What's the Pattern in This Number Sequence?

  • Context: High School 
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Discussion Overview

The discussion revolves around identifying the next number in a specific number sequence. Participants explore various interpretations and patterns within the sequence, which includes both numerical values and potential mathematical relationships. The scope includes mathematical reasoning and exploratory analysis.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant presents the sequence 1, 1, 6561, 2197, 289, 21, 1, 707281 and asks for the next number.
  • Another participant suggests an alternative sequence of 1, 1, 94, 133, 172, 211, 250, 294, expressing confusion about the initial '1' in the context of their interpretation.
  • A later reply proposes a mathematical representation of the sequence using exponents, suggesting that the next number could be 33 raised to the power of 3, based on their derived pattern.
  • One participant indicates that the exponents are determined by the bases mod 5, presenting a formula for the kth term as (4k+1)^{4k+1 MOD 5}.

Areas of Agreement / Disagreement

Participants express differing interpretations of the sequence and its components, with no consensus reached on a single correct pattern or next number. Multiple competing views remain evident throughout the discussion.

Contextual Notes

Some assumptions about the sequence's structure and the significance of the initial '1' are not fully explored. The mathematical steps leading to the proposed formula and the interpretation of exponents are also not resolved.

holomorphic
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What's the next number?

1, 1, 6561, 2197, 289, 21, 1, 707281, ...?
 
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hmm …

well, that's 1, 1, 94, 133, 172, 211, 250, 294 …

i don't see where the 1 at the start fits in :confused:
 
tiny-tim said:
hmm …

well, that's 1, 1, 94, 133, 172, 211, 250, 294 …

i don't see where the 1 at the start fits in :confused:

Well, that would be:
[tex]1^{...}, 5^{0},9^{4},13^{3}, 17^{2}, 21^{1}, 25^{0}, 29^{4}[/tex]
So, the next would be [itex]33^{3}[/itex], whatever the exponent of the first number in the sequence.
 
oh, i get it now! … it starts 11, 50 :rolleyes:
 
arildno said:
Well, that would be:
[tex]1^{...}, 5^{0},9^{4},13^{3}, 17^{2}, 21^{1}, 25^{0}, 29^{4}[/tex]
So, the next would be [itex]33^{3}[/itex], whatever the exponent of the first number in the sequence.

Yup :)

FYI, the exponents are the bases mod 5.

So the formula for the kth term is:
[tex](4k+1)^{4k+1 MOD 5}[/tex]
 

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