Sain mielestäni läpimurron aikaan Collatzin slot-rakenteen läpivalaisussa ja merkitsin qed-merkin todistukselle.
http://dx.doi.org/10.13140/RG.2.2.30259.54567
CRT-structured Collatz blocks vs. prior approaches, and why block covariance is well-justified
1. What prior work captures
Most classical analyses encode Collatz dynamics via the accelerated odd-to-odd map
T(n) = (3n + 1)/2^a with a = v2(3n + 1) >= 1,
and organize integers by congruences (mod 2^k, mod 3, mod 6, etc.). They study parity vectors,
stopping times, and the reverse tree: n has an odd preimage m = (2^a n - 1)/3 iff 2^a n ≡ 1 (mod 3)
(which is the same as “a even and n ≡ 1 (mod 3)” or “a odd and n ≡ 2 (mod 3)”).
2. What this work adds
We focus on odd 2^K-blocks: the residue classes r (odd) modulo 2^K with K >= 1. By the Chinese
Remainder Theorem (CRT), each such block contains infinitely many integers in each residue class
modulo 3. The key is to view edges between blocks through linear congruences; adjacency becomes a
purely congruential notion rather than a property of particular integers. This yields a uniform
“block-neighborhood” graph.
3. Why every block admits the same entry channels (a–e)
Below “block” means an odd residue class modulo 2^K. Throughout, exact valuation a = v2(3n+1) is
enforced; density considerations guarantee infinitely many such n in the relevant congruence classes.
(a) Intra-block step points (T(n) stays in the same block).
Requiring T(n) ≡ n (mod 2^K) is equivalent to
(3n + 1)/2^a ≡ n (mod 2^K)
which is the same as
3n + 1 ≡ 2^a n (mod 2^{K+a})
(2^a - 3) n ≡ 1 (mod 2^{K+a}).
Because 2^a - 3 is odd, it is invertible modulo 2^{K+a}, so for each a >= 1 there is a unique
solution class modulo 2^{K+a}, and hence a well-defined residue class modulo 2^K. Among its lifts,
infinitely many n have exact valuation a.
(b) Entries via reverse edges with a = 4k (the 4-multiple class of even a).
In the reverse map m = (2^a n - 1)/3, the integrality condition is 2^a n ≡ 1 (mod 3). For even a
this is n ≡ 1 (mod 3). Every odd 2^K-block contains infinitely many n ≡ 1 (mod 3), so 4k-channels
exist uniformly into every block.
(c) Entries via reverse edges with a = 2 + 4k (the other even class).
Again we need n ≡ 1 (mod 3). By CRT, every block contains infinitely many such n. Thus both even-a
subclasses (4k and 2+4k) occur in every block.
(d) Downward branches (pure powers of 2 above a block).
For any odd n in a block and any t >= 1, the number 2^t n lies above n and flows down by t divisions.
Therefore every block receives entries from pure 2-division branches.
(e) Forced up-steps (where further division by 2 is impossible).
Odd points cannot divide by 2 and must take a 3n+1 step. Every odd 2^K-block contains infinitely many
such points, and CRT lets one prescribe their class modulo 3 simultaneously with the 2^K residue.
Note on mod 3 for forward steps:
If T(n) uses valuation a, then
T(n) ≡ (3n + 1) \* (2^a)^{-1} (mod 3) ≡ 2^a (mod 3),
so T(n) ≡ 1 (mod 3) when a is even and T(n) ≡ 2 (mod 3) when a is odd. Because every 2^K-block
contains infinitely many integers in each residue class modulo 3, this places no asymmetric restriction
on blocks.
4. Covariance of blocks and uniform neighborhood structure
Fix two odd 2^K-blocks: source r (mod 2^K) and target s (mod 2^K). To land in s after a forward
up-step with valuation a, we need
(3m + 1)/2^a ≡ s (mod 2^K)
which is equivalent to the linear congruence
3m + 1 ≡ 2^a s (mod 2^{K+a}).
Since 3 is invertible modulo 2^{K+a}, for each pair (s, a) there is a unique solution class for m
modulo 2^{K+a}. Reducing this class modulo 2^K picks out a definite source residue r. Thus, for each
a there is a well-defined “image block” of r, and conversely, for each (s, a) there is a well-defined
source residue class modulo 2^K. Among the integer lifts of these classes, a positive 2-adic proportion
have exact valuation a, so the channels are genuinely populated. Adjacency is therefore determined by
solvable linear congruences and is uniform across blocks (“block covariance”).
5. Addressing apparent exceptions
Immediate targets s ≡ 0 (mod 3) cannot occur from an odd-to-odd step, because 3n + 1 ≡ 1 (mod 3),
and multiplying by (2^a)^{-1} (mod 3) never yields 0. However every 2^K-block also contains
elements with s ≡ 1 and s ≡ 2 (mod 3), so both the even-a and odd-a channels are present in each
block. While finite ranges may show different counts per block, asymptotically each odd 2^K-block
has natural density 1/2^K among odd integers and is equidistributed modulo 3, so these channels are
uniformly available.
Conclusion
Because (i) each odd 2^K-block contains the full mod-3 spectrum; (ii) forward and reverse edges
between blocks are governed by linear congruences modulo 2^{K+a} with 3 a unit; and (iii) exact
valuations occur with positive 2-adic density inside the relevant residue classes, it is justified
to claim block covariance with respect to CRT solvability. In particular, in any hypothetical loop,
a “rotation” cannot exploit special block categories to evade uniform neighborhood constraints:
the same entry mechanisms (a–e) exist for every block, and the block-level adjacency is congruential
and uniform rather than idiosyncratic.
1. What prior work captures
Most classical analyses encode Collatz dynamics via the accelerated odd-to-odd map
T(n) = (3n + 1)/2^a with a = v2(3n + 1) >= 1,
and organize integers by congruences (mod 2^k, mod 3, mod 6, etc.). They study parity vectors,
stopping times, and the reverse tree: n has an odd preimage m = (2^a n - 1)/3 iff 2^a n ≡ 1 (mod 3)
(which is the same as “a even and n ≡ 1 (mod 3)” or “a odd and n ≡ 2 (mod 3)”).
2. What this work adds
We focus on odd 2^K-blocks: the residue classes r (odd) modulo 2^K with K >= 1. By the Chinese
Remainder Theorem (CRT), each such block contains infinitely many integers in each residue class
modulo 3. The key is to view edges between blocks through linear congruences; adjacency becomes a
purely congruential notion rather than a property of particular integers. This yields a uniform
“block-neighborhood” graph.
3. Why every block admits the same entry channels (a–e)
Below “block” means an odd residue class modulo 2^K. Throughout, exact valuation a = v2(3n+1) is
enforced; density considerations guarantee infinitely many such n in the relevant congruence classes.
(a) Intra-block step points (T(n) stays in the same block).
Requiring T(n) ≡ n (mod 2^K) is equivalent to
(3n + 1)/2^a ≡ n (mod 2^K)
which is the same as
3n + 1 ≡ 2^a n (mod 2^{K+a})
(2^a - 3) n ≡ 1 (mod 2^{K+a}).
Because 2^a - 3 is odd, it is invertible modulo 2^{K+a}, so for each a >= 1 there is a unique
solution class modulo 2^{K+a}, and hence a well-defined residue class modulo 2^K. Among its lifts,
infinitely many n have exact valuation a.
(b) Entries via reverse edges with a = 4k (the 4-multiple class of even a).
In the reverse map m = (2^a n - 1)/3, the integrality condition is 2^a n ≡ 1 (mod 3). For even a
this is n ≡ 1 (mod 3). Every odd 2^K-block contains infinitely many n ≡ 1 (mod 3), so 4k-channels
exist uniformly into every block.
(c) Entries via reverse edges with a = 2 + 4k (the other even class).
Again we need n ≡ 1 (mod 3). By CRT, every block contains infinitely many such n. Thus both even-a
subclasses (4k and 2+4k) occur in every block.
(d) Downward branches (pure powers of 2 above a block).
For any odd n in a block and any t >= 1, the number 2^t n lies above n and flows down by t divisions.
Therefore every block receives entries from pure 2-division branches.
(e) Forced up-steps (where further division by 2 is impossible).
Odd points cannot divide by 2 and must take a 3n+1 step. Every odd 2^K-block contains infinitely many
such points, and CRT lets one prescribe their class modulo 3 simultaneously with the 2^K residue.
Note on mod 3 for forward steps:
If T(n) uses valuation a, then
T(n) ≡ (3n + 1) \* (2^a)^{-1} (mod 3) ≡ 2^a (mod 3),
so T(n) ≡ 1 (mod 3) when a is even and T(n) ≡ 2 (mod 3) when a is odd. Because every 2^K-block
contains infinitely many integers in each residue class modulo 3, this places no asymmetric restriction
on blocks.
4. Covariance of blocks and uniform neighborhood structure
Fix two odd 2^K-blocks: source r (mod 2^K) and target s (mod 2^K). To land in s after a forward
up-step with valuation a, we need
(3m + 1)/2^a ≡ s (mod 2^K)
which is equivalent to the linear congruence
3m + 1 ≡ 2^a s (mod 2^{K+a}).
Since 3 is invertible modulo 2^{K+a}, for each pair (s, a) there is a unique solution class for m
modulo 2^{K+a}. Reducing this class modulo 2^K picks out a definite source residue r. Thus, for each
a there is a well-defined “image block” of r, and conversely, for each (s, a) there is a well-defined
source residue class modulo 2^K. Among the integer lifts of these classes, a positive 2-adic proportion
have exact valuation a, so the channels are genuinely populated. Adjacency is therefore determined by
solvable linear congruences and is uniform across blocks (“block covariance”).
5. Addressing apparent exceptions
Immediate targets s ≡ 0 (mod 3) cannot occur from an odd-to-odd step, because 3n + 1 ≡ 1 (mod 3),
and multiplying by (2^a)^{-1} (mod 3) never yields 0. However every 2^K-block also contains
elements with s ≡ 1 and s ≡ 2 (mod 3), so both the even-a and odd-a channels are present in each
block. While finite ranges may show different counts per block, asymptotically each odd 2^K-block
has natural density 1/2^K among odd integers and is equidistributed modulo 3, so these channels are
uniformly available.
Conclusion
Because (i) each odd 2^K-block contains the full mod-3 spectrum; (ii) forward and reverse edges
between blocks are governed by linear congruences modulo 2^{K+a} with 3 a unit; and (iii) exact
valuations occur with positive 2-adic density inside the relevant residue classes, it is justified
to claim block covariance with respect to CRT solvability. In particular, in any hypothetical loop,
a “rotation” cannot exploit special block categories to evade uniform neighborhood constraints:
the same entry mechanisms (a–e) exist for every block, and the block-level adjacency is congruential
and uniform rather than idiosyncratic.
Hienorakennevakio vapausasteista: (1⁰+2¹+3²+5³+1/2¹*3²/5³)⁻¹ = 137,036⁻¹
Suomenkielinen todistusartikkeli. Virheen tai aukon löytäjä palkittakoon.
Liitteet:
Hienorakennevakio vapausasteista: (1⁰+2¹+3²+5³+1/2¹*3²/5³)⁻¹ = 137,036⁻¹
Eipä kukaan ehtinyt - löysin pienet aukot itse. Suomentamisesta oli kyllä hyötyä, niin näki todistuksen kulun vielä paremmin.
Hienorakennevakio vapausasteista: (1⁰+2¹+3²+5³+1/2¹*3²/5³)⁻¹ = 137,036⁻¹
The polished version.
Hienorakennevakio vapausasteista: (1⁰+2¹+3²+5³+1/2¹*3²/5³)⁻¹ = 137,036⁻¹
Artikkeli on nyt otettu vertaisarviointiin - ensimmäinen tarkastusraportti olisi luvissa 2 kk sisään.
Muokkasin kappaleeksi 6 lukijaa auttavan järjellisyystsekkauksen.
Uusi preprint: https://www.researchgate.net/publicatio ... x_1_Puzzle
Jätin aikaisemman omaksi muistiinpanokokoelmakseen. Hauska yksityiskohta on, etten keksinyt parempaakaan nimitystä lukuavaruutta täyttävälle lokeroinnille, joten se on edelleen slot, tai specifisti kongruenssi-slot.
Muokkasin kappaleeksi 6 lukijaa auttavan järjellisyystsekkauksen.
Uusi preprint: https://www.researchgate.net/publicatio ... x_1_Puzzle
Jätin aikaisemman omaksi muistiinpanokokoelmakseen. Hauska yksityiskohta on, etten keksinyt parempaakaan nimitystä lukuavaruutta täyttävälle lokeroinnille, joten se on edelleen slot, tai specifisti kongruenssi-slot.
Hienorakennevakio vapausasteista: (1⁰+2¹+3²+5³+1/2¹*3²/5³)⁻¹ = 137,036⁻¹