How to construct a (2,6)NBIBD(v)

We can use the results for (2,6)GWhD(v)'s, OBIBD(v,3,2;2)'s and PMD(v,6,1)'s. For v = 6n, 6n+1 the (2,6)GWhD paper gives the initial results. Note that solutions for small designs (v ≤ 36) may be found here or (in more detail, for v ≤ 16) here. The presentation given below uses only very easy and well-established methods.

v=6n+1

A (2,6)GWhD(6n+1) design exists for all n ≥ 1, so the same is true for (2,6)NBIBD(6n+1)'s.

v=6n

There are five values of v for which no (2,6)GWhD(v)has been published. These are: 18, 108, 132, 174, 264. For the last two, (2,6)NBIBDs can be derived from PMD(v,6,1)'s which exist. (2,6)NBIBD(v)'s exist for the remaining three values, as follows.

  • 18: a solution is given here.
  • 108=12.9: inflate blocks of (2,6)NBIBD(12) with TD(6,9) and fill holes of resulting (2,6)HNBIBD with (2,6)NBIBD(9)'s.
  • 132=6.19+18: drop 1 point from a TD(7,19) and inflate blocks of resulting PBD with (2,6)NBIBDs.
v=6n+3

The methods used by Greig & Rees can be adapted. The following section is the translation.

A (2,6)NBIBD(v) exists provided v≡ 3 mod 6, and v≡ m mod 36 , and v≥vm where vm is as follows.

m 3 9 15 21 27 33
vm 255 549 87 129 171 213

Apply Dropping points method to get a PBD(6n+r,{6,7,n,r},1) where n≡1 mod 6, and r∈{33,75,9,15,21,27}. Solutions for these last are as follows:

  • A solution for v∈{9,15,21,27} is given here .
  • 75=7.(12-1)+(9-1)+1: (dropping points method) drop 3 points from one group of a TD(7,11), inflate blocks with (2,6)NBIBD(7)'s and (2,6)NBIBD(6)'s, to give a (2,6)HNBIBD(74) of type 11681. Fill holes with (2,6)INBIBD(12,1)'s and a single (2,6)NBIBD(9).
  • 33=(33-6)+6: an IPMD(33,6;6,1) is given in Abel et al; convert this to a (2,6)INBIBD(33,6) and fill the hole.

This leaves unsolved the 29 values of v below.

39 45 51 57 63 69
81 93 99105 111 117
135141 147 153 177183
189 219225261 297333
369 405441 477 513 

Solutions for many of these may be derived from PMD(v,6,1)'s; the 19 exceptions are:

39 45 51 57 63 69
81 93 99105 111 117
135153177 183 219 297
513      

Solutions for most of these follow.

If v=45, then there exists a (2,6)NBIBD(v).

  • 45=(45-6)+6

Fill the hole in a (2,6)INBIBD(45,6), obtainable from an IPMD.

If v ∈ {57}, then there exists a (2,6)NBIBD(v).

  • 57=7.(9-1)+1

Use the Singular Direct Product method.

If v ∈ {63,81,135,513}, then an (2,6)NBIBD(v) exists.

  • 63=7.9
  • 81=9.9
  • 135=15.9
  • 513=19.27

Apply the Direct Product method.

If v ∈ {93,105,219}, then a (2,6)NBIBD(v) exists:

  • 93=6.(16-3)+(15-3)+3
  • 105=6.16+9
  • 219=6.(36-1)+(9-1)+1 (a TD(7,35) exists)

Drop points from a TD(k+1,n)and inflate blocks of the result (considered as a PBD) to give a (2,6)HNBIBD, where k=6. Then fill the holes, using extra points if necessary.

If v ∈ {183}, then a (2,6)NBIBD(v) exists.

  • 183=156+27

Add extra points to a RBIBD with 156 points, then inflate the blocks of the resulting PBD with (2,6)NBIBDs.

A (2,6)NBIBD(111) exists. Inflate the blocks of a BIBD(111,6,1) (considered as a PBD) with (2,6)NBIBDs.

A (2,6)NBIBD(v) exists for: v ∈ {99,117,153,297}:

  • 99=(9-3).16+3
  • 117=(9-3).19+3
  • 153=(9-3).25+3
  • 297=(9-3).49+3

Use the spike method: take k=6 and k+t=9.

A (2,6)HNBIBD(69) of type 9761 exists. This is derived from the HOBIBD given here, and the initial blocks are as follows:
((0,0), (1,15) | (2,6), (0,12) | (1,10), (1,11))
((0,0), (1,18) | (2,3), (0,6) | (1,19), (1,2))
((0,0), (1,9) | (2,8), (0,3) | (1,13), (1,8))
((0,0), (1,19) | (1,2), ∞1 | (0,15), (0,3))
((0,0), (1,13) | (1,8), ∞2 | (0,18), (0,12))
((0,0), (1,10) | (1,11), ∞3 | (0,9), (0,6))
((0,0), (2,13) | (0,12), ∞4 | (2,11), (2,3))
((0,0), (2,10) | (0,6), ∞5 | (2,2), (2,12))
((0,0), (2,19) | (0,3), ∞6 | (2,8), (2,6))
((0,4), (1,5) | (0,2), (1,6) | (0,15), (1,10))
((0,11), (1,10) | (1,2), (2,19) | (2,8), (0,13))
mod (3,21)
The (2,6)NBIBD(69) is then obtained by filling in the hole.

Summarizing the results for v≡3 mod 6, we have:

If v ∉ {39,51} and v≡3 mod 6, v≥6, then a (2,6)NBIBD(v) exists.

v=6n+4

The following is also an adaptation of Greig & Rees.

A (2,6)NBIBD(v) exists provided v ≡ 4 mod 6, and v ≡ m mod 36, and v ≥ vm, where vm is given below.

m 4 10 16 22 28 34
vm 508 298 88 130 172 214

Apply dropping points method to get a PBD(6n+r,{6,7,n,r},1) where n ≡ 1 mod 6, and r ∈ {70,40,10,16,22,28}.Solutions for these last are as follows:

  • a solution for v∈{10,16,22} is given here.
  • a solution for v∈{28,40} can be obtained by adapting PMDs here.
  • 70=10.7: (DP) inflate blocks of (2,6)NBIBD(10) with TD(6,7)'s, to give (2,6)HNBIBD(70) of type 710. Fill the holes with (2,6)NBIBD(7)'s.

This leaves unsolved the 30 values of v below.

34 4652 58 6476
82 94 100 106 112 118
136 142 148 154 178184
190 220 226 256 262292
328 364 400 436 472  

Solutions for many of them may be obtained from PMD(v,6,1)s; the 15 exceptions are:

34 52 58 64 7682
94 100 106 112 118 136
142 148        

Solutions for many of these follow.

If v∈{64,76,94,100,106,136}, then there exists a (2,6)NBIBD(v).
  • 64=7.(10-1)+1
  • 76=15.(6-1)+1
  • 94=7.(16-3)+3
  • 100=9.(12-1)+1
  • 106=7.(16-1)+1
  • 136=9.(16-1)+1

Use the SDP method.

If v ∈{112}, then a (2,6)NBIBD(v) exists.

  • 112=7.16

Use the DP method.

If v∈ {142}, then a (2,6)NBIBD(v) exists.

  • 142=126+16

Add 16 points to a RBIBD with v=126 then inflate blocks of the resulting PBD with (2,6)NBIBD's.

A (2,6)NBIBD(v) exists for:

  • v ∈{58,82,118}
    • 58=(10-4).9+4
    • 82=(10-4).13+4
    • 118=(10-4).19+4
  • v∈ {148}
    • 148=(13-4).16+4

Use the spike method with, respectively, (k,t)=(6,4),(9,4).

If v∉{34,52} and v ≡ 4 mod 6, v≥6, then a (2,6)NBIBD(v) exists.