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A332027 Savannah problem: number of distinct possible populations after n weeks, allowing populations after the empty set. 2

%I #15 Feb 08 2020 09:22:48

%S 3,7,10,15,19,23,29,34,39,44,51,57,63,69,75,83,90,97,104,111,118,127,

%T 135,143,151,159,167,175,185,194,203,212,221,230,239,248,259,269,279,

%U 289,299,309,319,329,339,351,362,373,384

%N Savannah problem: number of distinct possible populations after n weeks, allowing populations after the empty set.

%C 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:

%C 1. All hungry lions (if any) die.

%C 2. All fed lions (if any) become hungry.

%C 3. One animal enters the savannah. This can be a zebra, a fed lion or a hungry lion.

%C 4. Let m = min (number of zebras, number of hungry lions); then m hungry lions eat m zebras and become m fed lions.

%C 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).

%C 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.

%F a(n) = A060432(n) + A002024(n) + n.

%e 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.

%Y Cf. A002024, A332026, A060432, A332028.

%K nonn

%O 1,1

%A _Jan Ritsema van Eck_ and _Ali Sada_, Feb 05 2020

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Last modified July 13 15:08 EDT 2024. Contains 374284 sequences. (Running on oeis4.)