NUMBER THEORETIC FUNCTIONS AND THE DIRICHLET GENERATING FUNCTION ADRIAN PACURAR˘ Contents 1. Introduction 2 2. Primer On Generating Functions 3 3. The Dirichlet Generating Function 6 4. Multiplicative Number-Theoretic Functions 7 5. The M¨obiusFunction and Euler's Totient Function 8 6. References 10 1 1. Introduction Questions about the divisors d of an integer n are at the heart of number theory. We can define many functions f : N ! R that give us information about the divisors of n. Definition. The divisor function τ : N ! N counts the number of divisors of n. We have X τ(n) = 1 djn where the sum is taken over all positive divisors d of n. Example. τ(8) = 4, since 8 has 4 divisors f1; 2; 4; 8g Example. τ(12) = 6, since 12 has 6 divisors f1; 2; 3; 4; 6; 12g. Definition. The divisor sum function σ : N ! N is the sum of the divisors of n, given by X σ(n) = d djn where the sum is again taken over all positive divisors d of n. Example. • σ(2) = 1 + 2 = 3 • σ(3) = 1 + 3 = 4 • σ(8) = 1 + 2 + 4 + 8 = 15 The following table computes τ and σ for 1; 2; 3;:::; 12. n 1 2 3 4 5 6 7 8 9 10 11 12 τ(n) 122324243426 σ(n) 1 3 4 7 6 12 8 15 13 18 12 28 We immediately see that if p is prime, τ(p) = 2 and σ(p) = 1+p, as the only positive divisors of p are 1 and p. We can generalize this for powers of p. pk+1 − 1 Lemma 1. For p prime and k > 0, τ(pk) = k + 1 and σ(pk) = . p − 1 Proof. Let D(pk) = 1; p; p2; p3; : : : ; pk be the set of all positive divisors of pk. Then τ(pk) = jD(pk)j = k + 1 by definition. To compute σ(pk), we use geometric sums: X pk+1 − 1 σ(pk) = r = 1 + p + p2 + ··· + pk = p − 1 r2D(pk) which concludes our proof. Notice how τ(6) = τ(2)τ(3), τ(10) = τ(2)τ(5), but this pattern doesn't always hold: τ(12) = 6 6= τ(2)τ(6) = 2 · 4 = 8. It turns out that if m; n are relatively prime, τ(mn) = τ(m)τ(n). We have the following theorem: 2 Theorem 1. For distinct primes p1; p2; : : : pr, and positive integers k1; k2; : : : ; kr, we have k1 k2 kr k1 k2 kr τ(p1 p2 ··· pr ) = τ(p1 )τ(p2 ) ··· τ(pr ) k1 k2 kr Proof. We use a simple counting argument. Let N = p1 p2 ··· pr , and consider the sets n 2 ki o Di = 1; pi; pi ; : : : ; pi for i = 1; 2; : : : ; r. Any divisor of of N will be a product of up to ki elements from each Di, and so the set D(N) of all positive divisors of N has size r Y k1 k2 kr τ(N) = jD(N)j = jD1j · jD2j · · · jDrj = (ki + 1) = τ(p1 )τ(p2 ) ··· τ(pr ) i=1 where the last equality follows from Lemma 1. A closer look at our table suggests that, similarly to τ, in general we do not have σ(mn) = σ(m)σ(n), but it turns out that this result does hold for m and n are relatively prime. Lemma 2. Let m and n be relatively prime. Then σ(mn) = σ(m)σ(n). Proof. Let D(m) = f1; m1; m2; : : : ; mg be the set of all positive divisors of m, and D(n) = f1; n1; n2; : : : ; ng be the set of all positive divisors of n. Observe that any divisor of mn is of the form minj, where mi 2 D(m) and nj 2 D(n). The sum of the divisors of m and n, respectively, are σ(m) = 1 + m1 + m2 + ··· + m and σ(n) = 1 + n1 + n2 + ··· + n It is easy to see that multiplying these two expression gives a sum of all possible combinations minj, and so X σ(mn) = minj = (1 + m1 + m2 + ··· + m)(1 + n1 + n2 + ··· + n) = σ(m)σ(n) i;j as desired. Theorem 2. For distinct primes p1; p2; : : : pr, and positive integers k1; k2; : : : ; kr, we have r Y pki+1 − 1 σ(pk1 pk2 ··· pkr ) = σ(pk1 )σ(pk2 ) ··· σ(pkr ) = i 1 2 r 1 2 r p − 1 i=1 i Proof. Follows immediately from lemmas 1 and 2. 2. Primer On Generating Functions In the introduction, we saw that the sequences 1 fτ(n)gn=1 = f1; 2; 2; 3; 2; 4; 2; 4; 3; 4; 2; 6;:::g and 1 fσ(n)gn=1 = f1; 3; 4; 7; 6; 12; 8; 15; 13; 18; 12; 28;:::g are quite irregular, even though we were able to obtain some formulas for them. One use- 1 ful tool for studying sequences of real numbers fangn=0 = fa0; a1; a2;:::g is a generating function. 3 1 Definition. The ordinary generating function (OGF) of the sequence fangn=0 is a formal power series in the variable x whose coefficients are the terms in the sequence: 1 X n 2 3 A(x) = anx = a0 + a1x + a2x + a3x + ::: n=0 Example. The OGF of the sequence f1; 1; 1 :::g is the function 1 1 + x + x2 + x3 + ··· = 1 − x Usually, we are not interested in the question of convergence for these series. We are simply interested in the coefficients it encodes. Manipulating the power series corresponds to performing certain operations on the sequence of coefficients we are working with. Example. Starting with two sequences fangn≥0 = fa0; a1; a2;:::g and fbngn≥0 = fb0; b1; b2;:::g, with corresponding OGFs A(x) and B(x), respectively, we have the following: (a) Removing the first element of a sequence: A(x) − a fa ; a ; a ;:::g ! 0 1 2 3 x (b) Adding two sequences together, i.e. fan + c · bngn≥0 where c is a constant: fa0 + cb0; a1 + cb1; a2 + cb2;:::g ! A(x) + cB(x) (c) 0 f(n + 1)an+1gn≥0 ! A (x) (d) Z x n a1 a2 a3 o 0; a0; ; ; ::: ! A(t)dt 2 3 4 0 Pn Example. The convolution of two sequences, i.e. cn = k=0 akbn−k, has OGF fcngn≥0 ! A(x)B(x) Pn Example. Given a sequence an, observe that the sequence of partial sums cn = k=0 ak is given by A(x) fc g ! n n≥0 1 − x Example. It is widely known that the n-th harmonic number 1 1 1 H = 1 + + + ··· + n 2 3 n is never an integer for n ≥ 2. What is the OGF of the sequence fHng? By the previous 1 1 example, it suffices to find the OGF of 1; 2 ; 3 ;::: and divide it by 1−x to get the sequence of partial sums. From calculus, we know that X xn = − log(1 − x) x n≥1 4 and so the OGF must be X 1 1 H xn = log n 1 − x 1 − x n≥1 There are other types of generating functions, some more useful than others depending on the properties of the sequence we are dealing with. 1 Definition. The exponential generating function (EGF) of the sequence fangn=0 is 1 X xn A(x) = a = a + a x + a x2 + a x3 + ::: n n! 0 1 2 3 n=0 P1 xn x Example. The EGF of the sequence f1; 1; 1;:::g is n=0 n! = e . P1 xn Example. The EGF of the factorial sequence f1; 1; 2; 6; 24;:::g is n=0 n! n! = 1=(1 − x), which is the same as the OGF of f1; 1; 1;:::g. Example. Starting with a sequence fangn≥0 = fa0; a1; a2;:::g with EGF f(x), removing 0 the first element (i.e. looking at the sequence fa1; a2;:::g) is equivalent to computing f (x), since 1 1 1 X nxn−1 X xn−1 X xn f 0(x) = a = a = a n n! n (n − 1)! n+1 n! n=1 n=1 n=0 Example. The Fibonacci numbers satisfy the recurrence Fn+2 = Fn+1 + Fn, with initial conditions F0 = 0 and F1 = 1, and by the previous example the EGF of Fn must satisfy the differential equation f 00(x) = f 0(x) + f(x). I will skip the details here, but solving this differential equation gives the EGF of the Fibonacci sequence p 1 1 ± 5 (r+)x (r−)x f(x) = p e + e ; r± = 5 2 This gives a much easier method of solving for Binet's formula, which could be obtained using the OGF, but it involves partial fraction decompositions. Example. Given sequence fang and fbng with EGF A(x) and B(x), what is the sequence whose EGF is the product AB? (In the OGF case, we saw this was a convolution.) i j i+j ! n ! X x X x X x X X aibj X x X aibjn! AB = a b = a b = xn = i i! j j! i j i!j! i!j! n! i!j! i≥0 j≥0 i;j≥0 n≥0 i+j=n n≥0 i+j=n and so the coefficient of xn=n! in the product AB is X n c = a b n k k n−k k Hence the resulting sequence is another type of convolution of of an and bn. 5 3. The Dirichlet Generating Function While the OGF and EGF are useful in studying many sequences of numbers, they do not have the desired properties to help us in studying the sequences that arise from the number-theoretic functions τ(n) and σ(n).