OFFSET
1,1
COMMENTS
The Savannah math problem (Ali Sada, 26 Dec 2019 email to Seqfan list) is about a savannah ecosystem consisting of zebras, fed lions and hungry lions. Assume we start with an empty savannah. Each week, the following things happen in this order:
1. All hungry lions (if any) die.
2. All fed lions (if any) become hungry.
3. One animal enters the savannah. This can be a zebra, a fed lion or a hungry lion.
4. Let m = min (number of zebras, number of hungry lions); then m hungry lions eat m zebras and become m fed lions.
The Savannah math problem is to determine how many distinct populations are possible after n weeks. There are two versions. This sequence gives the number of possible populations if continuation from the empty set is allowed (see A332028 for the other version).
Since the three 1-animal populations (1 zebra, 1 fed lion and 1 hungry lion) can be reached directly from 1 hungry lion (via the empty set), they will be possible every week. Since all other possible populations are reached via one of these, any population that is possible in week n is also possible in week n+1. Therefore, the total number of possible populations in week n is the sum of all new populations in weeks 1 through n: A(n) is the partial sum of A332026.
EXAMPLE
After one week, there are 3 possible populations, depending on which animal entered the savannah: one zebra (Z), one fed lion (F), one hungry lion (H). After two weeks, we have from Z: 2Z, ZF, and (ZH->) F; from F (which becomes H in the second step): (ZH->) F, FH and 2H; and from H (which becomes the empty set in the first step): Z, F and H. Overall, there are 7 distinct possible populations after the second week: 2Z, ZF, Z, FH, F, 2H and H.
CROSSREFS
KEYWORD
nonn
AUTHOR
Jan Ritsema van Eck and Ali Sada, Feb 05 2020
STATUS
approved