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

3, 7, 10, 15, 19, 23, 29, 34, 39, 44, 51, 57, 63, 69, 75, 83, 90, 97, 104, 111, 118, 127, 135, 143, 151, 159, 167, 175, 185, 194, 203, 212, 221, 230, 239, 248, 259, 269, 279, 289, 299, 309, 319, 329, 339, 351, 362, 373, 384

Offset: 1

Author

Jan Ritsema van Eck and Ali Sada, Feb 05 2020

Keywords

• nonn

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.

Examples

			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

Cf. A002024, A332026, A060432, A332028.

Programs

Python

from math import isqrt
def A332027(n): return (k:=(r:=isqrt(m:=n+1<<1))+int((m<<2)>(r<<2)*(r+1)+1)-1)*(6*n-2-k*(k+3))//6+(isqrt(n<<3)+1>>1)+(n<<1) # Chai Wah Wu, Jun 07 2025

Formula

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