Prime Sum Graphs
Suryaprakash Rao
Powai, Mumbai-400076
Chetan S Rao
Penn, USA
Neha S Rao
CA, USA
(15 May 2011)
Abstract
The geome try 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 the oretical 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 labeli ng 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 K 1 , 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 f or 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 nu mber of edges in Pn is then given by: |E|=∑σ(i,n) over all
i≤n.
Prime Sum Graphs
Problems
Study the properties of the σ-function. What is the functional form of σ-function?
Fig.2 shows σ-Function for small values.
σ(n), Cum
Fig.2 σ-Function for small values
140
120
100
80
60
40
20
0
σ(n)
Cum
0
5
10
15
20
25
n
2
30
Prime Sum Graphs
Table-1 shows σ-function, pairs summing to a prime for n=2,…,32. The prime sum is shown in bold. Cu mulative
of σ-function values stand for the nu mber 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 R epresentation of Integers (CR I) 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 p ositive integers. The nodes on Cn
are equally spaced and the angle between two conse cutive integers is 360/n. Here, in the figures the circle radius is
taken arbitrary for convenience. We shall develop terminology required for CRI.
3
Prime Sum Graphs
We assume throughou t 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 j oining the nodes is a diameter. Diametric prime p d is a prime which is a sum of two integer s which are
diametric pair in Cn. That is a prime p is diametric if there exists n su ch 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, f or 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 corresp onds to the pairs of integers which are diametric or cl ose 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 conse cutive pair. Conversely, as the concentric circle
becomes smaller in diameter the pair diverges from a consecu tive 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 f ormed
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 anticlock wise directions.
Denoting these paths by j-k and k -j paths clockwise and letting l 1 and l 2 be their lengths, it then f ollows that
n=l 1 +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;7,1 6;
5
22
1,12;3,14;4,15;6,17;9,20;10,21
6
26
2,15;3,16;5,18;6,1 9;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;
42
1,22;4,25;5,26;8,29;10,31;11,32;13,34;16,37;19,40;20,41;
8
10
Theorem 1. Apparent concentric circles are formed in Pn iff n is even.
4
Prime Sum Graphs
Theorem 2. A set of chords form tangents to apparent concentric circle in Pn iff n is even and the set of chord s 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 foll ows 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
149
151
157
163
167
173
179
n/2+2
79
83
89
97
101
103
107
109
113
127
131
137
139
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
8
1,2;3,8
FP:3,11
1,6
IP:7
12
1,6;7,1 2
FP:7,19
1,12
CP:13
16
1,2;3,1 6
FP:3,19
1,10
IP:11
20
1,2;3,2 0
FP:3,23
1,12
24
1,4;5,2 4
FP:5,29
1,16;17,24
28
1,2;3,2 8
FP:3,31
32
1,12
IP:13
1,4
1,10;11,12
FP:11,23
1,4;5,1 2
FP:5,17
IP:13
4,15
IP:19
9,20
IP:29
FP:17,41
3,16;19,24
FP;19,43
1,6;9,2 2;
FP:7,31
1,16
IP:17
1,22;
IP:23
9,28
IP:37
1,28;29,32
FP:29,61
15,3 2
IP:47
1,30
IP:31
5
n
Pn
Orthogonal Pairs
Prime Sum Graphs
36
40
44
48
1,18
1,2
1,30
1,4
1,36
1,22
9,44
1,28
OP
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 f or 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 su ch 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’
6
Prime Sum Graphs
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 toge ther wi th j’=j+1 yields the points j,j+1,j+n/2+,j+n/2 and they form a
quadrangle.
Maximal chord s (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 ,y i )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 nu mber is utmost half of n, that is,
•
|Pp (x,y)|≤n/2.
Consider the pairs of integers which form a se t of parallel chords. Denote this set of pairs by P C(x,y). Let D(p 1 ,p 2,…,
p ), ≥1, be the distinct primes representing this set of parallel chords. Clearly, the maximum number of pairs
summing to p 1 ,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 f or 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,
7
Prime Sum Graphs
|{U Pp (x,y):pεD}|=n/2.
A prime whose chords form an incomplete matching in Pn is called an Incomplet e Pri me IP. It is possible that a se t
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 se t 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
p
n+p
n
p
n+p
4
3
7
16
13
29
5
2
7
17
2
19
6
5
11
18
5
23
8
3
11
18
11
29
8
5
13
18
13
31
9
2
11
20
3
23
10
3
13
20
11
31
10
7
17
20
17
37
11
2
13
21
2
23
12
5
17
22
7
29
12
7
19
22
19
41
12
11
23
24
5
29
14
3
17
24
7
31
14
5
19
24
13
37
15
2
17
24
17
41
16
3
19
24
19
43
16
7
23
24
23
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 su ch that n+p is prime.
The Lemma ab ove may be generalized as follows:
Lemma 10. Suppose that there are primes n≥p 1 >p 2>…>p>1, >0 such that n+p 1 , p 1 -1+p2 ,…,p-1 -1+p  are also
primes, then p 1 ,p 2,…,p  are FP in Pn.
Table-6 Complete Primes and Friendly Primes
FP
Pairs
n
CP
FP
24 Nil
7,17
3,4;3
8,9;2
5,29
1,4;2
5,17
2,3;2
8,9;4
7,31
1,6;3
7,19
3,4;3
9,10;3
13,3 7
1,12;6
11,1 3
5,6;5
11,1 2;1
17,4 1
1,16;8
14 Nil
3,17
1,2
3,14
19,4 3
1,18;9
5,19
1,4
5,14
23,4 7
1,22;11
16 17
3,17
9,10;7
1,2;1
26 Nil
3,29
1,2;1
7,23
3,4;3
11,1 2;5
5,31
1,4;2
13,2 9
6,7;6
14,1 5;2
11,3 7
5,6;5
18 19
5,23
5,18;7
1,4;2
17,4 3
1,16;8
11,2 9
1,10;5
11,1 8;4
28 29
3,31
1,2;1
13,3 1
13,1 8;3
1,12;6
13,4 1
6,7;6
22 23
7,29
1,6;3
14,1 5;8
19,4 7
9,10;9
19,4 1
1,18;9
19,2 2;2
23,4 7,5 3 11,1 2;1 1
3,23,43
1,2;1
3,20;9
21,2 2;1
* FP x,y;z means a pair (x,y) with x+y a prime and z number of parallel edges in Pn.
n
10
12
CP
11
13
Pairs
5,24;10
7,24;9
13,2 4;6
17,2 4;4
19,2 4;3
23,2 4;1
3,26;12
5,26;11
11,2 6;8
17,2 6;5
3,28;8
13,2 8;8
19,2 8;5
23,2 4;1
25,2 8;2
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.
8
Prime Sum Graphs
That is, knowing one smaller prime leads to a larger prime. Does this give an algorithm to find large r 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. Su ch 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 f or 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 p ower label?
Full degree vertex exists for n<4.
PS G raph is Bipartite
Theorem 12. Pn is bipartite.
The set of even labels form a partition and so is the se t 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 corresp onding 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).
9
Prime Sum Graphs
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 nu mbers so that consecutive
numbers sum to primes.
Theorem 15. Pn is a conne cted graph.
PSG Trees
Are all trees PSG? The answer is no. For example, K 1,4 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 home omorphic 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 whe ther n+1 is
prime or not and f(n) is the maximum nu mber of friendly primes.
10
Prime Sum Graphs
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
4
6
8
10
12
14
16
18
1,2,3,4 ,1;
1,4,3,2 ,5,6,1 ;
1,2,3,8 ,5,6,7 ,4,1;
1,2,9,4 ,7,6,5 ,8,3,1 0,1 ;
2
2
2
3
2
3
3
4
11
3,13;7,17;
4
3
4
4
5
5
6
7
13
5,17;7,19;11,23;
3,17;5,19
3,19;7,23;13,29
5,23;11,29;13,31
4
7
22
24
1,10,3,8,5 ,6,7,4 ,9,2,1 1,1 2;1
1,2,3,1 4,5 ,12,6,10,9,8 ,11,6,13,4,1 ;
1,2,15,4,1 3,6 ,11,8,9,1 0,7 ,12,5,14,3,1 6,1 ;
5,6,7,4 ,9,2,1 1,1 8,1 3,1 6,1 5,1 4,1 7,1 2,1 ,10,3,8,5 ;
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 ,9,2,1 1,2 0,1 7,1 4,1 5,1 6,1 3,1 8, 19,12,1,10,3,8 ,5,6;
1,2,3,4 ,7,6,5 ,8,9,2 0,1 1,1 8,1 3,1 6,1 5,1 4,1 7,1 2,1 9,1 0;
8,9,10,7,1 2,5 ,14,3,16,1,1 8,1 9,2 2,2 1,2 0,1 7,2 ,15,4,13,6,1 ,8;
2,3,4,1 ,6,23,8,2 1,1 0,1 9,1 2,1 7,1 4,1 5,1 6,1 3,1 8,1 1,2 0,9 ,22,7,24,5,2 ;
5
6
8
8
26
1,2,3,2 6,5 ,24,7,22,9,2 0,1 1,1 8,1 3,1 6,1 5,1 4,1 7,1 2,1 9,1 0,2 1,8 ,23,6,25,4,1 ;
6
8
n
Hamiltonian Path
Min
Max
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 ,11,8,9,1 0,7 ,6,5; 1,4,3,2 ,5,6,7 ,10,9,8,1 1;
3
4
13
1,4,3,2 ,5,6,7 ,12,11,8,9,1 0,1 3;
3
15
1,4,3,2 ,15,14,5,6,7 ,12,11,8,9,1 0;
4
17
17,2 ,15,4,13,6,1 1,8 ,9,10,7,1 2,5 ,14,3,16,1;
4
7
19
19,1 8,1 ,16,3,14,5,1 2,7 ,10,9,8,1 1,6 ,13,4.15.2.1 7;
4
7
20
17
19
23
FP
IP
7,11,13,23
11,3 1
17
7,29;19,41
5,29;17,41;19,43;23,47
11,1 3,1 7,3 1,3 7;
3,29;5,31;11,37;17,43
13,4 1;
CP
FP
IP
5
13
19,7
11;1 7
6
17
7,23;13,29;5,19;7,2 3
11
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) su ch 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
11
Prime Sum Graphs
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
50
51
52
101
103
53
54
55
107
109
4
8
9
10
17
19
9
5
14
15
16
29
31
10
Hamiltonian Pairs of Triples
Table-10 Hamiltonian Triples <500
n=4t
4
4
6
8
10
12
12
14
16
18
24
26
28
30
36
38
40
42
54
56
58
60
66
68
70
72
96
98
100
102
Triple-1
3
4
5
1
10
2
11
1
16
17
2
1
28
29
2
1
40
41
2
1
58
59
2
1
70
71
2
1
100
2
4t-1,4t,1
4
1
1
2
6
1
2
3
1
2
3
4
12
1
2
3
1
2
18
1
3
4
2
3
1
2
30
1
3
4
2
3
1
2
42
1
3
4
2
3
1
2
60
1
3
4
2
3
1
2
72
1
3
4
2
3
1
2
3
4
1
2
2
5
5
8
5
8
8
8
14
14
14
14
20
20
20
20
29
29
29
29
35
35
35
35
50
50
50
53
Triple-2
2
3
3
6
6
9
6
9
9
9
15
15
15
15
21
21
21
21
30
30
30
30
36
36
36
36
51
51
51
54
3
4
4
7
7
10
7
10
10
10
16
16
16
16
22
22
22
22
31
31
31
31
37
37
37
37
52
52
52
55
n=4t
102
104
106
108
132
134
136
138
144
146
148
150
174
176
178
180
186
188
190
192
192
194
196
198
222
224
226
228
234
236
Triple-1 4t-1,4t,1
101
102
1
1
2
3
106
1
2
107
108
1
2
3
4
1
2
3
136
1
2
137
138
7
2
3
4
1
2
3
148
1
2
149
150
10
2
3
4
1
2
3
178
1
2
179
180
17
2
3
4
1
2
3
190
1
2
2
3
4
191
192
20
1
2
3
196
1
2
197
198
22
2
3
4
1
2
3
226
1
2
227
228
29
2
3
4
1
2
3
50
53
53
53
68
68
68
68
74
74
74
74
89
89
89
89
95
95
95
98
95
98
98
98
113
113
113
113
119
119
Triple-2
51
54
54
54
69
69
69
69
75
75
75
75
90
90
90
90
96
96
96
99
96
99
99
99
114
114
114
114
120
120
52
55
55
55
70
70
70
70
76
76
76
76
91
91
91
91
97
97
97
100
97
100
100
100
115
115
115
115
121
121
n=4t
238
240
264
266
268
270
276
278
280
282
306
308
310
312
342
344
346
348
414
416
418
420
426
428
430
432
456
458
460
462
Triple-1 4t-1,4t,1
238
1
2
239
240
32
2
3
4
1
2
3
268
1
2
269
270
40
2
3
4
1
2
3
280
1
2
281
282
43
2
3
4
1
2
3
310
1
2
311
312
50
2
3
4
1
2
3
346
1
2
347
348
59
2
3
4
1
2
3
418
1
2
419
420
77
2
3
4
1
2
3
430
1
2
431
432
80
2
3
4
1
2
3
460
1
2
461
462
88
119
119
134
134
134
134
140
140
140
140
155
155
155
155
173
173
173
173
209
209
209
209
215
215
215
215
230
230
230
230
Triple-2
120
120
135
135
135
135
141
141
141
141
156
156
156
156
174
174
174
174
210
210
210
210
216
216
216
216
231
231
231
231
121
121
136
136
136
136
142
142
142
142
157
157
157
157
175
175
175
175
211
211
211
211
217
217
217
217
232
232
232
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.
12
Prime Sum Graphs
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
11
3
12
7
13
7
13
9
15
7
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
19
9
22
3
22
7
22
9
26
9
27
7
28
3
30
7
33
7
34
9
35
3
37
9
38
3
39
7
40
9
43
9
46
3
46
7
48
7
503
509
523
2p+
3
401
449
457
461
541
557
569
617
677
701
709
761
769
797
821
881
929
937
977
100
9
102
1
104
9
p
3
7
13
19
31
37
61
67
73
79
97
103
109
139
2p+5
11
19
31
43
67
79
127
139
151
163
199
211
223
283
151
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 f or 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 f or n even.
If this conjecture is true then Pn is bipancyclic follows f or n even from heredity property.
Pn Problems
•
What is the maximum spanning:
o Regular subgraph?
o Eulerian subgraph?
o Planar subgraph?
•
•
•
•
•
•
Constru ct 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?
13
Prime Sum Graphs
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 su ch a chord is:
m = (sj -sk )/(cj -ck ).
A chord representing a diametric prime:
In this case the origin or the p oint (0,0) is on the chord. That is:
((-sk )/(-ck ))-(sj -sk )/(cj -ck )=0
or
(sk -sj )/(ck -cj ) = sk /ck
or
cjsk=ck sj.
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.
14
Prime Sum Graphs
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 ,
sj 2 – sj sk = - cj 2 – cj ck
sjsk-cjc k = 1.
Two chords with a common point on the circle:
Suppose j=j’ then
(sk -sj )/(ck -cj ) = - (ck ’-cj )/(sk ’-sj ).
Simplifies to:
sks k’+ckc k’-(s ksj+c kcj)-(sk’sj+ck’c k)=1.
If the intersection p oint is origin that is the primes are diametric then:???
sks k’+ckc k’-(s ksj+c kcj)-(sk’sj+ck’c k)=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), j2,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 b oth j and k with the node labels reduced (mod n) where necessary. Alternately, the chord (j,k) is rotated in
anticlock wise or clockwise direction one step at a time resulting in equal chord s. Note that all these chords may or
may not be edges in Pn.
15
Prime Sum Graphs
Table-11 Chords of equal Length and Tangents to Smallest concentric circle in Pn
n
1
4
2
3
4
5
6
7
2,4
3
4
8
4,6
5
8
1,7
8
12
6
9
8,10
9
12
1,11
12
16
10
9,11
10
13
12,14
13
16
20
10,12
11
14
13,15
14
17
16
24
12
15
14,16
15
18
17
28
16
15,17
16
19
18
32
16,18
17
20
19
8
9
10
11
12
13
14
15
16
No Chords
4
6
23
8
15
11
20
19
21
21
20,22
22
21,23
22
22,24
23
27
20
12
24
12
26
25
28
27
28
26
29
28,30
29
32
15
31
17
Chords are read as column number with a nu mber 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
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
n+n/2+1
16