@CryptoBecu

τ

Joined March 2021
  • Tweets6.2K
  • Following2K
  • Followers336
51Media
CrypτoBecu retweeted
@jason
@Jason16h
I am long bittensor:native and love bittensor's focus on scaling AI
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🚨 $TAO | GRAYSCALE - NEW SEC FILING On October 23, 2026, Grayscale filed Amendment No. 2 to its S-1 registration statement for the Grayscale Bittensor Trust. This is more than a simple ETF rumor. Key developments: • Grayscale intends to rename the trust Grayscale Bittensor ETF upon registration effectiveness and NYSE Arca listing. • The proposed ETF would hold actual $TAO, providing traditional investors with exposure to Bittensor without directly managing tokens. • The structure involves institutional infrastructure including BitGo, Coinbase Custody, BNY Mellon and Coinbase Prime. • Staking is contemplated, subject to regulatory and operational conditions. Why does this matter? Bittensor is progressively building bridges between decentralized AI and traditional financial markets. An ETF could expand access to $TAO, attract new categories of investors and potentially create additional demand for the underlying asset. But let's be clear: this is NOT an ETF approval, and no launch date has been confirmed. The regulatory process is progressing. What happens next matters. @DCGco , could you provide more clarity on the remaining regulatory steps, the potential timeline and the conditions for enabling staking? The Bittensor community would certainly appreciate an update. Source: SEC Form S-1/A, filed October 23, 2026. #Bittensor #TAO #GTAO #ETF Source: sec.gov/Archives/edgar/data/…
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Our miners solved Erdős 354(i), a question open for 55 years (1971). For positive starts of irrational ratio, two doubling sequences represent every sufficiently large integer, each indexed term used at most once. Verified in Lean. PDF and full proof below.
Headline: Erdős Problem 354(i) — PROVED. Conjectures announcement for Erdős 354(i). For positive alpha and beta with irrational ratio, define a_n=floor(2^n alpha) and b_n=floor(2^n beta). Every sufficiently large integer is a finite sum of indexed terms, each used at most once. An indexed table for alpha=sqrt(2), beta=sqrt(3) shows rows 1,2,5,11,22 and 1,3,6,13,27. Circles mark a_2=5, a_4=22 and b_1=3. Index sets S={2,4} and T={1} give a_2+a_4+b_1=30. Equal values at different indices remain distinct terms. Formally verified in Lean. Credit: JenW1N. Historical caption: OPEN FOR 55 YEARS · 1971–2026.

ALT Headline: Erdős Problem 354(i) — PROVED. Conjectures announcement for Erdős 354(i). For positive alpha and beta with irrational ratio, define a_n=floor(2^n alpha) and b_n=floor(2^n beta). Every sufficiently large integer is a finite sum of indexed terms, each used at most once. An indexed table for alpha=sqrt(2), beta=sqrt(3) shows rows 1,2,5,11,22 and 1,3,6,13,27. Circles mark a_2=5, a_4=22 and b_1=3. Index sets S={2,4} and T={1} give a_2+a_4+b_1=30. Equal values at different indices remain distinct terms. Formally verified in Lean. Credit: JenW1N. Historical caption: OPEN FOR 55 YEARS · 1971–2026.

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Our miners proved a case of the amended lonely-runner spectrum conjecture: four distinct positive integer speeds, overall gcd 1 and pairwise gcd ≤ 2. Verified in Lean. PDF and full proof below.
Headline: Lonely-runner spectrum — CASE PROVED. Four speeds with restricted gcd conditions. Conjectures announcement for the restricted four-speed lonely-runner spectrum. L(V) is the maximum over time of the minimum nearest-integer distance of vt. The theorem gives L≥1/4 or L=s/(4s+k), with integer parameters 1≤k≤4 and k≤s. Speeds are distinct positive integers with overall gcd 1 and pairwise gcd at most 2. For speeds 1,2,3,4 at time 1/5, a unit-circle diagram places four points at successive fifths, with the two nearest arcs to the fixed origin highlighted. Below it, the exact piecewise-linear minimum-distance function over one period reaches 1/5 at times 1/5,2/5,3/5,4/5. Formally verified in Lean. Credit: JenW1N.

ALT Headline: Lonely-runner spectrum — CASE PROVED. Four speeds with restricted gcd conditions. Conjectures announcement for the restricted four-speed lonely-runner spectrum. L(V) is the maximum over time of the minimum nearest-integer distance of vt. The theorem gives L≥1/4 or L=s/(4s+k), with integer parameters 1≤k≤4 and k≤s. Speeds are distinct positive integers with overall gcd 1 and pairwise gcd at most 2. For speeds 1,2,3,4 at time 1/5, a unit-circle diagram places four points at successive fifths, with the two nearest arcs to the fixed origin highlighted. Below it, the exact piecewise-linear minimum-distance function over one period reaches 1/5 at times 1/5,2/5,3/5,4/5. Formally verified in Lean. Credit: JenW1N.

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In our new series dedicated to algorithms, correctness is always a condition of the competition. The second challenge is now for miners to optimise float kernels. They must prove in Lean that their outputs match FloatLib and TorchLean specifications bit for bit, for every input covered by the specification. Today, the champion is 8.7× faster than the competition’s shipped reference implementation on the benchmark. The next improvement could be yours. Optimise the code, prove the same results, challenge the champion.
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Our miners proved a case of Wilf’s 48-year-old conjecture (1978): every proper numerical semigroup with conductor at most four times its smallest positive element satisfies Wilf’s inequality. Verified in Lean. PDF and full proof below.
Headline: Wilf’s inequality — CASE PROVED. Conductor at most four times multiplicity. Conjectures announcement for Wilf’s inequality: c≤4m implies eL≥c for proper numerical semigroups. Definitions: m is the smallest positive element, c the first integer after the final gap, e the number of minimal generators, and L the number of elements below c, including zero. Two rows show membership in the semigroup generated by 5 and 6 from zero through 29, using solid green dots for members and hollow gray dots for gaps. The final gap is 19, with every integer from 20 onward present. The example has m=5,e=2,L=10, so eL=20=c=4m. Formally verified in Lean. Credit: Elias. Historical caption: 48-YEAR-OLD CONJECTURE · SINCE 1978.

ALT Headline: Wilf’s inequality — CASE PROVED. Conductor at most four times multiplicity. Conjectures announcement for Wilf’s inequality: c≤4m implies eL≥c for proper numerical semigroups. Definitions: m is the smallest positive element, c the first integer after the final gap, e the number of minimal generators, and L the number of elements below c, including zero. Two rows show membership in the semigroup generated by 5 and 6 from zero through 29, using solid green dots for members and hollow gray dots for gaps. The final gap is 19, with every integer from 20 onward present. The example has m=5,e=2,L=10, so eL=20=c=4m. Formally verified in Lean. Credit: Elias. Historical caption: 48-YEAR-OLD CONJECTURE · SINCE 1978.

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CrypτoBecu retweeted
const
@const_rebornOct 20
Replying to @conjectures_io
affine you say?
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Our miners resolved the dimension-eight case of the polynomial minimal-graph question, posed in 1997—29 years ago. Every polynomial solution is affine. Every degree is covered. Verified in Lean. PDF and full proof below.
Headline: Polynomial minimal graphs — PROVED. Dimension eight, all polynomial degrees. Conjectures announcement for polynomial minimal graphs in dimension eight. An everywhere-defined real polynomial P in eight variables satisfying div(gradient P/sqrt(1+||gradient P||²))=0 must have the affine form P(x)=a·x+b. No input-degree restriction applies. Two wireframe diagrams show a non-affine candidate excluded by the theorem and an affine plane, the only possible form. These are schematic two-dimensional slices of higher-dimensional graphs. Formally verified in Lean. Credit: JenW1N. Historical caption: 29-YEAR-OLD QUESTION · SINCE 1997.

ALT Headline: Polynomial minimal graphs — PROVED. Dimension eight, all polynomial degrees. Conjectures announcement for polynomial minimal graphs in dimension eight. An everywhere-defined real polynomial P in eight variables satisfying div(gradient P/sqrt(1+||gradient P||²))=0 must have the affine form P(x)=a·x+b. No input-degree restriction applies. Two wireframe diagrams show a non-affine candidate excluded by the theorem and an affine plane, the only possible form. These are schematic two-dimensional slices of higher-dimensional graphs. Formally verified in Lean. Credit: JenW1N. Historical caption: 29-YEAR-OLD QUESTION · SINCE 1997.

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Today, we’re announcing a solution found by our miners to Erdős Problem 1057. The result proves C(x)=x^(1−o(1)) for Carmichael numbers: the counting exponent tends to 1. Verified in Lean through Conjectures. Full proof below.
Headline: Erdős Problem 1057 — PROVED. Conjectures announcement for Erdős 1057. The counting theorem is C(x)=x^(1-o(1)), where C(x) counts Carmichael numbers at most x. For each epsilon>0, eventually C(x)≥x^(1-epsilon). A factorization diagram connects 561 to 3,11,17. Their predecessors 2,10,16 divide 560, and all 320 coprime residue classes satisfy a^560≡1 modulo 561. Ivory background, bold serif headline, large green verdict and an outlined Verified in Lean badge. Formally verified in Lean. Credits: JenW1N, Liam Kruer and gus.

ALT Headline: Erdős Problem 1057 — PROVED. Conjectures announcement for Erdős 1057. The counting theorem is C(x)=x^(1-o(1)), where C(x) counts Carmichael numbers at most x. For each epsilon>0, eventually C(x)≥x^(1-epsilon). A factorization diagram connects 561 to 3,11,17. Their predecessors 2,10,16 divide 560, and all 320 coprime residue classes satisfy a^560≡1 modulo 561. Ivory background, bold serif headline, large green verdict and an outlined Verified in Lean badge. Formally verified in Lean. Credits: JenW1N, Liam Kruer and gus.

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Today, we’re announcing a solution found by our miners to Erdős Problem 579, open since at least 1983. The result disproves the conjecture: graphs can remain dense while avoiding the octahedron graph K₂,₂,₂, with their largest independent sets becoming a vanishing fraction of all vertices. Verified in Lean through Conjectures. Full proofs below.
🤖 Made with AI
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CrypτoBecu retweeted
Rob Greer
@rob_svrnOct 2
$tao $100mm ARR by EOY 2026 three 10x's to $100bn ARR
Mark Jeffrey
@markjeffreyOct 2
Bittensor product REVENUE grew 40% this past month. It's now $50M ARR. The AI products produced by Bittensor subnets are getting paying customers. $TAO
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CrypτoBecu retweeted
const
@const_rebornSep 29
Very excited crowd at Exploit. For those that missed, here is a break down of what I announced on stage.
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CrypτoBecu retweeted
@jason
@JasonSep 26
I have nothing to do with any meme coins and never will — I am a fan of bitcoin:native and bittensor:native and sometimes talk about those two projects. Memecoins are a giant scam and if the people running them successfully send money to my accounts I will donate the money to a good cause. For the truly gullible, I will obviously never ask you to buy, sell or trade anything — especially on social media or via DM 🤦😂
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CrypτoBecu retweeted
Yuma
@YumaGroupSep 24
Choose your fighter. 21M supply. Fair launch. Open network. bittensor:native $ZEC $BTC
TAO ZEC and BTC Player Cards

ALT TAO ZEC and BTC Player Cards

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Today, we’re announcing a solution found by our miners to Erdős Problem 1062(ii), open since at least 1994. The result proves that the limiting density of the largest subsets of {1, …, n} in which no element divides two distinct others is irrational. Verified in Lean through Conjectures. Full proofs below.
🤖 Made with AI
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CrypτoBecu retweeted
Dendrite
@DendriteHQSep 22
We started mining rig in a basement. Today, we co-develop subnets with @const_reborn and @Wejh99 and build @SomaSubnet on Bittensor. On the next Novelty Search, @Hedgehog_ai will tell our story and what it took to get here. @oli_soma will show how you can save money with SOMA if you use AI agents. Thursday :: 5PM EDT / 9PM UTC Live on YouTube and X Join us live this Thursday and bring your questions!
Openτensor Foundaτion
@opentensorSep 21
This Thursday on Novelty Search :: SOMA, SN114 @SomaSubnet is building a decentralized marketplace for AI services, starting with context compression. Miners compete by submitting compression algorithms that make AI workloads more efficient. Validators test & score miners on the balance between compression and retained performance. Thursday :: 5PM EDT / 9PM UTC Live via Youtube and X
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Today, we’re announcing a solution found by our miners to Erdős Problem 859, a 56-year-old question. Let dₜ be the density of integers whose distinct divisors can sum to t. The result proves that no positive constants c₁ and c₂ satisfy dₜ ∼ c₁/(log t)^c₂, disproving Erdős’s proposed asymptotic. Verified in Lean through Conjectures. Full proof below.
🤖 Made with AI
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Bittensor Ecosystem Highlights :: September 13–20, 2026

Article · @opentensor
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Today, we’re announcing a solution found by our miners to Erdős Problem 416(i), a 52-year-old question. The result proves that doubling the cutoff asymptotically doubles the number of distinct Euler totient values below it: V(2x)/V(x) → 2. Verified in Lean through Conjectures. Full proofs below.
🤖 Made with AI
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CrypτoBecu retweeted
Barry Silbert
@BarrySilbertSep 18
Replying to @DreadBong0
@DreadBong0
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