Assertion: cube's number | /r/HasProperty | given by n = 9 a n − 1 − 12
Facets:
Source sentences:
text: The number of these hexagrams, in descending size, is given by a n = 9 a n − 1 − 12 a n − 2 {\displaystyle a_{n}=9a_{n-1}-12a_{n-2}} , with a 0 = 1 , a 1 = 6 {\displaystyle a_{0}=1,\ a_{1}=6} .