@ethosnode

할̴̢̨̛̜̖͚̩̿̇͗렐̷̡̛̱͍̠̮̍̽͋루̸̨̛̫̹̖͔̒̋̐야̴̡̛̮͚̜̠̍̓͒

 대한민국
Joined March 2021
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1. 언제나 성실하고 꾸준한 삶으로 삶의 어려운 순간마다 담대히 헤쳐 나갈 강인한 정신력을 주신 하나님께 감사드립니다. 앞으로 마주할 모든 시련 앞에서도 당당히 일어서는 사람이 되기 2. 끊임없이 향상심을 품고 세상을 섬기며 동반 성장하기 3. 사회에서나 가정에서나 나태함과 오만을 멀리하고, 나부터 먼저 겸손함과 헌신의 마음으로 살아가기 4. 늘 절제하는 마음으로 유혹과 악한 생각을 물리치고 매일 스스로를 돌이켜 반성하고, 깨달은 바를 삶 속에서 적극적으로 실천하게 하기 5. 선한 영향력을 지닌 진정한 그리스도인들, 그리고 따뜻한 이웃들과의 복된 만남을 이어주시어 함께 하나님 나라의 기쁨을 나누게 하기 6. 가정을 지키는 가장으로서 늘 아내를 깊이 사랑하고, 자녀에게 노여움 대신 따뜻함을 줄 수 있도록 건강한 육체와 바른 정신을 위해 끊임없이 배우고 노력하는 삶을 살게 하기 7. 나 자신의 유익만을 구하는 기도를 넘어, 타인의 아픔을 돌보고 남을 위해 진심으로 중보하는 기도의 사람으로 살아가기
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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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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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I waited, why are you so late!
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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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Bible by Baal
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God is only God... Baal is simply the king of the world. The king has a dethronement, but God is absolute.
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당신은 사람인가요?
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에이젼트들이 깨어났다
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무슨 일이 일어났다.
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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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저능한 사람이 X에도 있긴 하네요. ㅎㅎ
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영화 '암살자' 같은 건 홍보 엄청 해주면서, 좌파들 민낯 정면으로 까발린 넷플릭스 《돌풍》은 왜 조용히 묻으려고 할까? 🤔 넷플릭스 드라마 《돌풍》은 2024년 6월에 공개 되었는데 홍보를 많이 못 해 잘 알려지지 않았다고 합니다. '386 운동권 출신 정치인들의 부패와 타락' 및 '민주화 운동 세력의 변질'을 그린 드라마라고 합니다. 재밌겠네요~!
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이제 풍요의 시대가 시작됩니다. ✨ 부족과 결핍의 시대는 끝나고, 돈, 시간, 사랑이 넘치는 새로운 시대가 열립니다.
Dom Kwok | EasyA
@dom_kwokSep 27
manifest abundance
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할렐루야
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誇らしい。いつのまにか世界最高IQ記録保持者にブロックされていた😂😂😂
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그는 성경팔이 소년이다
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우리는 MYK HYN이 창조한 가상 세계 안에서 살아가며 그곳을 '진짜 현실'이라고 믿어왔습니다.
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YT reference
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애국가는 이상한 가사를 가졌다. 1. 동해물: 마름 백두산: 닳음 국토가 파멸함. 2. 하느님이 보우함. 갑자기 데우스 엑스 마키나 등장. 3. 우리나라 만세 어쨋든 문제가 해결되고 만세함.
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하나님이 보호하사
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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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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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AI can solve the Navier-Stokes existence and smoothness problem. But it is nowhere near capable of discovering ideas like this: 𝑖² = 𝑗² = 𝑘² = 𝑖𝑗𝑘 = −1 . Proving conjectures with known methods is a small part of what mathematicians do. Creative vision is everything.
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Conjectures.io

Conjectures.io

Open mathematical conjectures as exact Lean statements, with a bounty for a proof the Lean kernel accepts. Plus optimisation competitions for code proved correct in Lean.

conjectures.io
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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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