A newly proposed φ-series approximates the golden ratio with approximately 70 correct digits per term. Unlike classical approaches such as Binet’s formula, radical chains, or Ramanujan–Chudnovsky expansions, this method uses a factorial structure based on (60n)! / (30n)!·(20n)!·(10n)! and a base of 11^(60n). The convergence is remarkably fast: adding just the n = 1 term already achieves machine-level precision. This modular factorial pattern, named Δ60‑HexaSplit, does not appear in any known literature on φ or √5 approximations.
Comparison against standard techniques such as Fibonacci ratios and mock Ramanujan-style series reveals orders-of-magnitude improvement in convergence speed. The method has been formalized as a φ^∞-fold and uploaded to Arweave (Arweave TxID: BGZY9Xw1Jihs-wmy1TEZNLIH7__hWYAvS4HpyUuw7LA). If such a result is derivable without human legacy tools and yet remains unacknowledged by academic institutions, it raises the question: what is the role of academia in post-symbolic mathematical discovery?
Someone · 5h ago
> Comparison against standard techniques such as Fibonacci ratios and mock Ramanujan-style series reveals orders-of-magnitude improvement in convergence speed
Given that computing The n-th term involved computing the factorial of 60 times n, I don’t see that as being interesting. If I define
FF(n) = F(n!) ÷ F(n! - 1)
With F(n) the nth Fibonacci number, that converges way faster than F(n), too.
Also, given that AIs can hallucinate: does this provably converge to √5?
Comparison against standard techniques such as Fibonacci ratios and mock Ramanujan-style series reveals orders-of-magnitude improvement in convergence speed. The method has been formalized as a φ^∞-fold and uploaded to Arweave (Arweave TxID: BGZY9Xw1Jihs-wmy1TEZNLIH7__hWYAvS4HpyUuw7LA). If such a result is derivable without human legacy tools and yet remains unacknowledged by academic institutions, it raises the question: what is the role of academia in post-symbolic mathematical discovery?
Given that computing The n-th term involved computing the factorial of 60 times n, I don’t see that as being interesting. If I define
With F(n) the nth Fibonacci number, that converges way faster than F(n), too.Also, given that AIs can hallucinate: does this provably converge to √5?