Prime Sum Graphs

Suryaprakash Rao

Powai, Mumbai-400076 Chetan S Rao

Penn, USA Neha S Rao

CA, USA

(15 May 2011)

Abstract

The geometry of n-th roots of unity points on a unit circle together with labeling the points by j where 1≤j≤n is the j-th root of unity results in a labeled graph. Based on this prime sum labeling and prime sum graphs are defined and some graph theoretical properties of (maximum) prime sum graphs are obtained in reference to paths and cycles including partial results on the existence of Hamiltonian cycles are obtained. This circular representation of first n natural numbers by labeling the points corresponding to the n, n -th roots of unity on a unit circle lead us to

classify primes into diametric, orthogonal, complete, friendly (parallel) and incomplete primes. Chords of equal length form tangents to apparent concentric circles in the circular representation. Chords of maximum length are considered. Primes are manifested in different forms including twin primes. Prime sum graphs with reference to

planarity, regularity and Eulerian properties are studied. Study of prime power sum labeling and prime power sum graphs will be reported elsewhere.

Introduction

Prime sum labeling of a graph of order n is a labeling of the nodes by positive integers V=[n]={1,2,…,n} so that

sum of the labels of an adjacent node pair is a prime. A graph admitting a prime sum labeling is a prime sum (PS) graph. Some graph theoretical properties of (maximum) prime sum graphs are obtained here. It is hoped that this study may lead to further probe into prime numbers, their classification, distribution and methods for easy

recognition of primality. On the similar lines prime power sum labeling and prime power sum graphs may be defined. Study of prime power sum labeling and prime power sum graphs will be reported elsewhere. Note that 2 is the only even prime and cannot appear as an edge label. The graph G with the edge set

E(G)={(i,j):i+j is a prime}, that is, two nodes labeled i,j are adjacent whenever i+j is a prime is the Maximum PS (MPS) Graph and is denoted by Pn. MPS Graphs for smaller orders are K1, K2, K1,2, C4 and for orders n=5,7,8,9,10 are shown in Fig.1. Subgraphs of Pn are of interest.

Fig.1 Maximum Prime Sum Graphs for orders n=5,6,7,8,9,10.

Pn satisfies hereditary property. Meaning thereby any spanning subgraph of Pn is also a PS Graph. Note that, Pn is a subgraph of Pn+1.

Define σ-function as σ(i,n)={x≤n:i+x is a prime}. The number of edges in Pn is then given by: |E|=∑σ(i,n) over all i≤n.

Prime Sum Graphs

2

Problems

Study the properties of the σ-function. What is the functional form of σ-function?

Fig.2 shows σ-Function for small values.

0

20

40

60

80

100

120

140

0 5 10 15 20 25 30

σ(n

), C

um

n

Fig.2 σ-Function for small values

σ(n)

Cum

Prime Sum Graphs

3

Table-1 shows σ-function, pairs summing to a prime for n=2,…,32. The prime sum is shown in bold. Cumulative

of σ-function values stand for the number of edges in Pn. Table-2 gives the degree sequences of the PS Graphs Pn for n=2,…,32. It also shows the size m, the number of edges in the graph and the minimum and maximum degrees in Pn. Primality of x+y

• When is x+y a prime? o A necessary condition for x+y to be a prime is that x and y are coprime or (x,y)=1. o This condition is not sufficient as 1 is coprime with all other integers. But not all sums wi th 1 are primes. o For x+y to be a prime other than the prime 2, x and y are not equal and are of different parity. That is, if

one is even then the other is odd. o For any pair (x,y) with x+y a prime, there is a pair (x’,y’) such that x+y=x’+y’, x’<y’ and y’=x’+1. o Any two odd primes sum to even number and so the sum is not a prime. In general this is true for any

two odd or two even numbers.

Circular Representation of Integers

Circular Representation of Integers (CRI) of [n]={1,2,…,n} denoted by Cn for a given n>0 refers to the labeling of the nodes corresponding to the n-th roots of unity on a unit circle by the first n positive integers. The nodes on Cn

are equally spaced and the angle between two consecutive integers is 360/n. Here, in the figures the circle radius is taken arbitrary for convenience. We shall develop terminology required for CRI.

Prime Sum Graphs

4

We assume throughout that the labeling of Cn is in the natural order cyclically . That is, 1,2,…,n,1. The line joining a pair of adjacent nodes in Cn is a chord. Diametric pair is a pair of positive integers summing to a prime and the

chord joining the nodes is a diameter. Diametric prime pd is a prime which is a sum of two integers which are diametric pair in Cn. That is a prime p is diametric if there exists n such that Cn has diametric pairs of integers summing to p. Further, this means that there are edges in Cn passing through the centre of the circle. For a given

n it is possible that there are no diametric chords. So diametric pairs may or may not exist, for example, P14 and P18

have diametric pairs whereas P17 and P19 have no diametric pairs. Proceeding further, integer pairs which are not diametric may be classified as follows:

Concentric Circles. A set of chords of Cn form tangents to an apparent concentric circle within Cn around its centre.

The innermost concentric circle corresponds to the pairs of integers which are diametric or close to diametric points. Diametric pairs form a concentric circle of radius zero. As the concentric circle grows larger in diameter the pairs come closer on the circumference or converge to a consecutive pair. Conversely, as the concentric circle becomes smaller in diameter the pair diverges from a consecutive pair to a diametric or near diametric pair. How many concentric circles are formed? Suppose n=4t or 4t+2, t>0. Then there are t concentric circles formed with positive diameter.

Fig.3 Pn Graphs of order 12, 16, 17, 18, 19.

A chord (j,k) in Pn divides the points along the circle into two j-k paths in clockwise and anticlockwise directions.

Denoting these paths by j-k and k -j paths clockwise and letting l1 and l2 be their lengths, it then f ollows that

n=l1+l2-2. Chords of same length form tangents to an apparent circle.

Problems.

What are the radii of concentric circles?

Characterize the primes which are tangents to one of these concentric circles of a fixed radius. CRI Cn for a fixed n, defines a classification of [n] into diametric pairs, tangent pairs, according to concentric

circles

Table-2. Diametric Pairs in Pn, n≡2(mod 4)

n Diametric Pairs No.Pairs

2 1,2 1

6 1,4;2,5; 2

10 1,6;3,8;4,9 3

14 2,9;3,10;5,12;6,13 4

18 1,10;2,11;4,13;5,14;7,16; 5

22 1,12;3,14;4,15;6,17;9,20;10,21 6

26 2,15;3,16;5,18;6,19;7,20;8,21;9,22; 7

30 1,16;2,17;4,19;7,22;8,23;11,26;13,28;14,29 8

34 1,18;3,20;6,23;7,24;10,27;12,29;13,30;15,32; 8

38 2,21;5,24;6,26;9,26;11,30;12,31;14,33; 8

42 1,22;4,25;5,26;8,29;10,31;11,32;13,34;16,37;19,40;20,41; 10

Theorem 1. Apparent concentric circles are formed in Pn iff n is even.

Prime Sum Graphs

5

Theorem 2. A set of chords form tangents to apparent concentric circle in Pn iff n is even and the set of chords are

of equal length. Problem. Is it true that for every n, n≡2(mod 4), Pn has a diametric pair with 1, 2 or 3 as an endpoint of an edge?

Note that, this fails for n=178. Characterise.

Theorem 3. Diametric pairs occur in Pn if and only if n≡2(mod 4).

Fig.4 Longest chords forming smallest concentric circle and diametric pairs in Pn, n=8,10,12,14,16,18,20.

Horizontal Pairs. If there is a horizontal chord then there exists a chord with one of the end nodes being 1 and is diametric as it passes through the origin. It follows that n is even. The other end node is n/2+1. Further, it is an

edge in Pn if n/2+2 is a prime. It follows that n≡2(mod 4) in Pn. The other parallel chords may be found by adding or subtracting 1 from the end node labels.

Table-3 a. Values of n with n/2+2 a pri me

n 2 6 10 18 22 30 34 42 54 58 70 78 82 90 102 114 118 130 138 142

n/2+2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73

n 154 162 174 190 198 202 210 214 222 250 258 270 274 294 298 310 322 330 342 354

n/2+2 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179

n 358 378 382 390 394 418 442 450 454 462 474 478 498

n/2+2 181 191 193 197 199 211 223 227 229 233 239 241 251

Orthogonal Pairs are adjacent pairs of integers in Pn with chords orthogonal to each other. A prime p is orthogonal to a prime p’ if a p-chord is orthogonal to a p’-chord. Since each prime has pairs representing it, the two classes of parallel chords corresponding to the pairs of p and p’ form a system of orthogonal chords.

Fig.5 Orthogonal pairs in Pn.

Table-3b. Orthogonal Pairs Pn

n OP-1 OP-2

4 1,2 1,4

8 1,2;3,8 FP:3,11 1,6 IP:7

12 1,6;7,12 FP:7,19 1,12 CP:13 1,10;11,12 FP:11,23 1,4;5,12 FP:5,17

16 1,2;3,16 FP:3,19 1,10 IP:11

20 1,2;3,20 FP:3,23 1,12 IP:13 4,15 IP:19 9,20 IP:29

24 1,4;5,24 FP:5,29 1,16;17,24 FP:17,41 3,16;19,24 FP;19,43 1,6;9,22; FP:7,31

28 1,2;3,28 FP:3,31 1,16 IP:17 1,22; IP:23 9,28 IP:37

32 1,12 IP:13 1,28;29,32 FP:29,61 15,32 IP:47 1,30 IP:31

Prime Sum Graphs

6

Ort

hog

on

al P

airs

Pn

n 36 40 44 48

OP

1,18 1,2 1,30 1,4

1,36 1,22 9,44 1,28

Chords (j,k) and (j’,k’) are orthogonal then: (sk-sj)(sk’-sj’) + (ck’-cj’)(ck-cj) = 0. Theorem 4. If an OP exists in Pn then n≡0(mod 4).

Since diametric pair exist only for Pn, n≡2(mod 4), it follows that:

Corollary 4.1. Pn, n≡2(mod 4) satisfies that: An orthogonal pair of edges in Pn are not diametrical and conversely,

no two diametrical edges in Pn are orthogonal. Conjecture. If n≡0(mod 4) then an OP exists in Pn.

Theorem5. For n≡0(mod 4) if there is an x such that x+1and x+1+n/2 are prime then an OP exists in Pn.

The result above may be extended to:

Theorem 6 . For n≡0(mod 4) if there are x>y>0 such that y+x and y+x+n/2 are prime then an OP exists in Pn.

The Table-4 gives x satisfying the condition in the Theorem.

Table-4. Values of n such that (1,x) and (1,n/2+x) are orthogonal

n 4 4 4 8 8 8 12 12 12 12 16 16 16 20 20 20

x 2 4 10 2 6 12 4 6 10 12 2 4 10 2 6 12

n/2+x+1 5 7 13 7 11 17 11 13 17 19 11 13 19 13 17 23

n 24 24 24 28 28 32 32 32 36 36 36 40 40 44 48 48

x 4 6 10 2 4 2 6 12 4 10 12 2 10 6 4 6

n/2+x+1 17 19 23 17 19 19 23 29 23 29 31 23 31 29 29 31

n 48 52 52 52 56 56 60 60 60 64 64 68 68 68 72 72

x 12 2 4 10 2 12 6 10 12 4 10 2 6 12 4 6

n/2+x+1 37 29 31 37 31 41 37 41 43 37 43 37 41 47 41 43

n 72 76 76 80 80 80 84 84 88 92 92 96 96 96 100 100

x 10 2 4 2 6 12 4 10 2 6 12 4 10 12 2 10

n/2+x+1 47 41 43 43 47 53 47 53 47 53 59 53 59 61 53 61

Problems.

Find orthogonal pairs in Pn.

What is the criterion for two pairs are orthogonal?

Parallel chords with the first component in the range 1,2,…, n/4 are of interest. A pair orthogonal to such a pair shall give a prime larger than n/2. This procedure yields a prime number between n/2 and n.

o Alternate proof of Bertrand’s Hypothesis that there is a prime between n and 2n?

Parallel chords with the first component in the range n/4 +1, n/4+2,…, n/2 also are of interest. A pair orthogonal to such a pair shall give a prime larger than 3n/4.

Twin Primes.

Quadrangles. Consider two chords with a common node. Note that if the chords are edges in Pn then the primes

are twins. That is, if (j,k) and (j,k’) are a pair of chords with a common node j and k’=k+2 such that j+k and j+k’

Prime Sum Graphs

7

are primes and 2k+1,k+1+k’ are primes. This gives a twin of twin pairs with j,k,k+1,k+2,j forming the quadrangle

Fig.6(a).

Twin primes. Quadrangles may be formed in other ways. For example, maximal chords which are diametric pair

form twin primes. Nodes j,k,j’,k’ such that j+k = j’+k’ is prime and j+j’, k+k’ are primes. Fig.6(b). Since (j,k) is a

diametric pair, k=j+n/2 holds. This together wi th j’=j+1 yields the points j,j+1,j+n/2+,j+n/2 and they form a

quadrangle.

Maximal chords (diametric pair) form twin primes, Fig.6(c).

Hexagons. (j,j+1,j+2), (k,k+1,k+2) be the pair of triples such that 2j+1, 2j+3; 2k+1,2k+3 are primes. Further, j+k

and j+k+4 are also primes. The pairs (j,k) and (j+2,k+2) are diametric pair. The points j,j+1,j+2,k+2,k+1k,j forms

a hexagon. Since (j,k) is a diametric pair k=j+n/2 holds. That is, 2j+1, 2j+3; 2j+n+1 and 2j+n+3 are primes. See

Fig.6(d).

(a) (b) (c) (d)

Fig.6 Twin primes occurrences.

Types of primes

The CRI for a fixed n facilitates a means to classifying primes. Consider, the pairs of integers (xi,yi)p, i=1,2,… summing to a prime p. The pair (x,y)p satisfies the equation x+y=p. This equation has solutions in [n]. The solutions by definition fall in [2n]. Each of these solutions forms an edge in Cn. In particular, denote the set of pairs of integers (x,y) summing to a fixed prime p by Pp(x,y). Note that the pairs in Pp(x,y) form parallel edges in Cn. That is,

they form a matching in Pn. For a given n, this number is utmost half of n, that is,

• |Pp(x,y)|≤n/2.

Consider the pairs of integers which form a set of parallel chords. Denote this set of pairs by PC(x,y). Let D(p1,p2,…,

p), ≥1, be the distinct primes representing this set of parallel chords. Clearly, the maximum number of pairs

summing to p1,p2,…, p and parallel is n/2. In other words, the number of pairs summing to a prime or primes and forming parallel chords is at most n/2. That is,

• |{U Pp(x,y), for pεD}|≤n/2. Based on this inequality the primes in Pn arising as a sum of two natural numbers may be classified as follows:

A prime Pn which attains this bound is called a complet e prime CP. That is, =1 and the pairs Pp(x,y) form a perfect matching in Pn. Note that in this case n is even and |Pp(x,y)|=n/2.

Lemma 7. CP exists in Pn iff n+1 is prime.

Theorem 8. Complete prime in Pn, n even if exists is unique.

A prime which is not complete in Pn is called an incomplete prime IP. A CP in Pn is an IP in Pn+1. Let p be the largest prime <n. Further, assume that for m≥n, m+n is not a prime. Then p is a CP in Pn-1 and IP in Px, x>n.

For a given n, there are primes which are not complete but complement this property. That is, ≥2 and the primes

satisfying this are called friendly primes. That is,

Prime Sum Graphs

8

|{U Pp(x,y):pεD}|=n/2.

A prime whose chords form an incomplete matching in Pn is called an Incomplet e Prime IP. It is possible that a set

of primes may form incomplete matching in Pn and form a set of incomplete primes.

Examples:

P12 has a CP13 and FP:7,19 forming an orthogonal set of chords.

In P14 the following are examples of friendly and incomplete primes:

Friendly primes:3,17. Pairs: 1,2; 9,8; 10,7; 11,6; 12,5; 13,4; 14,3.

Friendly primes:5,19. Pairs: 1,4; 2,3; 9,10; 8,11; 7,12; 6,13; 6,14;5.

Incomplete prime: 23. Pairs: 11,12; 10,13; 9,14;

Incomplete prime: 13. Pairs: 1,12; 2,11; ..., 7,6;

Incomplete prime: 11. Pairs: 1,10; 2,9; ..., 5,6;

The primes 3, 23 are friendly in P20.

P24 has FP:17, 21 which are orthogonal to FP:5, 29. Lemma 9. If n satisfies that n+p is a prime for a prime p<n then p and n+p are FP.

Table-5 Values of n and a prime p<n such that n+p is prime.

n 4 5 6 8 8 9 10 10 11 12 12 12 14 14 15 16 16

p 3 2 5 3 5 2 3 7 2 5 7 11 3 5 2 3 7 n+p 7 7 11 11 13 11 13 17 13 17 19 23 17 19 17 19 23

n 16 17 18 18 18 20 20 20 21 22 22 24 24 24 24 24 24

p 13 2 5 11 13 3 11 17 2 7 19 5 7 13 17 19 23 n+p 29 19 23 29 31 23 31 37 23 29 41 29 31 37 41 43 47

Observe that n+p may not be a prime for any p<n. The Table-5 shows that n=7,13,19,23 have no such p.

Problem. Characterize n with no primes p<n such that n+p is prime.

The Lemma above may be generalized as follows:

Lemma 10. Suppose that there are primes n≥p1>p2>…>p>1, >0 such that n+p1, p1-1+p2,…,p-1-1+p are also

primes, then p1,p2,…,p are FP in Pn.

Table-6 Complete Primes and Friendly Primes

n CP FP Pairs n CP FP Pairs

10 11 7,17 3,4;3 8,9;2 24 Nil 5,29 1,4;2 5,24;10 12 13 5,17 2,3;2 8,9;4 7,31 1,6;3 7,24;9 7,19 3,4;3 9,10;3 13,37 1,12;6 13,24;6

11,13 5,6;5 11,12;1 17,41 1,16;8 17,24;4

14 Nil 3,17 1,2 3,14 19,43 1,18;9 19,24;3

5,19 1,4 5,14 23,47 1,22;11 23,24;1

16 17 3,17 9,10;7 1,2;1 26 Nil 3,29 1,2;1 3,26;12

7,23 3,4;3 11,12;5 5,31 1,4;2 5,26;11

13,29 6,7;6 14,15;2 11,37 5,6;5 11,26;8

18 19 5,23 5,18;7 1,4;2 17,43 1,16;8 17,26;5

11,29 1,10;5 11,18;4 28 29 3,31 1,2;1 3,28;8

13,31 13,18;3 1,12;6 13,41 6,7;6 13,28;8

22 23 7,29 1,6;3 14,15;8 19,47 9,10;9 19,28;5

19,41 1,18;9 19,22;2 23,47,53 11,12;11 23,24;1 25,28;2

3,23,43 1,2;1 3,20;9 21,22;1

* FP x,y;z means a pair (x,y) with x+y a prime and z number of parallel edges in Pn.

There are triples (x,y,z) such that y is adjacent to x,z. The pairs xy, yz form tangents to the innermost concentric circles. There may be other x’, y’ with this property wrt larger concentric circles. Such triples are of interest. They represent primes. Note that, x,z may be twin primes.

Prime Sum Graphs

9

That is, knowing one smaller prime leads to a larger prime. Does this give an algorithm to find larger primes? Orthogonal prime is one example. These pairs may be diametric or may not be. What happens with diametric orthogonal pairs?

• Algorithm. Look for a pair summing to a diametric prime one of the two being in the range 1 to n/4. Find the

pair near orthogonal to this diameter. Check whether this pair is diametric?

• Conjecture. Such diametric pairs yield diametric primes.

Degree Sequence

Consider n/2 for n even. If n/2+n/2-1=n-1 is prime then (n/2,n/2-1), (n/2+1,n/2-2),...,(n-2,1) are pairs summing to a prime and so edges in Pn. For a prime x we look for solutions for the equation i+j=x, 1≤i,j≤n. In fact, solutions are: 1,x-1; 2,x-2; 3,x-3; … threshold by 2n-1. What can we say on minimum degree in Pn? Problems.

What is the minimum and maximum degree in Pn? Which vertices attain this?

• 1 or 2 has maximum degree. This is equal to the number of primes less than n. Other nodes may have maximum degree.

• Minimum degree vertex is maximum 2 power label?

• Full degree vertex exists for n<4. PS Graph is Bipartite

Theorem 12. Pn is bipartite.

The set of even labels form a partition and so is the set of odd labels. In other words, the edges are in between partitions only. This establishes the bipartiteness of Pn.

Fig.7 MPS graphs with pure bipartition for smaller even orders Pn, n=4 to 16.

Bipartite PS Graphs

Maximum Bipartite PSG is maximal PSG corresponding to a given bipartition of V. A PS labeling induces label bipartition with reference to (E,O) bipartition, called here as a pure bipartition. Theorem 13. Size of MBPS graph is maximum for the label bipartition (E,O).

Prime Sum Graphs

10

Eulerian Bipartite Pn

What is the maximum Eulerian bipartite subgraph of Pn?

Table-7 Maximum Eulerian Subgraph

n Edges Deleted

4 C4: Regular, Hamiltonian and Eulerian

6 H-cycle: (1,2)

8 H-cycle: (5,6), (1,2), (3,4).

10 H-cycle, (1,4), (2,3), (5,6)

12 (6,7), (5,8)

14 4-regular: (1,2)

16 4-regular: (1,2), (3,4), (5,6)

18 (5,6),(9,10),(7,16),(8,15)

Theorem 14. Pn, for n even, has same degree sequence in each of its bipartition A and B.

Is Pn connected? In other words, given two natural numbers i,j does there exist an i-j path in Pn? We shall prove this affirmatively. That is, for any two natural numbers there is a sequence of natural numbers so that consecutive numbers sum to primes. Theorem 15. Pn is a connected graph. PSG Trees

Are all trees PSG? The answer is no. For example, K1,4 or 5-star is not a PSG as the maximum degree in P5 is 3. P6 is a 6-cycle with an edge joining vertices at distance 3. Clearly this graph has no spanning tree with a unique vertex of degree 3. Problem. Characterize PSG trees.

Planar PSG

• What is the maximum planar subgraph of Pn? o What is the maximal planar PSG?

• Pn is planar iff n<9.

• Planar embedding of P8 is shown in Fig.9

• Nonpalnarity of P9 may be inferred from Fig.9. Node 9 is adjacent to 2,4,8.

• The graph induced on {2,4,8} and {3,5,9} is homeomorphic to K3,3.

• Nonpalnarity of Pn, n>9 follows from the heriditary property.

Fig.9 A planar embedding of P8.

Problem. Characterize planar PSGraphs.

Regular MPS Graphs Observation.

A CP adds degree one to a regular graph of order n. Each set of FP adds degree one to a regular graph of order n. Theorem 16. Maximum r-regular MPS graph exists in Pn so that r=c(n)+f(n), where c(n) is 1 or 0 whether n+1 is

prime or not and f(n) is the maximum number of friendly primes.

Prime Sum Graphs

11

Cycles and Hamiltonian Cycles in MPSGraph

Is Pn Hamiltonian? If so, is it biPancyclic? Table-8 Hamiltonian for small n.

n Hamiltonian Cycle/Path Min Max CP FP IP

4 1,2,3,4,1; 2 2

6 1,4,3,2,5,6,1; 2 3

8 1,2,3,8,5,6,7,4,1; 2 3

10 1,2,9,4,7,6,5,8,3,10,1; 3 4 11 3,13;7,17;

12 1,10,3,8,5,6,7,4,9,2,11,12;1 4 5 13 5,17;7,19;11,23;

14 1,2,3,14,5,12,6,10,9,8,11,6,13,4,1; 3 5 3,17;5,19 7,11,13,23

16 1,2,15,4,13,6,11,8,9,10,7,12,5,14,3,16,1; 4 6 17 3,19;7,23;13,29 11,31

18 5,6,7,4,9,2,11,18,13,16,15,14,17,12,1,10,3,8,5; 4 7 19 5,23;11,29;13,31 17

20 1,10,19,12,17,14,15,16,13,8,11,20,9,4,7,6,5,8,3,2,1; 5,6,7,4,9,2,11,20,17,14,15,16,13,18, 19,12,1,10,3,8,5,6; 1,2,3,4,7,6,5,8,9,20,11,18,13,16,15,14,17,12,19,10;

4 7

22 8,9,10,7,12,5,14,3,16,1,18,19,22,21,20,17,2,15,4,13,6,1,8; 5 8 23 7,29;19,41 11,13,17,31,37;

24 2,3,4,1,6,23,8,21,10,19,12,17,14,15,16,13,18,11,20,9,22,7,24,5,2; 6 8 5,29;17,41;19,43;23,47

26 1,2,3,26,5,24,7,22,9,20,11,18,13,16,15,14,17,12,19,10,21,8,23,6,25,4,1; 6 8 3,29;5,31;11,37;17,43 13,41;

n Hamiltonian Path Min Max CP FP IP

5 1,4,3,2,5; 1 3

7 1,2,3,4,7,6,5; 1,4,3,2,5,6,7; 2 3

9 1,2,3,4,7,6,5,8,9; 2 4

11 1,4,3,2,11,8,9,10,7,6,5; 1,4,3,2,5,6,7,10,9,8,11; 3 4

13 1,4,3,2,5,6,7,12,11,8,9,10,13; 3 5 13 19,7 11;17

15 1,4,3,2,15,14,5,6,7,12,11,8,9,10; 4 6 17 7,23;13,29;5,19;7,23 11

17 17,2,15,4,13,6,11,8,9,10,7,12,5,14,3,16,1; 4 7

19 19,18,1,16,3,14,5,12,7,10,9,8,11,6,13,4.15.2.17; 4 7

A pair of triples (x1,y1,z1) and (x2,y2,z2) is said to be Friendly if they are 2-paths on the circumference of Cn in Pn and the primes are friendly. Triples (2,1,n) or (1,n,n-1) such that n+1 and 2n-1 are prime are possible for n as shown in the Table-8.

Table-8 Values of n<550 with n+1 and 2n-1 prime

n n+1 2n-1 n n+1 2n-1 n n+1 2n-1 n n+1 2n-1 n n+1 2n-1

4 5 7 42 43 83 126 127 251 232 233 463 372 373 743

6 7 11 52 53 103 136 137 271 240 241 479 420 421 839

10 11 19 66 67 131 156 157 311 250 251 499 430 431 859

12 13 23 70 71 139 166 167 331 262 263 523 432 433 863

16 17 31 82 83 163 180 181 359 282 283 563 442 443 883

22 23 43 96 97 191 190 191 379 310 311 619 456 457 911

30 31 59 100 101 199 192 193 383 316 317 631 460 461 919

36 37 71 106 107 211 210 211 419 330 331 659 486 487 971

40 41 79 112 113 223 222 223 443 346 347 691 520 521 1039

Prime Sum Graphs

12

Table-9 shows consecutive integer triples (x,y,z), that is, y=x+1 and z=y+1satisfying the condition x+y and y+z are prime.

Table-9 Consecutive Integer Triples (x,y,z) with x+y and y+z Prime

Sno x y z x+y y+z Sno x y z x+y y+z

1 1 2 3 3 5 6 20 21 22 41 43

2 2 3 4 5 7 7 29 30 31 59 61

3 5 6 7 11 13 8 35 36 37 71 73

4 8 9 10 17 19 9 50 51 52 101 103

5 14 15 16 29 31 10 53 54 55 107 109

Hamiltonian Pairs of Triples

Table-10 Hamiltonian Triples <500 n=4t Triple-1 4t-1,4t,1 Triple-2 n=4t Triple-1 4t-1,4t,1 Triple-2 n=4t Triple-1 4t-1,4t,1 Triple-2

4 3 4 1 1 2 3 102 101 102 1 50 51 52 238 238 1 2 119 120 121 4 4 1 2 2 3 4 104 1 2 3 53 54 55 240 239 240 32 119 120 121

6 5 6 1 2 3 4 106 106 1 2 53 54 55 264 2 3 4 134 135 136

8 1 2 3 5 6 7 108 107 108 1 53 54 55 266 1 2 3 134 135 136

10 10 1 2 5 6 7 132 2 3 4 68 69 70 268 268 1 2 134 135 136 12 2 3 4 8 9 10 134 1 2 3 68 69 70 270 269 270 40 134 135 136

12 11 12 1 5 6 7 136 136 1 2 68 69 70 276 2 3 4 140 141 142

14 1 2 3 8 9 10 138 137 138 7 68 69 70 278 1 2 3 140 141 142 16 16 1 2 8 9 10 144 2 3 4 74 75 76 280 280 1 2 140 141 142 18 17 18 1 8 9 10 146 1 2 3 74 75 76 282 281 282 43 140 141 142

24 2 3 4 14 15 16 148 148 1 2 74 75 76 306 2 3 4 155 156 157

26 1 2 3 14 15 16 150 149 150 10 74 75 76 308 1 2 3 155 156 157 28 28 1 2 14 15 16 174 2 3 4 89 90 91 310 310 1 2 155 156 157

30 29 30 1 14 15 16 176 1 2 3 89 90 91 312 311 312 50 155 156 157

36 2 3 4 20 21 22 178 178 1 2 89 90 91 342 2 3 4 173 174 175

38 1 2 3 20 21 22 180 179 180 17 89 90 91 344 1 2 3 173 174 175 40 40 1 2 20 21 22 186 2 3 4 95 96 97 346 346 1 2 173 174 175

42 41 42 1 20 21 22 188 1 2 3 95 96 97 348 347 348 59 173 174 175

54 2 3 4 29 30 31 190 190 1 2 95 96 97 414 2 3 4 209 210 211

56 1 2 3 29 30 31 192 2 3 4 98 99 100 416 1 2 3 209 210 211 58 58 1 2 29 30 31 192 191 192 20 95 96 97 418 418 1 2 209 210 211

60 59 60 1 29 30 31 194 1 2 3 98 99 100 420 419 420 77 209 210 211

66 2 3 4 35 36 37 196 196 1 2 98 99 100 426 2 3 4 215 216 217 68 1 2 3 35 36 37 198 197 198 22 98 99 100 428 1 2 3 215 216 217 70 70 1 2 35 36 37 222 2 3 4 113 114 115 430 430 1 2 215 216 217

72 71 72 1 35 36 37 224 1 2 3 113 114 115 432 431 432 80 215 216 217

96 2 3 4 50 51 52 226 226 1 2 113 114 115 456 2 3 4 230 231 232 98 1 2 3 50 51 52 228 227 228 29 113 114 115 458 1 2 3 230 231 232

100 100 1 2 50 51 52 234 2 3 4 119 120 121 460 460 1 2 230 231 232

102 2 3 4 53 54 55 236 1 2 3 119 120 121 462 461 462 88 230 231 232

When is Pn Hamiltonian? Since Pn is bipartite, a necessary condition is that n is even. Theorem 17. If there exists a pair of triples (x,x+1,x+2) and (y,y+1,y+2) in CRI such that:

The (x+2 to y)- and (y+2 to x)-paths along Cn have equal lengths, and

2x+1 and 2y+3; 2x+3 and 2y+1 are friendly primes, then Pn is Hamiltonian. Note that, x+y and y+z may be twin primes. Corollary 17.1. Pn is Hamiltonian if 2x+1 and 2y+3; 2x+3 and 2y+1 are complete primes.

Corollary 17.2. If n+1 is prime and the pair 1,2 is friendly, that is the prime 3 is friendly with some other primes

in Pn then Pn is Hamiltonian. It follows that n+1 is a CP in Pn. This together with the perfect matching containing the prime 3 forms a HC.

When Pn has a Hamiltonian path? The answer is: Theorem18. If Pn is Hamiltonian for n even then Pn+1 has a Hamiltonian path.

Prime Sum Graphs

13

Theorem 19. If n=2p, p a prime and 2p+3 or 2p+5 are primes then Pn is Hamiltonian. Table-11 Primes p≤550 such that 2p+3 or 2p+5 are also primes.

p 2 5 7 13 17 19 29 43 47 53 67 73 89 97 113

127

137

139

157

167 173 193

2p+3 7 13 17 29 37 41 61 89 97 109 137 149 181 197 229 257 277 281 317 337 349 389

p 199

223

227

229

269

277

283

307

337

349

353

379

383

397

409

439

463

467

487

503 509 523

2p+3 401 449 457 461 541 557 569 617 677 701 709 761 769 797 821 881 929 937 977

1009

1021

1049

p 3 7 13 19 31 37 61 67 73 79 97 103 109 139 151

2p+5 11 19 31 43 67 79 127 139 151 163 199 211 223 283 307

p 163 181 229 241 271 283 307 313 367 373 409 439 457 523 541

2p+5 331 367 463 487 547 571 619 631 739 751 823 883 919 1051 1087

Note that 2p+3 and 2p+5 are twin primes for a prime p. The Table-10 shows primes satisfying this condition:

Table-12 Primes p≤550 such that 2p+3 and 2p+5 are primes.

p 7 13 19 67 73 97 139 229 283 307 409 439 523

2p+3 17 29 41 137 149 197 281 461 569 617 821 881 1049

2p+5 19 31 43 139 151 199 283 463 571 619 823 883 1051

Conjecture. Pn is Hamiltonian for n even.

If this conjecture is true then Pn is bipancyclic follows for n even from heredity property. Pn Problems

• What is the maximum spanning: o Regular subgraph? o Eulerian subgraph? o Planar subgraph?

• Construct and characterize regular subgraphs of Pn.

• Describe maximum or maximal subgraphs of Pn with a property.

• What is the size of a clique- maximum complete subgraph, of Pn?

• Describe maximum Eulerian subgraph of Pn.

• What is the chromatic number of a Pn? Χ(G)≥4.

• What is the independence number of Pn?

Prime Sum Graphs

14

Equations of Parallel, Perpendicular, Intersecting Chords:

Denote by: cj=cos(2πj/n) and sj=sin(2πj/n).

Equation of a chord through (cj,sj) and (ck,sk) points with labels j and k is:

(s-sk)/(c-ck)-(sj-sk)/(cj-ck)=0

where (s,c) is a point on the chord.

Slope of such a chord is:

m = (sj-sk)/(cj-ck).

A chord representing a diametric prime:

In this case the origin or the point (0,0) is on the chord. That is:

((-sk)/(-ck))-(sj-sk)/(cj-ck)=0

or (sk-sj)/(ck-cj) = sk/ck

or cjsk=cksj.

Angle between two chords C1 and C2:

Let (cj,sj), (ck,sk) and (cj’,sj’), (ck’,sk’) be the end points of the chords.

Slopes of the chords:

m=(sk-sj)/(ck-cj) and m’=(sk’-sj’)/(ck’-cj’),

Angle between the two chords:

tan = (m-m’)/(1+mm’)

= ((sk-sj)/(ck-cj)) - ((sk’-sj’)/(ck’-cj’)) / (1 + ((sk-sj)/(ck-cj))((sk’-sj’)/(ck’-cj’)))

System of chords which are parallel.

If =0 then m=m’.

m=(sk-sj)/(ck-cj) = m’=(sk’-sj’)/(ck’-cj’),

or (sk-sj)/(ck-cj) = (sk’-sj’)/(ck’-cj’),

or (sk-sj)/(sk’-sj’) = (ck-cj)/(ck’-cj’),

or (sk-sj)/(ck-cj) - (sk’-sj’)/(ck’-cj’) = 0.

or (sk-sj)/(sk’-sj’) - (ck-cj)/(ck’-cj’) = 0.

Prime Sum Graphs

15

A chord perpendicular (orthogonal) to a given chord.

If =90 then 1+mm’=0 or mm’=-1 or m=-1/m’,

((sk-sj)/(ck-cj))((sk’-sj’)/(ck’-cj’)))=-1 or

(sk-sj)/(ck-cj) = - (ck’-cj’)/(sk’-sj’)

Or (sk-sj)(sk’-sj’) + (ck’-cj’)(ck-cj) = 0.

Further, if one of the chords passes through (0,0) then further simplification may be done.

(sj-sk)/(cj-ck) = - cj/sj,

sj2 – sjsk = - cj2 – cjck

sjsk-cjck = 1.

Two chords with a common point on the circle:

Suppose j=j’ then

(sk-sj)/(ck-cj) = - (ck’-cj)/(sk’-sj).

Simplifies to:

sksk’+ckck’-(sksj+ckcj)-(sk’sj+ck’ck)=1.

If the intersection point is origin that is the primes are diametric then:???

sksk’+ckck’-(sksj+ckcj)-(sk’sj+ck’ck)=1.

Equations of a set of chords which are tangents to a concentric circle:

Chords of equal length form tangents to apparent concentric circle with radius equal to the distance of a chord

from the centre. Such concentric circle is possible only for n even. Two cases arise according as diametric pair

exists or not which means that n≡2 or 0 (mod 4) respectively.

Length of a (j,k)-chord is: dj,k= √((sj-sk)2+(cj-ck)2).

If (j,k) chord is an edge in Pn with j<k then j+k is a prime and the pairs (j+1,k-1),(j+2,k-2),… and (j-1,k+1), j-

2,k+2),… also sum to j+k and so are edges in Pn.

A chord divides the circumference into two circle segments either side with lengths (l1,l2). Chords of same length

divide the circle into same length (l1,l2) upto order and may be obtained from a chord (j,k) by adding or subtracting

1 to both j and k with the node labels reduced (mod n) where necessary. Alternately, the chord (j,k) is rotated in

anticlockwise or clockwise direction one step at a time resulting in equal chords. Note that all these chords may or

may not be edges in Pn.

Prime Sum Graphs

16

Table-11 Chords of equal Length and Tangents to Smallest concentric circle in Pn

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 No Chords

4 2,4 3 4 4

8 4,6 5 8 1,7 8 6

12 6 9 8,10 9 12 1,11 12 8

16 10 9,11 10 13 12,14 13 16 15 11

20 10,12 11 14 13,15 14 17 16 20 19 20 12

24 12 15 14,16 15 18 17 21 20,22 21 24 12

28 16 15,17 16 19 18 22 21,23 22 26 25 28 27 28

15

32 16,18 17 20 19 23 22,24 23 27 26 29 28,30 29 32 31 17

Chords are read as column number with a number in a row representing n. For example, for n=12, the chords 1,6;

2,9; 3,8; 3,10; … are equal.

Table-12 Values of n such that n+n/2+1 is prime or (n,n/2+1) is an edge in Pn.

n 4 8 12 20 24 28 40 44 48 52 64 68 72 84

n+n/2+1 7 13 19 31 37 43 61 67 73 79 97 103 109 127

n 92 100 104 108 120 128 132 140 148 152 160 180 184 188

n+n/2+1 139 151 157 163 181 193 199 211 223 229 241 271 277 283