CHARACTER SUMS AND RAMSEY PROPERTIES OF GENERALIZED PALEY GRAPHS. Nicholas Wage Appleton East High School, Appleton, WI 54915, USA.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 CHARACTER SUMS AND RAMSEY PROPERTIES OF GENERALIZED PALEY GRAPHS Nicholas Wage Appleton East High School, Appleton, WI 54915, USA wageee@gail.co Received: 8/3/05, Revised: 5/16/06, Accepted: 5/6/06, Published: 6/1/06 Abstract In a classic paper, Evans, Pulha, and Sheehan coputed the nuber of coplete graphs of size 4 for a special class of graphs called Paley Graphs. Here we consider analogous questions for generalized Paley-type graphs. 1. Introduction The Rasey nuber R is the sallest positive integer such that any graph G with R or ore vertices contains a coplete subgraph of size or its copleent G c contains a coplete subgraph of size. The only values of for which R is known are 4. We consider a natural etension of coputing and estiating R. For a graph G, let k G denote the nuber of coplete graphs of size contained in G. Let T n := ink G + k G c 1 as G ranges over all graphs with n vertices. Note that R is the sallest n for which T n > 0. In [Erd6], Erdős conjectured that T n n n = 1 1!. However, this is not quite correct. In [Tho97], Thoason showed that for = 4, this it is no ore than 1/33 4!, rather than 1/3 4! as the conjecture predicts. For a prie p 1 od 4, let Gp denote the Paley graph with p vertices, indeed 0 through p 1. A Paley graph contains an edge y if and only if φ y = 1, where φ denotes the Legendre sybol. This notation assues that the prie p will be clear p

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 fro contet. For eaple, the Paley graph G5 is a pentagon. The Legendre sybol is ultiplicative and φ 1 = 1 if p 1 od 4. Therefore, φy = φ y. This equation iplies that our definition is syetric in and y, so the edges are well defined. For pries p 3 od 4, φ 1 = 1, so φ y = φy. In this case we define a generalized directed Paley graph. Note that since φ 1 = 1 for every and y, there is eactly one directed edge, creating a type of graph called a tournaent. Paley graphs have any interesting properties. They are both self-copleentary and self-siilar. Also, for the function k G + k G c on graphs with p vertices, Paley graphs are inial in certain ways. For eaple, the Rasey nuber R4 = 18; the Paley graph G17 is the only graph up to isoorphis with 17 vertices such that k 4 G + k 4 G c = 0. This inspired Evans, Pulha, and Sheehan [EPS81] to copute the eact value of k 4 Gp + k 4 Gp c using character sus. For > 4, the character sus are difficult to deterine eplicitly. We prove an asyptotic result for all. Theore 1. For pries p 1 od 4 and positive integers k Gp 1 =. p p! Doubling this value to account for the coplete graphs in its isoorphic copleent, we see that the conjecture of Erdős, while false for general graphs, is true for the Paley graphs. Moreover, the conjectured bound is eactly achieved by these graphs. For all pries p and integers t > 1 we define two other types of generalized Paley graphs, G t p and G tp. The graph G t p has an edge y if and only if both φ ty = 1 and φy t = 1. The graph G tp has an edge y if and only if at least one of these stateents is true, that is φ ty = 1 or φy t = 1. Note that G t p is a subgraph of G tp. We now define two related generalized Paley graphs. Definition 1. For a prie p and integer t, let G t p denote the graph with p vertices, indeed 0 through p 1, which contains an edge y if and only if φ ty = 1 and φy t = 1. Siilarly, let G tp denote the graph with p vertices, indeed 0 through p 1, which contains an edge y if either φ ty = 1 or φy t = 1, and contains no edge y if neither of these stateents hold true. We prove a siilar result for these generalized graphs. Theore. Let be a positive integer and t be an integer such that t 1, 0, 1 od p. Then, for pries p, and we have k G t p + k G t p c 1 = 1 3 +. p p! 4 4

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 3 Siilarly, k G tp + k G tp c = 1 p p! 1 4 + 3. 4 Reark. After this work was copleted, it was brought to the author s attention that a paper of Graha, Chung, and Wilson, [CGW88], iplies these its because Paley graphs are eaples of quasi-rando graphs. However, we stress that the proofs in this paper are direct, and lead to stronger error ters in estiates. In other words, a careful analysis of these proofs leads to strong bounds for the errors between the iting values and the actual values. A key result in the work of Evans, Pulha and Sheehan is a forula that eactly coputes the nuber of coplete graphs of size 4 in a Paley graph, naely, they obtain k 4 Gp = p 1 p p 9 4 3 1536 where is an even integer such that p = + y. It is known that for p 1 od 4, that not only eists, but is unique up to sign. Their proof essentially follows fro an evaluation of the hypergeoetric su 3F t = φ 1 p Z/pZ y Z/pZ φφyφ1 φ1 yφ ty 4 when t = 1. One should note that while it can be shown that this is equivalent to the definition of 3 F t, it is not a priori the definition. Siilarly, we will require the su F 1 t = φ 1 p Z/pZ φφ1 φ1 t. 5 In this paper we copute a graph theoretic result siilar to that of Evans, Pulha, and Sheehan which would norally be intractable without the help of the special t evaluations of these character sus. Theore 3. Let p be a prie, and let t be a non-zero eleent of Z/pZ whose ultiplicative order is greater than 3. Let H t p represent the directed graph where y is an edge if and only if φ = φ ty = 1, and let C t p denote the nuber of 3-cycles in H t p y z. Then C t p = p 1 [ 90 3pφt + 1φt + t + p 14p + 48φ t + 7φt + 6φ1 t 19 + 6φ1 t 3 + 1φt t 3 + 1φ1 t + 1φ 1 + 1φ1 + t + 4φt t + 6φt t 4 + 1φt t + 3φt 4 tp 6pφ1 t 3φ tp + p 3F t 3 ] 6p1 + φ1 t F 1 t 3p1 + φt F 1 t 3.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 4 Although this forula is difficult to grasp at first glance, we note that it siplifies draatically for several special t. For eaple, we obtain forulas uch like 3 in the following cases: Corollary 1. If t =, and p is a prie congruent to 5 or 7 od 4, then { p 1 19 C t p = p 6p + 4 + 1 if p 5 od 4, + y = p, and odd, p 1 19 p 6p + 5 if p 7 od 4. Corollary. If t = 1, and p is a prie congruent to 7 or 13 od 4, then { p 1 19 C t p = p 6p + 5 if p 7 od 4, p 1 19 p + p 4 + 1 if p 13 od 4, + y = p, and odd. In Section we state the tools that will be necessary in Section 3, where we will prove the ain theores.. Preinaries We will need the following facts for our proofs. The deepest result we require is due to Weil. Weil s Theore. [BEW98], pg. 183 If p is a prie, and f Z[] is a polynoial with degree n which is not congruent odulo p to cg for any integer c and polynoial g with integer coefficients, then f φ < n 1 p. Z/pZ Reark. Fro this point onward, we oit the suation indices with the understanding that all sus are over Z/pZ unless indicated otherwise. For sall n, eplicit evaluations of these sus are siple. Clearly, if n = 0, φc = φcp. 6 Note that as runs over all the eleents of Z/pZ, so does a + b if a 0 od p. Using this fact, we can also copute the value for n = 1. Let q denote a non-zero, non-residue of

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 5 Z/pZ. Then φa + b = φy y=a+b = z=q 1 y φqz = φq φz z = y = φy φa + b, so φa + b = 0. 7 Finally, we prove a forula for the case n =. Proposition 1. For any prie p and polynoial f = a + b + c with roots r and s is Z/pZ, { φa p 1φa if r s od p + b + c = φa if r s od p. Proof. φa + b + c = = φa φ a r s φφ s r. Let t = s r. Note that when t 0 od p, it is clear that φ = p 1. When, t 0 od p, φφ t = φtφt 1 = φφ 1, t so the suation is equivalent to the suation when t 1 od p. Thus, this is equivalent for all p 1 non-zero values of t. Note that φ t = φ φ t t t = φ φu u= t = φ 0 = 0.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 6 However, we can also show that φ t t = φ + p 1 φφ t = p 1 1 + φφ t. Equating these two epressions, we obtain φa φφ t = φa. For n 3 there does not appear to be a general evaluation. However, with additional achinery, certain evaluations are possible. A character χ is a ultiplicative function that aps the eleents of Z/pZ to the cople roots of unity. Let χ denote the conjugate character of χ, that is, χ = 1/χ. By convention, we say that χ0 = 0. The Legendre sybol is a siple eaple of a character. Another siple character is ɛ, the trivial character, for which ɛ = 1 unless 0 od p. A Jacobi su, for characters χ and ψ is Jχ, ψ := χψ1. 8 Additionally, we define the noralized Jacobi su for characters A and B A := B 1 JA, B p B. 9 Then, we define n+1f n A0, A 1,...A n t := p B 1,...B n p 1 A0 χ χ χ A1 χ B 1 χ An χ χt, 10 B n χ where the su is over all χ with odulus p. Additionally, we define φ, φ,...φ n+1f n t := n+1 F n ɛ,...ɛ t. 11 Although these functions are epressed in ters of Jacobi sus, it is known that the definitions in the introduction, 5 and 4, are equivalent see [Ono04]. These objects allow us to find a general forula in ters of F 1 when n = 3. Proposition. For any prie p and third-degree polynoial with leading coefficient a and distinct roots q, r, s Z/pZ, we have r s φ a q r s = φ aφq s F 1 p. q s

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 7 Proof. The proof of this is straightforward: φ a q r s = φa = φaφq s φ q s r s q s = φ aφq s F 1 r s q s φφ1 φ 1 r s q s p. 3. Proof of the Theores Using Weil s Theore fro the previous section, we will prove Theore 1. Proof of Theore 1. The following is a proof by induction. For = 1, k 1 Gp = p, so the theore holds. Assue the theore holds for 1. For each prie p 1 od 4 we want to write k Gp as a character su. To count the nuber of coplete -subgraphs, we look at each possible set of distinct points, and check to see if it is a coplete subgraph. Since y is an edge if and only if φ y = 1, then assuing y the equation 1+φ y conveniently returns 1 if y is an edge, and 0 otherwise. Thus, if we take the product of this equation over all pairs selected fro our points, it will return 1 if we have a coplete subgraph, and 0 otherwise. We need only su this product over all possible sets of distinct points to get k Gp. Therefore k Gp = a 1 < <a 0<i<j 1 + φa i a j. We will now anipulate this equation to ake it easier to evaluate. First we can isolate the a ters into their own suation. Second, we can replace the condition that a is greater than a 1 and siply require it to be distinct fro each of the other a i. Note that this will count each set of vertices ties, once for each possible final ter, so we ust divide by. Then, we can separate out the ters of the product not involving a, and pull out a 1 ter. The result of all these anipulations is k Gp = a 1 <...<a 1 [ 0<i<j< 1 + φa i a j 1 1 1 a a i k=1 1 + φa a k ].

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 8 Note that the beginning of this equation resebles k 1 Gp. If we bound 1 a a i k=1 1 + φa a k in ters of p and, then we can pull it outside of the suation. Specifically, if we show that then p p + 1 a a i k=1 p p + k 1 1 Gp k Gp and by the inductive hypothesis we know that p p + k 1 Gp p p 1 1 + φa a k p + p + 1, 1 p + p + 1 1 p p + = p 1 p = 1!. The right hand bound evaluates to the sae it, and so k Gp 1 =, p p! which proves our result. k 1 Gp, k 1 Gp p 1 All that reains is to prove the bound for 1. We start by ultiplying out each product, getting 1 ters. The first ter is 1, which we su over the p 1 possible values of a. Now, consider an arbitrary ter fro the reaining 1 ters. It is a product of Legendre sybols of the for φa a i. Since the Legendre sybol is ultiplicative, we can for a single Legendre sybol, that is, write the ter as φfa where f is a polynoial with coefficients in Z/pZ. We know the roots of f are aong the different a i values, that f contains at least one such root, and, since all the a i are distinct, f has no repeated roots. This eans that it is not congruent od p to cg for any integer c and polynoial g with integer coefficients. Moreover, f has degree at ost 1 so by Weil s theore, we have p φ fa p. a Z/pZ

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 9 Note that this su, however, includes 1 ters which are not in the su we wish to evaluate, naely a 1,..., a 1. Each of these ters is between 1 and 1, so reoving the changes the su by at ost 1. Therefore, p + 1 1 p + 1 fa 1 p + 1 p + 1. a a i φ This bound holds for each of the 1 1 ters involving Legendre sybols, so p + 1 1 p + 1 which iplies 1. 1 a a i k=1 1 + φa a k p + 1 + 1 p + 1, The proof of Theore is siilar to the proof of Theore 1. This tie we eaine k G t p as a character su and use Weil s Theore to estiate it. Proof of Theore. We first prove that k G t p = 1 p p! 1. 13 4 First consider and y such that y, ty, t 1 y, t y, t y od p. There is an edge between and y in G t p if and only if φ ty = φy t = 1. Thus, { 1 + φ ty 1 + φy t 1 if y is an edge, = 14 0 otherwise. As such, we define h p := a 1 <...<a a i ta j,t a j 0<i,j i j 1 + φa i ta j. 15 This counts the nuber of coplete subgraphs in G t p that do not contain any edges where t y, t 1 y, t 1 y, t y od p. However, each Z/pZ has at ost 4 such edges, so there are at ost p of these edges total since each edge is counted twice. Each such edge is a eber of p subgraphs of size, and so we are issing at ost p p coplete subgraphs of size. Thus h p k G t p h p + p p.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 10 We divide this equation by p and take the it as p tends to infinity. The p p ter has degree p 1, so it vanishes, aking the left hand side and right hand side its equal. Therefore k G t p p p = p h p p, and all that reains is to evaluate 15. In particular, we will show that h p p p = 1. 4! This will be proved by induction. If = 1, it is clearly true. Otherwise, assue the clai is true for 1. Perforing the sae anipulations as in the proof of Theore 1, we get h p = a 1<...<a a i ta j,t a j 0<i,j< i j 1 4 1 1 + φa i ta j a t e a i e 1 k=1 1 + φa ta k 1 + φa k ta. So, if we show that 1 1 + φa ta k 1 + φa k ta p 4 1 p + 5, 16 a t e a i k=1 then e p p 4 1 p + 5 4 1 Applying the inductive hypothesis, we obtain p 4 1 p + 5 h 1 p h p p + 4 1 p + 5 h 4 1 1 p. h 1 p p 4 1 p + 5 = 4 1 p p 4 1 p = 1 4!. The right hand side it evaluates to the sae value and thus h 1p p 1 p h p p = 1 4!. 17

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 11 All that reains is to prove the bounds for 16. We ultiply out the product, getting 4 1 ters. The first ter is siply 1. There are at ost p choices for a, and at least p 5, so suing it over all of the returns a value between p 5 and p. Now, consider an arbitrary ter out of the reaining 4 1 1 ters. It is a product of Legendre sybols of the for φa ta i and φa i ta. Since the Legendre sybol is ultiplicative, we can for a single Legendre sybol, that is, write the ter as φfa where f is a polynoial with coefficients in Z/pZ. We know f has at least one root, and we know that each of the roots of f is either of the for ta i or t 1 a i for soe i. Also, since t ±1, ta i t 1 a i, so if f had a repeated root r, we could write r = ta i = t 1 a j. However, this iplies that a j = t a i and our suation requires that we choose a i and a j such that this is false. Thus, f has no repeated roots, and this eans that it is not congruent odulo p to cg for any integer c and polynoial g with integer coefficients. Moreover, f has degree at ost, so by Weil s theore we have φ fa 3 p. a Note that the above su includes at ost 5 ters that the suations in 16 ecludes. Each of these ters is between 1 and 1, so reoving the changes the su by at ost 5. Therefore, a t e a i e φ fa p + 5. This bound holds for each of the 4 1 1 ters involving Legendre sybols, so 16 holds. The other three calculations of p k /p are analogous. For and y where t y, t 1 y, y, ty, t y, we replace 14 with 3 φ ty φy t φ tyφy t 4 3 + φ ty + φy t φ tyφy t 4 1 φ ty 1 φy t The calculations follow as above. for G t p c ; for G tp; for G tp c. We now prove Theore 3, the result epressing C t p in ters of hypergeoetric functions. Proof of Theore 3. The first step is to once again write our epression for C t p as a character su. No cycle can contain 0 because φ0 = 0, and so 0 has no outgoing edges. So we

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 1 look at each set of three distinct, non-zero points, y, and z, and check for a cycle. Also, we ay assue that ty, y tz, and z t, because if one of these is true, there is no cycle. Note that any such group of, y, z the epression 1 + φ 1 + φ y 1 + φ z 1 + φ ty 1 + φ y tz 1 + φ z t returns 6 if, y, and z for a cycle, and 0 otherwise. If we su this over each possible set, we count each cycle three ties. Thus 3 6 C t p = 1 + φ 1 + φ y 1 + φ z z 0 0,y, y 0,z, tz z,ty,t 1 z 1 + φ ty 1 + φ y zt 1 + φ z t. Since z 0, z runs over Z/pZ as does, and yz runs over Z/pZ as y does, so we can replace by z and y by yz. We can pull the 1 + φz ter to the very front of the suation. Note that φz 0, and when φz = 1 and 1 + φz = 0, the whole inner ter is ultiplied by 0, so its accuracy is irrelevant. Thus, assuing that φz = 1 is always true within the inner suations does not change the value of the epression as a whole. Using this fact, we can entirely rid the inner suation of z and write 3 6 C t p = c 1 + φ z = p 1c. z 0 where c = y 0,1, t 0,1, t 1,y,ty 1 + φ 1 + φ y 1 + φ y t 1 + φ 1 t 1 + φ ty.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 13 The 5 values does not take on are all distinct unless y = t 1 or y = t. Thus c = 1 + φ 1 + φ y 1 + φ y t 1 + φ 1 t 1 + φ ty y 0,1, t,t 1,t + 0,1 t 1 + 0,1 t 1,t y 0,t y 0,t y 0,t y 0,t y 0,t 1 + φ 1 + φ t 1 + φ 1 t 1 + φ t t 3 1 + φ 1 1 + φ 1 + φ 1 t 1 + φ 1 t 3 1 + φ t 1 1 + φ y 1 + φ y t 1 + φ ty 1 + φ y 1 + φ y t 1 + φ 1 t 1 + φ 1 ty 1 + φ y 1 + φ y t 1 + φ t 1 + φ t 1 ty 1 + φ y 1 + φ y t 1 + φ 1 ty 1 + φ y ty 1 + φ y 1 + φ y t 1 + φ ty 1 + φ 1 t y = S 0 + S 1 + S S 3 S 4 S 5 S 6 S 7, where S i corresponds to the ter in the ith row. Note that since the order of t is greater than three, t 1, t 1, t so there is no double counting. Now we wish to evaluate these 8 ters. Key to the siplification will be the following trick: if we have an epression of the for 1 + φα..., where we know α 0, then we ay assue throughout the... that φα = 1, because either this will be true, or the entire epression will be ultiplied by 0 so it is irrelevant. We will also ake use of equations 6 and 7, as well as propositions and outlined in the preinaries. With these tools it is not difficult to reduce the epressions to closed fors in ters of F 1 and 3F values. Fro this point on, all equations have been verified as accurate for sall pries less than 00 by direct coputer calculations in order to prevent arithetic errors during the anipulations. For S 0, we pull the two ters involving just y out of the inner suation, and evaluate it. Then we switch the order of suation and solve the inner one. Thus

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 14 S 0 = = = y 0,1, t,t 1,t y 0,t 1 + φ 1 + φ y 1 + φ y t 1 + φ 1 t 1 + φ ty 1 + φ y p 1 φ t 1 + φ y t p 1 φ t + φφ1 tφ ty p 8 φt φ1 t φ1 t 3 φ t φt t 3 φ1 t + p 3F t 3 + 3 + φt 4 tp + φ t + φ 1 + φt 3 1 p 1 + φ t 1 + φ 1 t F 1 t p 1 + φ 1 t F 1 t p 1 + φ t F 1 t 3. To evaluate S 1, we pull the ters independent of to the front of the suation. Then since the epression is ultiplied by 1 + φt, we ay use the trick entioned above to assue that φt = 1 throughout the rest of the epression. Net, we take the suation over all and subtract out the specific cases 0, 1, and t 1. Then we ultiply out the inside of the suation and evaluate it ter by ter. This shows that S 1 = = 1 + φ t 1 + φ t t 3 0 t 1,1 1 + φ 1 + φ 1 t 1 + φ 1 1 + φ t 1 + φ 1 t p + p F 1 t φ 1 4φ1 t φ t 7. The value of S is coputed by noting that after pulling out a φ t fro the third ter of the suand, it is siilar to the second ter and so we can pull a 1+φ t to the front using the trick above, and evaluate it in a fashion siilar to S 1. The resulting epression can be siplified further using the trick entioned above. Hence S = 1 + φ 1 t 3 0,t t 1,1 = 1 + φ 1 t 3 1 + φ t = 1 + φ 1 t 3 1 + φ t 1 + φ 1 + φ 1 t 1 + φ tφ1 t 0,t t 1,1 1 + φ 1 + φ 1 t p 8 4φ1 t φ 1. We can use a siilar trick in the evaluation of S 3. This shows that S 3 = y 0,t = 1 + φ t 1 + φ y 1 + φ y t 1 + φ ty p 9 φ 1 φ1 t 3 φ1 t φt 1 φ1 t.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 15 The evaluation of S 4 involves the sae anipulations as in S 1. These result in S 4 = 1 + φ 1 t 1 + φ y 1 + φ y t 1 + φ 1 ty Likewise, we obtain = 1 + φ 1 t S 5 = = 1 + φ t 1 + φ t y 0,t p + p F 1 t 15 6φ t φt φ1 + t + t φ t t t 3 φ1 + t φt + t. y 0,t 1 + φ y 1 + φ y t 1 + φ 1 t y p + p F 1 t 3 11 4φ 1 4φ1 t 3 4φ1 t 4φ1 + t 4φ1 t. To evaluate S 6, we siply pull ters to the front and notice a siilarity to a previous ter. We have S 6 = y 0,t 1 + φ y 1 + φ y t 1 + φ y 1 + φ 1 ty 1 + φyφ1 t = 1 + φ 1 t = S 4. Siilarly, we have S 7 = = S 5. Adding this all up, we get C t p = y 0,t 1 + φ t and this is the desired forula. y 0,t 1 + φ y 1 + φ y t 1 + φ 1 ty 1 + φ y 1 + φ y t 1 + φ 1 t y p 1 [ 90 3pφt + 1φt + t + p 14p + 48φ t + 7φt + 6φ1 t 19 + 6φ1 t 3 + 1φt t 3 + 1φ1 t + 1φ 1 + 1φ1 + t + 4φt t + 6φt t 4 + 1φt t + 3φt 4 tp 6pφ1 t 3φ tp + p 3F t 3 ] 6p1 + φ1 t F 1 t 3p1 + φt F 1 t 3, Proof of Corollaries 1 and. We begin by stating two well known facts about quadratic residues. First, φ is 1 when p 1 or 7 od 8 and 1 when p 3 or 5 od 8. Second, φ3 is 1 when p = 1 or 11 od 1 and is 1 when p 5 or 7 od 1. With these it is siple to show that if p 5 or 7 od 4 and t = or p 7 or 13 od 4

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 006, #A18 16 and t = 1/, then φt = φ1 t = 1. When these two facts hold, Theore 3 siplifies to C t p = p 1 [ p + p 3F t 3 5p + 3pφ 1 1 + φ 1 + t + t ] + 13 1φ 1. 19 The clais follow fro the fact that see [Ono04], pg. 19 for p, 7 we have { 1 if p 3 od 4, p 3F 8 = 4 p if p 1 od 4, + y = p, and odd, p and that for p, 3 we have { 1 φ if p 3 od 4, p 3F = φ4 8 p if p 1 od 4, + y = p, and odd. p Note that we cannot apply these facts in the single case p = 7 in Corollary 1. However, it is siple to check by hand that the forula still holds in this case. Acknowledgents. This research was supported by the University of Wisconsin-Madison NSF Vigre REU progra. The author would like to thank Ken Ono and Jerey Rouse for their invaluable guidance. The author would also like to thank Karl Mahlburg for his help in proofreading, Sharon Garthwaite for her patience in helping e to use L A TEX, and the referee for nuerous helpful suggestions. References [BEW98] [CGW88] [EPS81] [Erd6] [Ono04] Bruce C. Berndt, Ronald J. Evans, and Kenneth S. Willias, Gauss and Jacobi sus, Canadian Matheatical Society Series of Monographs and Advanced Tets, John Wiley & Sons Inc., New York, 1998, A Wiley-Interscience Publication. MR MR165181 99d:1109 F. R. K. Chung, R. L. Graha, and R. M. Wilson, Quasirando graphs, Proc. Nat. Acad. Sci. U.S.A. 85 1988, no. 4, 969 970. MR MR98566 89a:05116 R. J. Evans, J. R. Pulha, and J. Sheehan, On the nuber of coplete subgraphs contained in certain graphs, J. Cobin. Theory Ser. B 30 1981, no. 3, 364 371. MR MR64553 83c:05075 P. Erdős, On the nuber of coplete subgraphs contained in certain graphs, Magyar Tud. Akad. Mat. Kutató Int. Közl. 7 196, 459 464. MR MR0151956 7 #1937 Ken Ono, The web of odularity: arithetic of the coefficients of odular fors and q-series, CBMS Regional Conference Series in Matheatics, vol. 10, Published for the Conference Board of the Matheatical Sciences, Washington, DC, 004. MR MR00489 005c:11053 [Tho97] Andrew Thoason, Graph products and onochroatic ultiplicities, Cobinatorica 17 1997, no. 1, 15 134. MR MR1466580 99i:05087