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A332028 Savannah problem: number of distinct possible populations after n weeks, not allowing new populations after the empty set. 2
3, 5, 7, 11, 14, 17, 22, 26, 30, 34, 40, 45, 50, 55, 60, 67, 73, 79, 85, 91, 97, 105, 112, 119, 126, 133, 140, 147, 156, 164, 172, 180, 188, 196, 204, 212, 222, 231, 240, 249, 258, 267, 276, 285, 294, 305, 315, 325, 335, 345 (list; graph; refs; listen; history; text; internal format)
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 not allowed (See A332027 for the other version).

Since all populations with one fed lion remain stationary as long as one zebra enters the savannah each week, and all populations with more than one lion follow directly or indirectly from those with one fed lion, we know for all population with one or more lions (except one hungry lion) that if they are possible in week n, they will also be possible in week n+1. The population with one hungry lion can only be reached via the empty set, so it is not possible after week 1. The same goes for the population of 1 zebra without lion. The population of k zebras without lion can only be reached from k-1 zebras without lion. So it is only possible in week k. This means that in week n, there are n populations that were possible in previous weeks but not any more: 1 hungry lion, and 1 through (n-1) zebras without lion. Therefore, the total number of possible populations in week n is equal to the number if continuation of the empty set is allowed (A332027), minus n (except in week 1, when it is 3).

LINKS

Table of n, a(n) for n=1..50.

FORMULA

a(n) = A060432(n) + A002024(n), for n>1.

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; and from F (which becomes H in the second step): (ZH->) F, FH and 2H. Population which follow from H (which becomes the empty set in the first step), are not allowed. Overall, there are 5 distinct possible populations after the second week: 2Z, ZF, FH, F and 2H.

CROSSREFS

Cf. A060432, A002024, A332026, A332027.

Sequence in context: A133954 A087325 A072151 * A280018 A029626 A111622

Adjacent sequences: A332025 A332026 A332027 * A332029 A332030 A332031

KEYWORD

nonn

AUTHOR

Jan Ritsema van Eck and Ali Sada, Feb 05 2020

STATUS

approved

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Last modified December 9 23:05 EST 2022. Contains 358710 sequences. (Running on oeis4.)