). The umber of sets of r primes that satisfy the coditio is at most π( x ) ad at least π( x ) r. r r Therefore, we have x x (π( ) r) f(1) + f() f(x) (π( )), r r r=1 r=1

5 AVERAGE VALUES OF ARITHMETIC FUNCTIONS 5 where is determied by p 1 + p p x < p 1 + p p +1 Usig the above iequality ad p r r log r we get x log x This implies that r = O(x). If we ca show that (π( x )) r=1 r=1 r x log, it will follow that f(1) + f() f(x) x log Usig the above relatioship ad the prime umber theorem, we have x x x r=1 (π( r )) r=1 ( r log x/r ) 1 r log x/r dr = [ x log(log x/r)] 1 x = x(log log x log log ) x(log log x log log x log x) x log. Notatio explaatio: Let f ad g be fuctios of x. The otatio f g deotes that f(x)/g(x) is bouded above ad below by positive umbers for large values of x. The otatio f = O(g) deotes that c such that f(x) cg(x). The otatio f g deotes that f(x) lim x, = 1 g(x) 5. Average value of primitive pythagorea triagles with hypoteuse equal to Let P() be the umber of primitive Pythagorea triagles with hypoteuse equal to. For example, P(5)=1 ad ad P(65)= because + = 5, + 56 = = 65 We would lie to determie whether the average value of P() coverges, ad if so, to what value. That is, P(1) P() lim = K Rather tha approachig this problem through mathematical argumets, we will use computer computatios to ifer a value. For each value of from 1 to 1000, we geerate ad graph the average values up to ad icludig. Looig at Figure 1, we see that the average values (measured o the vertical axis) oscillate, but as, the

6 ROSE WONG Figure 1. Average P() for =[1,1000] average values level off. We ca cojecture that the limit as of the average value is: P(1) P() lim = Average value of divisors of a atural umbers Let d() be the umber of divisors of the atural umber. We will show that Theorem 6.1. The average value of the umber of divisors of atural umbers grows lie log. d(1) + d() d() log Proof. Let be a fixed iteger. If we list the multiples of less tha or equal to :,,,...[ ], we fid that there are [ ] multiples, where [ ] deotes the floor fuctio. Each of those multiples cotributes 1 to the sum d(1) d(). If we examie multiples of all itegers, it follows that summig over gives [ ] = d(1) d() =1

7 AVERAGE VALUES OF ARITHMETIC FUNCTIONS 7 Now, we wat to prove that =1 [ ] lim = 1 log First, we establish the relatioship: 1 < [ ]. Summig over gives: ( 1) < [ ]. =1 =1 =1 We the factor out to get: ( ) < [ ] =1 =1 =1 The first ad last term i the above iequality ca be rewritte as the itegrals ( 1 ) d ad d. 1 Itegratig gives log + 1 < [ ] log. =1 So taig, we have [ =1 ] lim = 1. log Ad so, d(1) + d() d() log. Refereces [1] Moser, L., O the Sum of Cosecutive Primes, Notes o Number Theory III, Caadia Mathematical Bulleti 6, p , 196. [] Youg, Robert M., Excursios i Calculus, The Mathematical Associatio of America, USA, 199.

Average Values of Arithmetic Functions. Rose Wong December 5,

Average Values of Arithmetic Fuctios Rose Wog December 5, 006 8.04 Outlie. Itroductio. Average umber of represetatios of as a sum of two squares. Average umber of represetatios of as a sum of three squares