OFFSET
1,2
COMMENTS
a(4*n) = 0 and a(4*n-1) = 0 because these graphs have an even number of nodes.
For n >= 2, a(n) is a multiple of 8, because swapping the two non-corner vertices of any corner triangle preserves the magic sums but still yields a distinct configuration.
The triangular magic sum is (9*n*(n+1)+6)/4, and the hexagonal magic sum is (9*n*(n+1)+6)/2.
It seems that the sequence is likely finite. Considering each of the N(n)! possible labelings of the 3*n*(n+1)/2 nodes, and noting that for each of the C(n) = (3*n^2 - 3*n + 2)/2 triangular or hexagonal faces only one choice for the remaining vertex can give the required magic sums, we divide N(n)! by N(n)^(C(n)) to obtain an estimate of a(n). For n >= 9, N(n)! / N(n)^(C(n)) < 1.
LINKS
Donghwi Park, Source code for a(2) and overview.
Donghwi Park, Source code for a(5).
EXAMPLE
For n = 2, numbers 1..9 are placed in the node of triangles shown:
o
/ \
o-- o
/ \
o o
/ \ / \
o---o---o---o
Example solution for n = 5:
29
38 2
37 16
27 5 40 13
20 24 35
31 18 44 1 25 9
45 7 43 33
10 14 23 39 4 22 6 30
12 32 26 41 8
21 36 34 3 15 28 17 11 42 19
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Donghwi Park, Dec 04 2025
STATUS
approved