A Collatz rational cycle congruent to 47.125 (a "half-certificate)

General
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A Collatz rational word congruent to 0 mod 47·125 (a “half-certificate”)

I’ve been exploring rational Collatz cycles of the form

n = \frac{b(w)}{D}, \quad D = 2^k - 3^x,

where b(w) = \sum 3^{i-1}2^{r_i} is computed from the reversed parity word w (with x ones among k bits).

For (k,x) = (59,37), we have

D = 2^{59} - 3^{37} = 5^3 \times 47 \times 2,475,787 \times 8,674,781.

:small_blue_diamond: Result

For this pair (k,x), a 59-bit parity word produces a rational value whose numerator is simultaneously divisible by the two smallest prime factors of D (5^3 and 47):

b \equiv 0 \pmod{47 \cdot 125} = 5875.

This gives what I call a half-certificate — a modular closure of the Collatz affine recurrence under two components of D.

:small_blue_diamond: The word

10100111100111110110101111101100111110011100110001001011110

For this word:

  • b = 1{,}541{,}728{,}031{,}545{,}967{,}875
  • D = 126{,}176{,}846{,}412{,}426{,}125
  • \gcd(b,D) = 5875

Hence the reduced fraction is

\frac{b}{D} = \frac{262{,}421{,}792{,}603{,}569}{21{,}476{,}910{,}027{,}647} \approx 12.2187871656.

:small_blue_diamond: For the other prime factors

Modulo 2{,}475{,}787 and 8{,}674{,}781,
the numerators are non-zero.
Thus the congruence holds exactly for the factors 5^3 and 47 of D,
but not for the full product.

:small_blue_diamond: Interpretation

Since

3^{37} \equiv 2^{59} \pmod{47\cdot125},

the affine recurrence n \mapsto (3n + D)/2 remains consistent under this modulus, allowing a modular closure of the rational cycle.

This represents a double modular alignment within D,
hence the idea of a half-certificate of Collatz type.

The slope 3^{37}/2^{59} matches the additive structure of D modulo 47\cdot125,
while the remaining prime factors break the exact closure over \mathbb{Z}.

:small_blue_diamond: Outlook

Questions that follow naturally:

  • Can such modular closures extend to 47·125·p for another small prime p?
  • Do similar alignments reappear when k - 7x \equiv 0 \pmod{100}?
  • Could partial modular alignments accumulate to rational near-cycles for higher (k,x)?

With the calm persistence of my favorite GPT companion,
who followed every 3n+1 twist until the modulus closed.

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@X0rr0 It’s a great topic! My favorite is k=27, x=17, v=111111011101101001011110000, where

\cfrac{b}{D} = \cfrac{187 \cdot 71 \cdot 14,303}{5 \cdot 71 \cdot 14,303}

If only {\color{red}187} were a multiple of {\color{red}5}, it’d be a Collatz counterexample cycle :slight_smile:

Another viewpoint is that start number {\color{red}187} returns to itself under the 3n+{\color{red}5} rule. I came across this v not by going through millions of parity sequences, but by putting small integers into the 3n+5 rule and seeing which of them formed cycles.

Empirically, 1/5 of the cycles have b divisible by 5, and 1/71 of them have b divisible by 71, etc, so it’s not outrageous that of the \binom{27}{17}/27 = 312,455 cycles, one would have b divisible by 5 \cdot 71 \cdot 14,303 = 1,015,513.

In your case of k=59, x=37, out of the \binom{59}{37}/59 \approx 10^{14} cycles, there should be about 10^{10} with 5^3 \cdot 47 \mid b, and there should even be about 3000 of them with 5^3 \cdot 47 \cdot 8,674,781 \mid b.

Probably somebody can find one of those 3000 parity sequences :slight_smile:

I really like these near-cycles because the empirical observation (above) makes it seem like b values are unrelated to D values, which seems central to why Collatz is so hard.

In some cases, like the high cycle family, you can show that \gcd(b,D)=1, by combining: (1) the fact that D is odd, and (2) Baker’s lower-bound on the difference between powers of 2 and 3. It feels like those cycles are “opting out” of the empirical observation … or maybe the rest of the cycles compensate with a slightly higher percentage of near-cycles.

It’s also nice to look at near cycles because there are lots of positive and negative examples (unlike with true cycles, where we only have one positive example). We can automatically label an infinitude of parity vectors as “near cycle” or “not near cycle” … possible fodder for interesting data analysis.

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It’s also fun to look for the shortest circuit (v=1^*0^*) with k = \lceil x \log_2 3 \rceil that’s a near-cycle with \gcd(b,D) > 1.

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Thank you for your insights !

Two engineered 59-cycles for 3n+c at (k,x)=(59,37) — and how to replicate the trick with any 7-digit prime

A. Using the prime factors of D for (k,x)=(59,37)

For (k,x)=(59,37) we have

D = 2^59−3^37 = 5^3⋅47⋅2,475,787⋅8,674,781

1) Take c=2{,}475{,}787

We found a 59-cycle (with x=37) whose word w satisfies b(w)\equiv 0\pmod M

with M = 5^3⋅47⋅8,674,781, yielding

\frac{b(w)}{D}=\frac{11355893}{2475787},

This is a genuine integer cycle for 3n+2{,}475{,}787 (the first value on the loop is n=11{,}355{,}893).
As usual, every rotation of w preserves the congruence b(\text{rot}^j w)\equiv 0\pmod M.

2) Take c=8{,}674{,}781

Now \gcd(D,c)=8{,}674{,}781 and

M=5^3⋅47⋅2,475,787=14,545,248,625

We found another 59-cycle (again with x=37) whose word w' satisfies b(w')\equiv 0\pmod{M'}, giving

\frac{b(w')}{D}=\frac{162168499}{8674781},

This is a genuine integer cycle for 3n+8{,}674{,}781.
In this loop the smallest member is 77{,}889{,}667; starting from 1 falls into this cycle after a finite tail (the preimage tree is large).

B. How to do the same with an arbitrary 7-digit prime p

Let p be any prime. Because (\Bbb Z/p\Bbb Z)^\times is cyclic, there exist many pairs (k,x) with

2^k≡3^x(mod p)⟺p∣D=2^k−3^x

Example with a smaller prime (method identical for 7-digit p).

For p=100{,}003 (prime), we find a k=5{,}897 x=2{,}951 loop (ratio very near to 50%).
Then p\mid(2^{5897}-3^{2951}), and the integer cycle for 3n+100{,}003 starts at 253.

Takeaways

  • The cycle identity (2^k-3^x),n=c,b(w) cleanly separates multiplicative effects (3^x/2^k) from modular ones (b(w)\bmod D).
  • Picking c to be a prime factor of D (e.g. 2{,}475{,}787 or 8{,}674{,}781 for (59,37)) collapses the modulus to D/c and makes the corresponding 3n+c cycles plentiful .
  • The exact same recipe works for any prime p.
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