
EXAMPLE

For n = 1, there is only a(1) = 1 possibility, CH4.
For n = 2, one has C2H6 (ethane, H3CCH3), C2H4 (ethylene, H2C=CH2 with a double bond), C2H2 (ethyne, HC≡CH, triple bond), whence a(2) = 3.
For n = 3, one has C3H8 (H3CCH2CH3), C3H6 (H2C=CHCH3, propene), and two C3H4 (H2C=C=CH2, propadiene, and HC≡CCH3: methylacetylene), thus a(3) = 4. Cyclic molecules like cyclopropane C3H6 and cyclopropropene C3H4 are excluded. (End)
For n = 4, we have butane, isobutane, 1butene, cis/trans2butene, buta1,2diene, buta1,3diene, butatriene, isobutylene, but1yne, but2yne, diacetylene, but1en3yne.
For n = 5 we have:
 3 alkanes: pentane, methylbutane and neopentane.
 17 alkenes: 1pentene, (E/Z)2pentene, 1,2pentadiene, (E/Z)1,3pentadiene, 1,4pentadiene, 1,2,3petatriene, penta1,2,4triene, pentatetraene, 2methylbut1ene, 2methylbut2ene, 3methylbut1ene, isoprene, 3methylbuta1,2diene, (R/S)penta2,3diene.
11 alkynes: 1pentyne, 2pentyne, pent1en4yne, (E/Z)pent3en1yne, penta1,2dien4yne, penta1,4diyne, penta1,3diyne, pent1en3yne, 3methylbut1yne, 2methylbut1en3yne. (End)
