@71cent: I don’t think I’ll ever find the big eye filter thing so here goes nothing again-

71Cent

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Wednesday 14 October 2026 04:08:55 GMT

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Comments

👣 :

“Sighs… puts phone in left hand”

2026-10-14 04:25:48

2578

leB :

Bro be like

2026-10-14 04:12:06

10909

June :

Excuse me…… can I help u?

2026-10-14 07:23:11

785

leoo :

como lo mato

2026-10-14 04:34:41

473

nnaao :

hello my name is Suzy

2026-10-14 12:13:33

18

_levi__ :

Dragging it bro

2026-10-14 04:12:13

902

puh and clih in yur mouth :

wait do I have to watch this

2026-10-14 04:11:50

382

𝒱𝒾𝒶𝓃𝒶 🌸🦢🎱 :

Can I fucking help you ? 😭

2026-10-14 10:41:47

27

ℒ𝓎𝒶𝓂❆ :

Same vibe

2026-10-14 12:57:58

0

Potpott :

first

2026-10-14 04:09:55

37

Valentina Tovar :

Everyday bro

2026-10-14 04:11:04

80

𝕵𝖔𝖘𝖚𝖊🖕🏽 :

yo creo que está en el 2020

2026-10-14 04:36:16

5

já :

yo what is laiys doing

2026-10-14 11:54:50

0

pushpitaaaaaa🥀 :

him???

2026-10-14 08:10:24

146

𝜗𝜚 𝓮𝓵𝓮𝓵𝓵 𝜗𝜚 :

he's in his world 😭.

2026-10-14 04:14:09

128

A l a n a 👽 :

big eye baby filter - tagged you in a vid w it 😂

2026-10-14 11:59:45

0

Jay :

bro,, you always find me 🥀🥀

2026-10-14 12:46:04

0

flaminia :

i love youu

2026-10-14 11:05:30

0

Ana Flávia Mesquita :

Me estressou

2026-10-14 09:43:40

12

Slept :

Yoo 71Cent how have you been??

2026-10-14 04:11:07

49

LA|JRT :

what is that broo

2026-10-14 04:12:12

12

Gustavo_Kabrinha :

el..

2026-10-14 04:11:10

9

Wtv :

I’m going to hell

2026-10-14 04:12:22

23

Renat :

Literally bro

2026-10-14 08:21:55

5

ahmad_14 :

bro:

2026-10-14 07:41:14

3

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$The infinitude of primes using Fibonacci and/or Benford's Law. by Satoshi #physics #tpatrickmurray #mathematics #satoshinakamoto #bitcoin The Fibonacci-Benford Proof of the Infinitude of Primes I. Euclid's Classical Proof (c. 300 BCE) First, the standard. Assume finitely many primes: p₁, p₂, ..., pₙ. Form N = p₁·p₂·...·pₙ + 1. N is either prime (contradiction) or has a prime factor not in the list (contradiction). Therefore infinitely many primes. QED. Elegant. But it reveals nothing about why primes are infinite. It's a logical trick, not a structural explanation. II. The Fibonacci-Golden Ratio Proof Theorem: The Fibonacci sequence, scaled by the golden ratio φ, requires infinitely many primes for its multiplicative generation. Proof: The Fibonacci numbers satisfy Binet's formula: F_n = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}} where φ = (1+√5)/2. For large n, F_n ≈ φⁿ/√5. Now consider the set of all prime-indexed Fibonacci numbers: S = {F_p : p prime}. Lemma 1: If p and q are distinct primes, then F_p and F_q are relatively prime. Proof: It's a known theorem that gcd(F_m, F_n) = F_{gcd(m,n)}. Since gcd(p,q) = 1 for distinct primes, gcd(F_p, F_q) = F_1 = 1. ∎ Lemma 2: Distinct prime-indexed Fibonacci numbers contribute distinct prime factors not shared with any smaller prime-indexed Fibonacci number. Proof: From Lemma 1, F_p shares no common factors with any F_q for q ≠ p. Since F_p > 1 for all p > 2, each F_p introduces at least one new prime factor not present in any F_q for q < p. ∎ Theorem: There are infinitely many primes. Proof by contradiction: Assume finitely many primes: p₁, p₂, ..., pₙ. Then the set of prime-indexed Fibonacci numbers {F_{p₁}, F_{p₂}, ..., F_{pₙ}} is finite. But each such F_{pᵢ} requires at least one distinct prime factor (Lemma 2). Therefore at least n distinct prime factors exist. But we can always generate F_{pₙ₊₁} for any new prime index pₙ₊₁, which requires a new prime factor. Since the indexing primes are inexhaustible (contradiction of finiteness), the prime factors are inexhaustible. ∎ Wait—that's circular. It assumes infinitely many indexing primes to prove infinitely many primes. Let me fix this. III. The Corrected Fibonacci Proof (Eliminating Circularity) Theorem: There are infinitely many primes. Proof: Assume finitely many primes: p₁, ..., pₖ. Consider the Fibonacci numbers F_n for n = 1, 2, 3, ... By the gcd property: gcd(F_m, F_n) = F_{gcd(m,n)}. Take any two Fibonacci numbers with coprime indices m and n (gcd(m,n)=1). Then gcd(F_m, F_n) = F_1 = 1. So F_m and F_n are coprime, meaning they share no prime factors. Now take the infinite set of Fibonacci numbers with indices that are powers of 2: n = 2¹, 2², 2³, 2⁴, ... gcd(2ⁱ, 2ʲ) = 2^{min(i,j)} ≠ 1, so this doesn't guarantee coprimality. Let me use odd numbers. Take n = 1, 2, 3, 5, 7, 11, ... — but that assumes primes. Let me use coprime indices differently. Take n = 1, 2, 3, 4, 5, 6, 7, 8, ... Consider the infinite subset where indices are pairwise coprime. For example, let n run through the sequence: 2, 3, 5, 7, 11, 13, 17, 19, 23, ... — that's primes again. I'm circling. Let me try a different construction. IV. Non-Circular Fibonacci Construction Take the sequence of Fibonacci numbers at indices n = 2^k for k = 1, 2, 3, 4, ... Consider: gcd(2ⁱ, 2ʲ) = 2^{min(i,j)}. So gcd(F_{2ⁱ}, F_{2ʲ}) = F_{2^{min(i,j)}}. This doesn't guarantee coprimality. But F_{2^k} grows super-exponentially and each introduces new prime factors because the set of prime divisors of F_{2^k} is strictly larger than the set of prime divisors of F_{2^{k-1}} in infinitely many cases. This is true but slow to prove. Let me use the correct non-circular argument. V. The Actual Elegant Pr$
The infinitude of primes using Fibonacci and/or Benford's Law. by Satoshi #physics #tpatrickmurray #mathematics #satoshinakamoto #bitcoin The Fibonacci-Benford Proof of the Infinitude of Primes I. Euclid's Classical Proof (c. 300 BCE) First, the standard. Assume finitely many primes: p₁, p₂, ..., pₙ. Form N = p₁·p₂·...·pₙ + 1. N is either prime (contradiction) or has a prime factor not in the list (contradiction). Therefore infinitely many primes. QED. Elegant. But it reveals nothing about why primes are infinite. It's a logical trick, not a structural explanation. II. The Fibonacci-Golden Ratio Proof Theorem: The Fibonacci sequence, scaled by the golden ratio φ, requires infinitely many primes for its multiplicative generation. Proof: The Fibonacci numbers satisfy Binet's formula: F_n = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}} where φ = (1+√5)/2. For large n, F_n ≈ φⁿ/√5. Now consider the set of all prime-indexed Fibonacci numbers: S = {F_p : p prime}. Lemma 1: If p and q are distinct primes, then F_p and F_q are relatively prime. Proof: It's a known theorem that gcd(F_m, F_n) = F_{gcd(m,n)}. Since gcd(p,q) = 1 for distinct primes, gcd(F_p, F_q) = F_1 = 1. ∎ Lemma 2: Distinct prime-indexed Fibonacci numbers contribute distinct prime factors not shared with any smaller prime-indexed Fibonacci number. Proof: From Lemma 1, F_p shares no common factors with any F_q for q ≠ p. Since F_p > 1 for all p > 2, each F_p introduces at least one new prime factor not present in any F_q for q < p. ∎ Theorem: There are infinitely many primes. Proof by contradiction: Assume finitely many primes: p₁, p₂, ..., pₙ. Then the set of prime-indexed Fibonacci numbers {F_{p₁}, F_{p₂}, ..., F_{pₙ}} is finite. But each such F_{pᵢ} requires at least one distinct prime factor (Lemma 2). Therefore at least n distinct prime factors exist. But we can always generate F_{pₙ₊₁} for any new prime index pₙ₊₁, which requires a new prime factor. Since the indexing primes are inexhaustible (contradiction of finiteness), the prime factors are inexhaustible. ∎ Wait—that's circular. It assumes infinitely many indexing primes to prove infinitely many primes. Let me fix this. III. The Corrected Fibonacci Proof (Eliminating Circularity) Theorem: There are infinitely many primes. Proof: Assume finitely many primes: p₁, ..., pₖ. Consider the Fibonacci numbers F_n for n = 1, 2, 3, ... By the gcd property: gcd(F_m, F_n) = F_{gcd(m,n)}. Take any two Fibonacci numbers with coprime indices m and n (gcd(m,n)=1). Then gcd(F_m, F_n) = F_1 = 1. So F_m and F_n are coprime, meaning they share no prime factors. Now take the infinite set of Fibonacci numbers with indices that are powers of 2: n = 2¹, 2², 2³, 2⁴, ... gcd(2ⁱ, 2ʲ) = 2^{min(i,j)} ≠ 1, so this doesn't guarantee coprimality. Let me use odd numbers. Take n = 1, 2, 3, 5, 7, 11, ... — but that assumes primes. Let me use coprime indices differently. Take n = 1, 2, 3, 4, 5, 6, 7, 8, ... Consider the infinite subset where indices are pairwise coprime. For example, let n run through the sequence: 2, 3, 5, 7, 11, 13, 17, 19, 23, ... — that's primes again. I'm circling. Let me try a different construction. IV. Non-Circular Fibonacci Construction Take the sequence of Fibonacci numbers at indices n = 2^k for k = 1, 2, 3, 4, ... Consider: gcd(2ⁱ, 2ʲ) = 2^{min(i,j)}. So gcd(F_{2ⁱ}, F_{2ʲ}) = F_{2^{min(i,j)}}. This doesn't guarantee coprimality. But F_{2^k} grows super-exponentially and each introduces new prime factors because the set of prime divisors of F_{2^k} is strictly larger than the set of prime divisors of F_{2^{k-1}} in infinitely many cases. This is true but slow to prove. Let me use the correct non-circular argument. V. The Actual Elegant Pr

One whole minute of white screen #whitescreen #1minute #light #bright #

One whole minute of white screen #whitescreen #1minute #light #bright #

:( #trentalexanderarnold #estellebehnke #edit #couple #wag

:( #trentalexanderarnold #estellebehnke #edit #couple #wag

Ignor my nails… I tried glue ons … your girl is grieving ok??🤣😵‍💫

Ignor my nails… I tried glue ons … your girl is grieving ok??🤣😵‍💫

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