n = 5 … 100 · k = 5 · source A = 0⁵1n−5 · t(n) = C(n,5) · asymptote m/t → 7/32 = 0.21875
t(n) = C(n, 5) (total vertices)
m(n) = C(⌊n/2⌋, 5) + C(⌈n/2⌉, 5) + C(⌊n/2⌋, 4)·⌈n/2⌉ (mismatch vertices)
| All 5 zeros at even positions | Δ = 3 | C(⌊n/2⌋, 5) ways |
| All 5 zeros at odd positions | Δ = 2 | C(⌈n/2⌉, 5) ways |
| Exactly 4 even + 1 odd zero | Δ = 2 | C(⌊n/2⌋, 4)·⌈n/2⌉ ways |