Similar results also holds true for semi-finite genralized hexagons containing the known generalized hexagons of order 3, 4 as full subgeometries (unpublished, and mostly computational).

Proving something like this for generalized hexagons without the subgeometry assumption seems extremely difficult.

]]>Let X={1,2,…,8} and

f_1 = g_1 = (5,7),

f_2 = g_2 = (6,8),

f_3 = g_3 = (2,6)(4,8)(5,7),

f_4 = g_4 = (1,5)(3,7)(6,8),

f_5 = g_5 = (1,3),

f_6 = g_6 = (2,4),

f_7 = g_7 = (1,3)(2,6)(4,8),

f_8 = g_8 = (1,5)(2,4)(3,7).

Then all the conditions you mentioned are satisfied. However, f_x=f_y if and only if x=y. ]]>

1, 10, 30, 36, 20, 7, 1, 1. ]]>

http://www.secondshoutout.com/product/vintage-french-lottobingo-cards-and-numbers

]]>Is there some standard old source of lotto cards people keep copying?

]]>http://www.jeugdwerker.be/downloads/team-bingo-formulieren.pdf

The middle row finds this quote from a book, which could well be a coincidence, and I cannot find the rest of the book to see the wider context unfortunately.

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