It can be shown that increasing the number of circles on each edge gives the following recurrence relation:

H (n) =

1 if n = 1

H (n - 1) + 6n - 6 if n > 1.

Calculate

H (9).

Method 1:H (1) = 1

H (2) = 1+6 (2) - 6=7

H (3) = 7+6 (3) - 6=19

H (4) = 19+6 (4) - 6=37

H (5) = ... = 61

H (6) = ... = 91

H (7) = ... = 127

H (8) = ... = 169

H (9) = 169+6 (9) - 6 = 217

Method 2:

H (n) = H (n-1) + 6n-6

H (n-1) = H (n-2) + 6 (n-1) - 6

H (n-2) = H (n-3) + 6 (n-2) - 6

...

H (3) = H (2) + 6 (3) - 6H (2)

=H (1) + 6 (2) - 6

Sum all lines, and cancel common terms on either side

H (n) = H (1) + 6 (n+n-1+n-2 + ... + 3+2) - 6 (n-1)

=H (1) + 6 (n+2) (n-1) / 2-6 (n-1)

=H (1) + 3n^2-3n [ after expanding and simplifying ]

This mean that H (9) = H1+3 (9^2) - 3 (9) = 217

if needed, for example, H (123) = 1+3 (123^2) - 3 (123) = 45019