CC BY
11
УДК 519.17
On Distance-Regular Graphs with Л = 2
Alexander A. Makhnev*
Institute of Mathematics and Mechanics, Ural Branch of the Russian Academy of Sciences, Kovalevskaja, 16, Ekaterinburg, 620990,
Russia
Marina S. Nirova
Kabardino-Balkarian State University, Chernyshevskogo, 28, Nalchik, 360000,
Russia
Received 24.12.2013, received in revised form 25.01.2014, accepted 26.02.2014 V.P. Burichenko and A.A. Makhnev have found intersection arrays of distance-regular graphs with Л = 2, ^ > 1, having at most 1000 vertices. Earlier, intersection arrays of antipodal distance-regular graphs of diameter 3 with Л < 2 and ^ = 1 were obtained by the second author. In this paper, the possible intersection arrays of distance-regular graphs with Л = 2 and the number of vertices not greater than 4096 are obtained.
Keywords: distance-regular graph, nearly n-gon.
Introduction
We consider undirected graphs without loops and multiple edges. Given a vertex a in a graph r, we denote by ^(a) the subgraph induced by r on the set of all vertices, that are at a distance i from a. The subgraph [a] = ri(a) is called the neighborhood of the vertex a.
We denote by ka the degree of a vertex a, i. e. the number of vertices in [a]. A graph r is said to be regular with degree k, if ka = k for every vertex a of r. A graph r is called a strongly regular graph with parameters (v,k, A, if r is regular with degree k on v vertices, in which every edge is placed in precisely A triangles, and for any two non-adjacent triangles and any non-adjacent vertices a, b one has |[a] n [b]| = A graph with a diameter d is called antipodal, if the relation on the set of its vertices - to coincide or to be at a distance d - is an equivalence relation. Classes of this relation are called the antipodal classes.
If vertices u, w are at a distance i in r, then we denote by bj(u, w) (by cj(u, w)) the number of vertices in the intersection of ri+1(u) (of ri_1(u)) with [w]. A graph r of diameter d is said to be distance-regular with the intersection array (bo, b1,..., bd-1; c1,..., cd}, if the values of bj(u, w), cj(u, w) do not depend on the choice of vertices u and w separated by a distance i in r, and are equal to bj, cj for i = 0,..., d. Let a» = k — b — cj. Note that a distance-regular graph is amply regular with k = b0, A = k — b1 — 1 and ^ = c2, by definition c1 = 1. Further, we denote by pjj (x, y) the number of vertices in the subgraph ^(x) n r (y) for vertices x,y that are at a distance l in the graph r. In a distance-regular graph, the numbers pj (x, y) are independent of the choice of the vertices x,y; they are denoted by pj and are called the intersection numbers of the graph r.
V. P. Burichenko and A. A. Makhnev found [1] the intersection arrays for distance-regular graphs with A = 2, ^ > 1, such that the number of vertices is not greater than 1000.
* makhnev@imm.uran.ru © Siberian Federal University. All rights reserved
Note here that the arrays {9, 6,3; 1,2,3} of Hemming's graph H(3,4) with v = 64, and {19,16,15, 9; 1,2,3,4} of Hemming's graph H(4,4) with v = 256, and the array {45,42,1; 1,14,45} were omitted from the consideration of [1]. However, there is an additional array {13,10,7; 1, 2, 7} (according to [2], a graph with such an intersection array should not exist). In [3], there were found intersection arrays for antipodal distance-regular graphs of diameter 3 with A < 2 and ^ = 1. In the present paper, the possible intersection arrays of distance-regular graphs with A = 2 and 4096 vertices at most are obtained.
Theorem. Let r be a distance-regular graph with A = 2, ^ =1, having 4096 vertices at most. Then r has one of the following intersection arrays:
(1) {21,18; 1,1} (v = 400);
(2) {6, 3, 3, 3; 1,1,1, 2} (r is a generalized octagon of order (3,1), v = 160), {6, 3, 3; 1,1, 2} (r is a generalized hexagon of order (3,1), v = 52), {12, 9, 9; 1,1, 4} (r is a generalized hexagon of order (3, 3), v = 364), {6, 3, 3, 3, 3, 3; 1,1,1,1,1, 2} (r is a generalized dodecagon of order (3,1), v = 1456);
(3) {18,15, 9; 1,1,10} (v = 1 + 18 + 270 + 243 = 532, Ts is a strongly regular graph); {21,18,12,4; 1,1, 6, 21} (v = 1 + 21 + 378 + 756 + 144 = 1300, 4 = 0).
Corollary. Let r be a distance-regular graph of diameter greater than 2, with A = 2, and having at most 4096 vertices. Then one of the following assertions holds:
(1) r is a primitive graph with the intersection array
{6, 3, 3; 1,1, 2}, {9, 6, 3; 1, 2, 3}, {12, 9, 9; 1,1, 4}, {15,12, 6; 1, 2,10}, {18,15, 9; 1,1,10}, {19,16, 8; 1, 2, 8}, {24, 21, 3; 1, 3,18}, {33, 30,15; 1, 2,15}, {35, 32, 8; 1, 2, 28}, {42, 39,1; 1,1, 42}, {51, 48, 8; 1, 4, 36};
(2) r is an antipodal graph with ^ = 2 and the intersection array
{2r +1, 2r - 2,1; 1, 2, 2r +1}, r e {3,4,..., 44} - {10,16, 28, 34, 38} and v = 2r(r + 1);
(3) r is an antipodal graph with ^ ^ 3 and the intersection array
{15,12,1; 1, 4,15}, {18,15,1; 1, 5,18}, {27, 24,1; 1, 8, 27}, {35, 32,1; 1,4, 35},
{45,42,1; 1, 6, 45}, {42, 39,1; 1, 3, 42}, {63, 60,1; 1, 4, 63}, {75, 72,1; 1,12, 75},
{99, 96,1; 1, 4, 99}, {108,105,1; 1, 5,108}, {143,140,1; 1, 20,143}, {147,144,1; 1,16,147},
{171,168,1; 1,12,171};
(4) r is a primitive graph with the intersection array
{6, 3, 3, 3; 1,1,1, 2}, {19,16,15, 9; 1, 2, 3,4}, {21,18,12,4; 1,1, 6, 21},
{15,12, 9, 6, 3; 1, 2, 3,4, 5}, {6, 3, 3, 3, 3, 3; 1,1,1,1,1, 2}, {18,15,12, 9, 6, 3; 1, 2, 3,4, 5, 6}.
We note that only arrays of some generalized polygons, Hemming's graphs H(n, 4), two graphs with ^ =1, the array {33,30,15; 1,2,15}, and arrays of antipodal graphs of diameter 3 have been added to the list of Burichenko and Makhnev.
Now we prove the Theorem. Let r be a distance-regular graph of diameter d with A = 2, ^ = 1, having 4096 vertices at most. Let a be a vertex in the graph r and k = |Fj(a)|. Then [a] is the union of t + 1 isolated 3-cliques, k = 3(t + 1) and t ^ 20. Otherwise, v > 1 + 66 + 66 • 63, a contradiction.
Lemma 1. The following assertions hold:
(1) if the diameter of r is 2, then r possesses the parameters (400, 21, 2,1);
(2) if r is a generalized 2n-gon, then r has the intersection array from the Corollary.
Proof. If the diameter of r is equal to 2, then, according to [5], r has the parameters (400,21,2,1). Assume that the diameter of r is greater than 2.
Let r be a regular almost n-gon. Then s = 3, and in accordance with [4, Theorem 6.4.1] we have bj = k — 3cj for i = 0,1,..., d — 1, k ^ 3cd, here n = 2d if k = 3cd, and n = 2d +1 if not. If A is a pointwise graph of a generalized polygon of order (s,t), then k = sjtj-1(t + 1)/cj. In the case of n = 6, the number of its vertices is (s + 1)(s212 + st + 1). Therefore v = 4(9t2 + 3t +1) and t ^ 10. If t > 1, then, in view of [4, Theorem 6.5.1], the number st is a square, hence t = 3. If n = 8 and t > 1, then, according to [4, Theorem 6.5.1], the number 2st is a square, and so t > 6 and v > 4096, a contradiction. If n = 12, then t =1 and v = 1 + 6 + 18+----= 1456. □
Lemma 2. Let r be not a generalized 2n-gon. Then the following assertions hold:
(1) if the diameter of r is 3, then r has the intersection array {18,15, 9; 1,1,10};
(2) if the diameter of r is greater than 4, then k < 45.
Proof. Let the diameter of r be equal to 3.
If k = 63, then r has the intersection array {63,60, b2; 1,1, c3}, b2 ^ 4 and c3 divides 33140b2. In any case, there is no valid intersection array. In a similar way one considers the cases 57 < k < 30.
If k = 27, then r has the intersection array {27,24, b2; 1,1,c3}, c3 divides 348b2. Here arise interesting intersection arrays {27, 24, 8; 1,1,16}, v = 1000 with integer eigenvalues 7, 2, —5, but 2 and —5 have fractional multiplicity, and {27,24,4; 1,1, 24}, v = 784 with integer eigenvalues 6, —1, —5, where 6 and —5 have fractional multiplicity. In all cases, there is no admissible intersection array.
If k = 24, then r has intersection array {24, 21, b2; 1,1,c3}, c3 divides 3256b2. Interesting intersection array {24, 21,11; 1,1,18}, v = 837 with integer eigenvalues 6, —3, —7 arise, but 6 and —7 have fractional multiplicity, and there is also {24, 21, 7; 1, 1, 18}, v = 725 with integer eigenvalues 6, —1, —5, but 6 and —5 have fractional multiplicity. In any case, there is no admissible intersection array.
If k = 21, then r has the intersection array {21,18, b2; 1,1, c3}, c3 divides 3314b2. There arises an interesting intersection array {21,18,10; 1,1,12}, v = 715 with integer eigenvalues 6, —1, —5, but —1 and —5 have fractional multiplicity. In any case, there is no admissible intersection array.
If k = 18, then r has the intersection array {18,15, b2; 1,1, c3}, c3 divides 3310b2. There arise interesting intersection arrays {18,15,13; 1,1, 6}, v = 874 with integer eigenvalues 6, —1, —5, having fractional multiplicity, {18, 15, 5; 1, 1, 18}, v = 364 with integer eigenvalues 5, —3, —6, but 5 and —6 have fractional multiplicity, and the array {18,15, 9; 1,1,10} with the spectrum 181, (1 + %/105)/2171, — 1189, (1 — %/105)/2171. There are no other admissible intersection arrays.
If k = 15, then r has the intersection array {15,12, b2; 1,1, c3}, c3 divides 3220b2. There arise interesting intersection arrays {15,12, 8; 1,1,10}, v = 340 with integer eigenvalues 5, —2, —5, but where 5 and —2 have fractional multiplicity, and {15, 12, 6; 1, 1, 10}, v = 304 with integer eigenvalues 5, —1, —4, but 5 and —4 have fractional multiplicity. In any case, there are no admissible intersection arrays.
If k = 12, then r has the intersection array {12,9, b2; 1,1,c3}, c3 divides 334b2. There arise interesting intersection arrays {12, 9, 3; 1, 1, 6}, v = 175 with integer eigenvalues 5, 2, —3, but 5 and —3 are with fractional multiplicity, and {12, 9,1; 1,1,12}, v = 130 with integer eigenvalues 4, —1, —3, but 4 and —3 have fractional multiplicity. In any case, there are no admissible intersection arrays.
If k = 9, then r has the intersection array {9,6, b2; 1,1, c3}, c3 divides 332b2. There arises an interesting intersection array {9,6,4; 1,1, 6}, v = 100 with integer eigenvalues 4, —1, —3, but 4 and —3 have fractional multiplicity. In any case, there are no admissible intersection arrays.
If k = 6, then r has the intersection array (6,3, b2; 1,1, c3}, c3 divides 322b2. An interesting intersection array (6, 3,1;1,1,6}, v = 28 with integer eigenvalues 3, —1, —2 arises here, but 3 and —2 have fractional multiplicity. In any case, there are no admissible intersection arrays.
Assertion (1) is proved.
Let now the diameter of r be greater than 4. Then b > c5-i and k3 > k2. It follows that 4096 > v > 2(1 + k + k(k — 3)), and taking into account the divisibility of k by 3, we see that k ^ 45. The Lemma is proved. □
Let the diameter of r be greater than 3, and r be not a generalized 2n-gon. Considering admissible intersection arrays with A = 2 from [4], we obtain only the array (21,18,12,4; 1,1, 6,21}. The Theorem is thus proved. □
Let us prove the Corollary. If r is not an antipodal graph of diameter 3, then considering admissible intersection arrays with A = 2 from [4], we obtain only the arrays from the Corollary.
Lemma 3. If r is an antipodal graph of diameter 3 with A = y = 2, then r has the intersection array (2r + 1, 2r — 2,1; 1, 2, 2r + 1}, r e (3, 4,..., 44} — (10,16, 28, 34, 38}.
Proof. By the assumption, r has the intersection array (2r + 1,2r — 2,1; 1,2,2r + 1} and v = r(2r + 2) vertices. If r ^ 45, then v ^ 4 • 45 • 23, a contradiction with v ^ 4096. In view of [4, Proposition 1.10.5], if r is even, then k = 2r + 1 is the sum of squares of two integers, therefore r e (3, 4,..., 44} — (10,16, 28, 34, 38}. The Lemma is proved. □
In Lemmata 4-9 it is supposed that r is an antipodal graph of diameter 3 with A = 2 < y. Therefore, r has the spectrum k1, nf, — 1k, —mg, where n, —m are integers, that are the roots of the equation x2 — (A — y)x — k = 0, f = m(r — 1)(k +1)/(m + n), g = n(r — 1)(k + 1)/(m + n) and r = (k + y — 3)/y. If r = 2, then r is Taylor's graph and y = k — 3. In this case, k = 6, n = 2, m = 3, a contradiction with the fact that f = 3 • 7/5. Consequently, r > 2, and the condition q33 > 0 gives m < n2.
Lemma 4. If y ^ 5, then r has one of the following intersection arrays:
(1) (42, 39,1; 1, 3, 42};
(2) (4u2 — 1, 4u2 — 4,1; 1,4,4u2 — 1}, u e (2, 3, 4, 5};
(3) (18,15,1; 1, 5,18} or (108,105,1; 1, 5,108}.
Proof. Let y = 3. Then 4k + 1 = (2n + 1)2, and so, k = n(n +1), m = n +1 and r = k/3. If n = 3s, then f = (3s + 1)(3s2 + s — 1)(9s2 + 3s + 1)/(6s + 1). In this case, (6s + 1, 9s2 + 3s + 1) divides 3 and (6s + 1, 3s2 + s — 1) = (6s + 1, s — 2) divides 13, therefore, s = 2 and r has the intersection array (42, 39,1; 1, 3,42}.
If n = 3s — 1, then f = 3s(3s2 — s — 1)(9s2 — 3s + 1)/(6s — 1). In this case, (6s — 1, 9s2 — 3s + 1) divides 3 and (6s — 1, 3s2 — s — 1) = (6s — 1, s + 2) divides 13, consequently, s = 11, a contradiction with the fact that 5 does not divide 33 • 351 • 1057.
Let y = 4. Then r = (k + 1)/4, k + 1 = 4u2, and so, k = 4u2 — 1, n = 2u — 1 and m = 2u + 1. Further, f = (2u + 1)4u2(u2 — 1)/(4u), g = (2u — 1)u(u2 — 1) and v = 4u4 < 4096, therefore, r has the intersection array (4u2 — 1, 4u2 — 4,1; 1,4,4u2 — 1}, u e (2,..., 5}.
Let y = 5. Then r = (k + 2)/5, 4k + 9 = (2u + 1)2, and hence, k = u2 + u — 2, n = u — 1 and m = u + 2. Further, f = (u + 2)((u2 + u)/5 — 1)(u2 + u — 1)/(2u + 1), (2u + 1,u + 2) divides 3 and (u2 + u — 1, 2u + 1) = (u — 2, 2u + 1) divides 5.
If u = 5s, then (10s + 1, 5s2 + s — 1) = (10s + 1, s — 2) divides 21. In this case, 10s + 1 divides 63, therefore, s = 2 and r has the intersection array (108,105,1; 1, 5,108}.
If u = 5s — 1, then (10s — 1, 5s2 — s — 1) = (10s — 1, s + 2) divides 21. In this case 10s — 1 divides 63, hence s = 1, and r has the intersection array (18,15,1; 1,5,18}. □
Lemma 5. If 6 ^ y ^ 8, then r has one of the following intersection arrays:
(1) {45, 42,1; 1, 6,45};
(2) {27, 24,1; 1, 8, 27}.
Proof. Let ^ = 6. Then r = (k + 3)/6, k + 4 = (2u + 1)2, and so, k = 4u2 + 4u - 3, n = 2u - 1 and m = 2u + 3. Further, f = (2u + 3)(2u2 + 2u - 1)((4u2 + 4u)/6 - 1)/(2u + 1), (2u + 1, 4u2 + 4u - 2) = (2u +1, 2u - 2) divides 3.
If u = 3s, then f = (6s + 3)(18s2 + 12s - 2)(6s2 +2s - 1)/(6s + 1). In this case, (6s +1, 6s2 + 2s - 1) = (6s + 1, s - 1) divides 7, therefore 6s + 1 divides 21, s = 1 and r has the intersection array {45, 42,1; 1, 6, 45}.
If u = 3s - 1, then f = (6s + 1)(18s2 - 6s - 1)(6s2 - 2s - 1)/(6s - 1). In this case (6s - 1, 6s2 - 2s - 1) = (6s - 1, s + 1) divides 7 and 6s - 1 divides 21, a contradiction.
Let ^ = 7. Then r = (k + 4)/7, 4k + 25 = (2u + 1)2, hence k = u2 + u - 6, n = u - 2 and m = u + 3. Further, f = (u + 3)((u2 + u - 2)/7 - 1)(u2 + u - 5)/(2u + 1), (2u +1,u + 3) divides 5 and (2u + 1, u2 + u - 5) = (2u + 1, u - 5) divides 11.
If u = 7s + 1, then (14s + 3, 7s2 + 3s - 1) = (14s + 3, 3s - 2) divides 37. In this case, 14s + 3 divides 5 • 11 • 37, a contradiction.
If u = 7s + 5, then (14s + 11, 7s2 + 11s + 3) = (14s + 11,11s + 6) divides 37. In this case 14s + 11 divides 5 • 11 • 37, a contradiction.
Let ^ = 8. Then r = (k + 5)/8, k + 9 = 4u2, therefore k = 4u2 -9, n = 2u -3 and m = 2u + 3. Further, f = (2u + 3)((u2 - 1)/2 - 1)(u2 - 2)/u, (u, 2u + 3) divides 3 and (u2 - 2,u) divides 2. Consequently, u = 2s + 1, (2s + 1,2s2 + 2s - 1) divides 3 and 2s + 1 divides 9, and so, s = 1 and r has the intersection array {27, 24,1; 1, 8,27}. □
Lemma 6. If 9 ^ ^ ^ 11, then there is no admissible intersection array.
Proof. Let ^ = 9. Then r = (k + 6)/9, 4k + 49 = (2u + 1)2, therefore k = u2 + u - 12, n = u - 3 and m = u + 4. Further, f = (u + 4)(u2 + u - 11)((u2 + u - 6)/9 - 1)/(2u + 1), (2u + 1, u + 4) divides 7, and (2u + 1, u2 + u - 11) = (2u + 1, u - 22) divides 45.
If u = 9s + 2, then (18s + 5, 9s2 + 5s - 1) = (18s + 5, 5s - 2) divides 61. In this case, 18s + 5 divides 35 • 61, a contradiction.
If u = 9s - 3, then (18s - 5, 9s2 - 5s - 1) = (18s - 5, 5s + 2) divides 61. In this case, 18s - 5 divides 35 • 61, a contradiction.
Let ^ = 10. Then r = (k + 7)/10, k + 16 = (2u +1)2, therefore k = 4u2 +4u - 15, n = 2u - 3 and m = 2u + 5. Further, f = (2u + 5)(2u2 + 2u-7)((2u2 + 2u-4)/5- 1)/(2u + 1), (2u +1, 2u + 5) divides 4, and (2u + 1,2u2 + 2u - 7) divides 15.
If u = 5s + 1, then (10s + 3,10s2 + 6s - 1) = (10s + 3, 3s - 1) divides 19. In this case, 10s + 3 divides 57, a contradiction.
If u = 5s + 3, then (10s + 7,10s2 + 14s + 3) = (10s + 7, 7s + 3) divides 19. In this case, 10s + 7 divides 57, s = 5, u = 28, a contradiction with v ^ 4096.
Let ^ = 11. Then r = (k + 8)/11, 4k + 81 = (2u + 1)2, therefore k = u2 + u - 20, n = u - 4 and m = u + 5. Further, f = (u + 5)((u2 + u - 12)/11 - 1)(u2 + u - 19)/(2u + 1), (2u + 1,u + 5) divides 9 and (u2 + u - 19, 2u + 1) = (u - 38, 2u + 1) divides 77.
If u = 11s + 3, then (22s + 7,11s2 + 7s - 1) = (22s + 7, 7s - 2) divides 93, and 22s + 7 divides 27 • 7 • 31, a contradiction with the fact that v < 4096.
If u = 11s - 4, then (22s - 7, 11s2 - 7s - 1) = (22s - 7, 7s + 2) divides 93, and so, 22s - 7 divides 27 • 7 • 31, that contradicts with v ^ 4096. □
Lemma 7. If 12 ^ ^ ^ 14, then r has either the intersection array {75, 72,1; 1,12, 75} or the intersection array {171,168,1; 1,12,171}.
Proof. Let ^ = 12. Then r = (k + 9)/12, k + 25 = 4u2, therefore k = 4u2 - 25, n = 2u - 5 and m = 2u + 5. Further, f = (2u + 5)((u2 - 4)/3 - 1)(u2 - 6)/u, (2u + 5, u) divides 5, and (u2 - 6, u) divides 6.
If u = 3s + 1, then (3s + 1, 3s2 + 2s - 2) = (3s + 1, s - 2) divides 7 and 3s + 1 divides 70, hence s = 2 and r has the intersection array {171,168,1; 1,12,171}.
If u = 3s — 1, then (3s — 1, 3s2 — 2s — 2) = (3s — 1, s + 2) divides 7 and 3s — 1 divides 70, and so, s = 2 and r has the intersection array {75, 72,1; 1,12,75}.
Let y = 13. Then r = (k + 10)/13, 4k +121 = (2u +1)2, therefore k = u2 + u — 30, n = u — 5 and m = u + 6. Further, f = (u + 6)((u2 + u — 20)/13 — 1)(u2 + u — 29)/(2u + 1), (2u +1,u + 6) divides 11, and (2u + 1, u2 + u — 29) = (2u + 1, u — 58) divides 117.
If u = 13s + 4, then (26s + 9,13s2 + 9s — 1) = (26s + 9, 9s — 2) divides 133 and 26s + 9 divides 99 • 133, a contradiction with that v < 4096.
If u = 13s — 5, then (26s — 9,13s2 — 9s — 1) = (26s — 9, s + 2) divides 61, and 26s — 9 divides 99 • 61, a contradiction with v ^ 4096.
Let y = 14. Then r = (k + 11)/14, k + 36 = (2u + 1)2, therefore k = 4u2 + 4u — 35, n = 2u — 5 and m = 2u + 7. Further, f = (2u + 7)((2u2 + 2u — 12)/7 — 1)(2u2 + 2u — 17)/(2u +1), (2u + 1, 2u + 7) divides 6 and (2u2 + 2u — 17, 2u + 1) = (u — 17, 2u + 1) divides 35.
If u = 7s + 2, then (14s + 5,14s2 + 10s — 1) = (14s + 5, 5s — 1) divides 39 and 14s + 5 divides 15 • 39, a contradiction with the condition v < 4096.
If u = 7s — 3, then (14s — 5,14s2 — 4s — 1) = (14s — 5, s — 1) divides 9, and so, 14s — 5 divides 135, s = 1, n = 3, m = 15, a contradiction with m < n2. □
Lemma 8. If 15 ^ y ^ 17, then r has the intersection array {147,144,1; 1,16,147}.
Proof. Let y = 15. Then r = (k + 12)/15, 4k + 169 = (2u + 1)2, therefore k = u2 + u — 42, n = u — 6 and m = u + 7. Further, f = (u + 7)((u2 + u — 30)/15 — 1)(u2 + u — 41)/(2u + 1), (2u + 1, u + 7) divides 13 and (u2 + u — 41, 2u + 1) = (u — 82, 2u + 1) divides 165.
If u = 15s, then (30s + 1,15s2 + s — 3) = (30s + 1, s — 6) divides 181 and 30s + 1 divides 11 • 13 • 181, a contradiction with the condition v ^ 4096.
If u = 15s — 1, then (30s — 1,15s2 — s — 3) = (30s — 1, s + 6) divides 181 and 30s — 1 divides 11 • 13 • 181, a contradiction with v ^ 4096.
If u = 15s + 5, then (30s + 11,15s2 + 11s — 1) = (30s + 11,11s — 2) divides 181 and 30s + 11 divides 11 • 13 • 181, a contradiction with the condition v ^ 4096.
If u = 15s — 6, then (30s — 11,15s2 — 11s — 1) = (30s — 11,11s + 2) divides 181 and 30s — 11 divides 11 • 13 • 181, a contradiction with v ^ 4096.
Let y = 16. Then r = (k + 13)/16, k + 49 = 4u2, therefore k = 4u2 — 49, n = 2u — 7 and m = 2u + 7. Further, f = (2u + 7)((u2 — 9)/4 — 1)(u2 — 12)/u, (2u + 7,u) divides 7 and (u, u2 — 12) divides 12. Consequently, u = 2s + 1, (2s + 1, s2 + s — 3) = (2s + 1, s — 6) divides 13 and 2s + 1 divides 21 • 13, hence, s = 3 and r has the intersection array {147,144,1; 1,16,147}.
Let y = 17. Then r = (k + 14)/17, 4k + 225 = (2u + 1)2, therefore k = u2 + u — 56, n = u — 7 and m = u + 8. Further, f = (u + 8)((u2 + u — 42)/7 — 1)(u2 + u — 55)/(2u + 1), (2u + 1, u + 8) divides 15 and (u2 + u — 55, 2u + 1) = (u — 110, 2u + 1) divides 221. Hence, u = 7s — 1, (14s — 1, 7s2 — s — 7) = (14s — 1, s + 14) divides 197 and 14s — 1 divides 15 • 221 • 197, a contradiction with the condition v ^ 4096. □
Lemma 9. If 18 ^ y ^ 20, then r has the intersection array {143,140,1; 1, 20,143}.
Proof. Let y = 18. Then r = (k + 15)/18, k + 256 = (2u + 1)2, hence k = 4u2 + 4u — 255, n = 2u — 6 and m = 2u + 8. Further, f = (u + 4)((2u2 + 2u —120)/9 — 1)(4u2 +4u — 255)/(2u +1), (2u + 1, u + 4) divides 7 and (4u2 + 4u — 255, 2u + 1) = (2u — 255, 2u + 1) divides 256.
If u = 9s + 2, then (18s + 5,18s2 + 10s — 13) = (18s + 5, 5s — 13) divides 259 and 18s + 5 divides 7 • 259, a contradiction with v ^ 4096.
If u = 9s — 3, then (18s — 5,18s2 — 10s — 13) = (18s — 5, 5s + 13) divides 259 and 18s — 5 divides 72 37, a contradiction with v ^ 4096.
Let ^ = 19. Then r = (k +16)/19, 4k + 289 = (2u + 1)2, therefore k = u2 + u — 72, n = u — 8 and m = u + 9. Further, f = (u + 9)((u2 + u — 56)/19 — 1)(u2 + u — 71)/(2u + 1), (2u + 1,u + 9) divides 17 and (2u + 1, u2 + u — 71) divides 285.
If u = 19s + 7, then (38s + 15,19s2 + 15s — 1) = (38s + 15,15s — 2) divides 301 and 38s + 15 divides 15 • 17 • 301, a contradiction with v ^ 4096.
If u = 19s — 8, then (38s — 15,19s2 — 15s — 1) = (38s — 15,15s + 2) divides 301 and 38s — 15 divides 15 • 17 • 301, a contradiction with v ^ 4096.
Let ^ = 20. Then r = (k + 17)/20, k + 81 = 4u2, hence k = 4u2 + 4u — 81, n = 2u — 9 and m = 2u + 9. Further, f = (2u + 9)((u2 + u — 16)/5 — 1)(u2 + u — 20)/u, (2u + 9,u) divides 9 and (u2 +u—20, u) divides 20. Consequently, u = 5s+2, (5s+2, 5s2+5s—3) = (5s+2, 3s—3) divides 21 and 5s + 2 divides 21 • 36, therefore s = 1 and r has the intersection array {143,140,1; 1, 20,143}. The Lemma is proven. □
Computer calculations show that there is no admissible intersection array in the case ^ > 21. The Theorem, and also the corresponding Corollary with it, are thus proven.
This work is partially supported by the Russian Foundation for Basic Research (project No. 12-01-00012), by the joint RFBR-GFEN (China) grant # 12-01-91155, by Programs of the Branch of Mathematics of the Russian Academy of Sciences (project No. 12-T-1-1003), by the Joint Research Programs of the Ural Branch and the Siberian Branch of the Russian Academy of Sciences (project No. 12-C-1-1018), and by a joint grant with the National Academy of Sciences of Belarus (project No. 12-C-1-1009).
References
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[2] K.Coolsaeet, Local structure of graphs with A = ^ = 2, a2 =4, Combinatorica, 15(1971), no. 4, 481-487.
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О дистанционно—регулярных графах с Л = 2
Александр А. Махнев Марина С. Нирова
В.П. Буриченко и А.А. Махнев нашли массивы пересечений дистанционно регулярных графов с Л = 2, ß > 1 и числом вершин не большим 1000. Ранее вторым автором найдены массивы пересечений антиподальных дистанционно-регулярных графов диаметра 3 с Л < 2 и ß = 1. В данной статье найдены возможные массивы пересечений дистанционно-регулярных графов с Л = 2 и не более 4096 вершинами.
Ключевые слова: дистанционно-регулярный граф, почти n-угольник.
