On the infinite hexagonal grid we start at stage 0 with no toothpicks, so a(0) = 0.
At stage 1 we place 5 toothpicks on the edges of the first hexagonal cell, so a(1) = 5.
At stage 2, from the last exposed endpoint, we place 4 other toothpicks on the edges of the second hexagonal cell, so a(2) = 5 + 4 = 9 because there are 9 toothpicks in the structure.
At stage 3, from the last exposed endpoint, we place 3 other toothpicks on the edges of the third hexagonal cell, so a(3) = 9 + 3 = 12 because there are 12 toothpicks in the spiral.
Illustration of initial terms:
. _ _ _ _
. _ _/ \ _/ \_ _/ \_ _/ \_
. _ _ / _ / _ / _ \ / _ \ / _ \
. / \ / \ \ / \ \ / \ \ / \ / \ / \ / \ / \ /
. _/ / _/ / _/ / _/ / _/ / _/ \ / _/ \
. \_/ \_/ \_/ \_/ \_/ _/ \_/ _/
. _/
.
. 5 9 12 15 18 21 23
.
(End)