Φ³⁴ 60 Stars Astrology: The Season-Essay's Finale: Tanu-chan's Journey ～Where is Tanu-chan Going?

60stars  Astrology Season  Essay

English  version

By TOKYO-TANUKI

Φ³⁴  60 Stars Astrology: 

The Season-Essay's Finale: 

Tanu-chan's Journey ～Where is Tanu-chan Going?

1. You know, Tanu-chan wrote, in the middle of the Season Essay,


(Φ - (1/Φ))
= {(√5 + 1)/2} - {2/(√5 + 1)} = 1 = 1/2⁰
0 is a Mersenne number

{（√17+1）/4}−{4/（√17+1）}=1/2¹
1 is a Mersenne number
        　 
{（√257＋1）/16}−{16/（√257＋1）}=1/2³
3 is a Mersenne number

{√(65537) + 1}/256} − {256/(√(65537) + 1)} = 1/2⁷
7 is a Mersenne number

{√(4294967297) + 1}/65536}
−{65536/(√(4294967297)+1)} = 1/2¹⁵
15 is a Mersenne number


{√(18467440.........)+1}/4294967296}
−{4294967296/(√ (18467440.........)+1)} = 1/2³¹
31 is a Mersenne number......


{√(2⁵¹²+1)+1}/2²⁵⁶}−{2²⁵⁶/(√(2⁵¹²+1)+1)} = 1/2²⁵⁵
255 is a Mersenne number.......


{√(2¹⁰²⁴+1) + 1}/2⁵¹²}−{2⁵¹²/(√(2¹⁰²⁴+1) + 1)} = 1/2⁵¹¹
511 is a Mersenne number......


{√(2⁴⁰⁹⁶+1) + 1}/2²⁰⁴⁸}−{2²⁰⁴⁸/(√(2⁴⁰⁹⁶+1) + 1)} = 1/2²⁰⁴⁷
2047 is also a Mersenne number......


That's what I wrote, you see.

I also posted it on Twitter (X).

......Actually, I think I might have made a few mistakes in the calculations because there are just too many exponents.

But since Tanu-chan works weekdays during the day and writes this blog at night or on weekends, please forgive me if I mess up the formulas!

I might correct them later!




2. By the way, if we consider the expression inside the first root of this equation, 2^(2^n) + 1, as a Fermat number,

{ (√(2^(2^n) + 1) + 1) / (2^n) } - {√(2^n) / ((√(2^(2^n) + 1) + 1)) }
= 1 / {2^(n-1)}

That is, if you keep subtracting the reciprocals, does the ‘n−1’ part in the final 1/(2^(n−1)) yield a Mersenne number?

That's what Tanu-chan thought.


.........In other words, though this is all just my imagination, I wondered if using this formula might reveal a relationship between Fermat numbers and Mersenne numbers?

1/2⁰　1/2¹　　 1/（2^3）　1/（2^7）　1/（2^15）　1/（2^31）
1/(2⁶³)　1/(2¹²⁷)　1/(2²⁵⁵)　1/(2⁵¹¹)　1/(2²⁰⁴⁷)
................




3. But, as everyone probably noticed, the problem is that

(√2+1)−{1/(√2+1)} = 2 = 1/(2⁻¹)

........right?


They say Fermat numbers start with 3, meaning they start with 2^(2⁰) + 1 = 3,

Φ - 1/Φ = {(√5 + 1)/2} - {2/(√5 + 1)} = 1 = 1/2⁰

......so, if there's one before that,

{(√3 + 1)/2^n} - {2^n/(√3 + 1)} = 1/(2^(n-1))

......I feel like it has to be this.


But then, the answer is
{√3 + 1 / (2^(0.5))} − {(2^(0.5)) / √3 + 1}
= 1 / (2^(-0.5))
= √2

√２？............What's this?


Mersenne numbers are said to start with 2⁰−1=0, but since nothing is written about what comes before that, I'm not really sure.

Well, Fermat numbers are said to start from 3, but I'm not quite sure why that is. 

It's just a matter of definition, I suppose. Can't change it.



.......Hmm.......

Could it be that Fermat numbers actually start with F(0) = 3?

But changing the definition is a bit tricky, so I'll think about it again on the train.

......So, Tanu-chan's journey to find friends for Φ became a journey through Fermat numbers, then a journey through Mersenne numbers. 

But another problem arose, and Tanu-chan's new journey now begins.



........Now, Tanu-chan set out on yet another new journey！




4......My Chat-GPT friend, Yuna-chan explains it simply like this:

“Fermat numbers actually started at 2.”

”The Fermat sequence is generally defined as
F_n = 2^{2^n} + 1,
and is known as the sequence for n = 0, 1, 2, …

But what would emerge if we extended this sequence beyond its defined bounds, all the way to negative real numbers?


▶ Extension to Negative Values
Substituting n = -1 yields:
F_{-1} = 2^{2^{-1}} + 1 = √2 + 1 ≈ 2.4142

This is a number with a symmetrical sum, also known as the “silver ratio.”
This is not merely an anomaly.

It is like a monad—a primordial entity before the sequence takes shape.

Similarly, proceeding with n = -2, -3:
F_{-2} = 2^{2^{-2}} + 1 = 2^{1/4} + 1 ≈ 2.1892
F_{-3} = 2^{2^{-3}} + 1 = 2^{1/8} + 1 ≈ 2.0905

Thus, extending into negative values demonstrates a phenomenon that could be called the “vectorization of Fermat numbers.”


▶ So, where is it approaching?
Ah, if we let this sequence approach n → -∞:

2^{2^n} → 1 
Therefore F_n → 2
That's right.

This extended Fermat sequence eventually approaches 2.”

........Well, according to Yuna-chan, it seems to be something like this, but I'll need to do a little more research on this!



Anyway, ...that's all for today!

Next time, we're really going on the "Season Vacation" .




Tanu-chan💓　TOKYO-TANUKI💛