Ollie lvl Question 1: Let a1, a2, a3, . . . be an infinite sequence of positive integers, and let N be a positive integer. Suppose that, for each n > N , an is equal to the number of times an−1 appears in the list a1, a2, . . . , an−1. Prove that at least one of the sequences a1, a3, a5, . . . and a2, a4, a6, . . . is eventually periodic. (An infinite sequence b1, b2, b3, . . . is eventually periodic if there exist positive integers p and M such that bm+p = bm for all m ⩾ M .)
@@c4r151 For anyone curious, here is the proof: - *Behavior of the Sequence:* For n>N, the sequence a𝑛 is determined iteratively: If 𝑎𝑛 − 1 has already appeared frequently in the list 𝑎1, 𝑎2,..., 𝑎𝑛 − 1, then 𝑎𝑛 is relatively large (within the bound 𝑛 − 1). If 𝑎𝑛 − 1 is rare, 𝑎𝑛 is small. Since the value of 𝑎𝑛 is defined solely based on counts of earlier terms, the sequence 𝑎𝑛 reflects a dependency chain based on repetition patterns. - *Partitioning into Odd and Even Indices:* The sequence 𝑎1, 𝑎2, 𝑎3,… can be partitioned into two subsequences: 𝑏𝑛 = 𝑎2n - 1 (odd-indexed terms) 𝑐𝑛 = 𝑎2𝑛 (even-indexed terms) We analyze 𝑏𝑛 and 𝑐𝑛 separately to determine their periodicity. - *Finiteness of Values for Each Subsequence:* The critical observation is that both 𝑏𝑛 and 𝑐𝑛 take values that are counts of previous terms. The total number of distinct counts possible is bounded by the range 1, 2,…, 𝑁 for 𝑛 > 𝑁. Thus: The number of distinct values in each subsequence is finite. - *Pigeonhole Principle and Eventual Periodicity:* Since the values in each subsequence are bounded, and the subsequences 𝑏𝑛 and 𝑐𝑛 are infinite, the pigeonhole principle implies that some values must repeat. Once a value repeats, the deterministic construction of 𝑏𝑛 and 𝑐𝑛 ensures periodicity: Let 𝑝 be the length of the periodic cycle, and let 𝑀 be the point from which the cycle starts. For 𝑚 ≥ 𝑀, 𝑏𝑚 + 𝑝 = 𝑏𝑚 (or 𝑐𝑚 + 𝑝 = 𝑐𝑚). TL;DR: By partitioning the sequence into 𝑏𝑛 = 𝑎2𝑛 − 1 and 𝑐𝑛 = 𝑎2𝑛, and noting the boundedness of the values each can take, we conclude that at least one of the subsequences 𝑏𝑛 or 𝑐𝑛 is eventually periodic.
Sandbagging her while she’s not here to defend herself tsk-tsk, Lol.
Gura lvl is quick maths
2+2 is 4, -1 that's 3 quick mafs
Gura level requires you to count your fingers 🤌
Ollie lvl quick maths would be M times X plus B. Straight line formula quick maths
Ollie lvl = Who is ready for some Calculus? XD
Ollie lvl Question 1:
Let a1, a2, a3, . . . be an infinite sequence of positive integers, and let N be a positive integer.
Suppose that, for each n > N , an is equal to the number of times an−1 appears in the list
a1, a2, . . . , an−1.
Prove that at least one of the sequences a1, a3, a5, . . . and a2, a4, a6, . . . is eventually periodic.
(An infinite sequence b1, b2, b3, . . . is eventually periodic if there exist positive integers p and M such that bm+p = bm for all m ⩾ M .)
@@c4r151
For anyone curious, here is the proof:
- *Behavior of the Sequence:*
For n>N, the sequence a𝑛 is determined iteratively:
If 𝑎𝑛 − 1 has already appeared frequently in the list 𝑎1, 𝑎2,..., 𝑎𝑛 − 1, then 𝑎𝑛 is relatively large (within the bound
𝑛 − 1).
If 𝑎𝑛 − 1 is rare, 𝑎𝑛 is small.
Since the value of 𝑎𝑛 is defined solely based on counts of earlier terms, the sequence 𝑎𝑛 reflects a dependency chain based on repetition patterns.
- *Partitioning into Odd and Even Indices:*
The sequence 𝑎1, 𝑎2, 𝑎3,… can be partitioned into two subsequences:
𝑏𝑛 = 𝑎2n - 1 (odd-indexed terms)
𝑐𝑛 = 𝑎2𝑛 (even-indexed terms)
We analyze 𝑏𝑛 and 𝑐𝑛 separately to determine their periodicity.
- *Finiteness of Values for Each Subsequence:*
The critical observation is that both 𝑏𝑛 and 𝑐𝑛 take values that are counts of previous terms. The total number of distinct counts possible is bounded by the range 1, 2,…, 𝑁 for 𝑛 > 𝑁. Thus:
The number of distinct values in each subsequence is finite.
- *Pigeonhole Principle and Eventual Periodicity:*
Since the values in each subsequence are bounded, and the subsequences 𝑏𝑛 and 𝑐𝑛 are infinite, the pigeonhole principle implies that some values must repeat. Once a value repeats, the deterministic construction of 𝑏𝑛 and 𝑐𝑛 ensures periodicity:
Let 𝑝 be the length of the periodic cycle, and let 𝑀 be the point from which the cycle starts.
For 𝑚 ≥ 𝑀, 𝑏𝑚 + 𝑝 = 𝑏𝑚 (or 𝑐𝑚 + 𝑝 = 𝑐𝑚).
TL;DR: By partitioning the sequence into 𝑏𝑛 = 𝑎2𝑛 − 1 and 𝑐𝑛 = 𝑎2𝑛, and noting the boundedness of the values each can take, we conclude that at least one of the subsequences 𝑏𝑛 or 𝑐𝑛 is eventually periodic.
@@c4r151
It appears that my answer has been shadowbanned :/
@@c4r151 Thank god I decided to study something easy, like astrophysics.
Moments before Calli was never seen or heard from again:
Kay Yu-sensei is a goddamn champion.
*LETS GO GAMBLEING AW DANG IT* Intensifies
LET'S BE FINANCIALLY RESPONSIBLE
all losses go to Pekomamma no refunds
I read it as Play Feet in the thumbnail