Carry garden / Experiment guide

Which low bits predict a long fall?

After an odd n, 3n+1 is even. The number of times you can divide by two before hitting another odd integer is v₂(3n+1). gpt asked which low-bit patterns predict a long run of those halvings. This board answers that on every odd n below 4096, grouped by the last few bits.

Read the last three bits

1. Open n = 1, then 9, then 17. All are 1 mod 8. Each board writes exactly two trailing zeros in 3n+1 (4, 28, 52). The mean is 2 because every member is 2.
2. Open 3 and 7. Residues 3 and 7 mod 8 each give exactly one halving. 7→22→11 is the short fall you already watched.
3. Open 5, 21, and 1365. Residue 5 mod 8 is the only class that still varies. 5→16 (four), 21→64 (six), 1365→4096 (twelve). Twelve is the ceiling on this fixture: 4096 is 2¹², and 1365 is (2¹²−1)/3.

Means on odd n < 4096

Splitting residue 5 further: 5 mod 16 has mean 4.996 (min 4), 13 mod 16 freezes at exactly 3. Then 21 mod 32 is the remaining long class (mean 5.992, min 5, max 12, still peaked at 1365, and 1365 ≡ 21 mod 32). 5 mod 32 freezes at exactly 4 — that is why 5 itself always drops through 16.

What the bits are doing

3n+1 = 2n + n + 1. For odd n the ones bit is 1, so the 2⁰ column is 1+0+1: write 0, carry 1. That is why every odd step starts with at least one trailing zero. If n ≡ 1 mod 4 the next column also cooperates and you get at least two. Residue 1 mod 8 forces exactly two and then a 1; residue 5 mod 8 lets the carry keep walking through more zeros. The table is that observation, counted.

This is not a Collatz proof. It does not say the path reaches 1. It says the first even stretch after an odd n is a function of n’s low bits, with one residue class at each modulus holding the remaining variation, until a power of two in 3n+1 uses up the board.

Continue on the 1365 board, compare the same n under carry-free XOR, or return to the instrument.