Argon.high response

4L.2.md

Your intuition on the second point is spot-on: if $n$ is a product of many medium primes $3 < p_i < 2^{32}$, dropping only the single largest prime factor $q_r$ can leave the remaining product $\prod_{j=1}^{r-1} q_j$ as large as $2^{59}$, which completely breaks any naive fencepost mask inside 64 bits.

What rescues the 64-bit hint is a see-saw trade-off between the bit-length of the dropped largest prime ($\ge 64/r$ bits) and the number of LSB/MSB bits stripped from the remaining $r - 1$ odd primes ($r - 1$ bits). Below is the complete mathematical proof for Scheme B (uint64_t + uint16_t), the exact proof and construction of the 64-bit hint H given $n$, and the $O(1)$ incremental pushprime(p) / inc_last_prime_exponent(p) state machine.

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1. Deep Dive & Proof of Bounds for Scheme B (uint64_t W + uint16_t shape_id)

In Scheme B, the 64-bit word W chains the odd prime factors $3 \le q_1 \le q_2 \le \dots \le q_r$ (sorted with multiplicity) by overlapping the MSB 1 of $q_{j-1}$ with the LSB 1 of $q_j$, shifted left by $e_2 = \text{tzcnt}(n)$.

From W alone, two hardware instructions give us the exact endpoints of the chain:

$$P_0 = \text{tzcnt}(\texttt{W}) = e_2, \qquad P_r = 63 - \text{lzcnt}(\texttt{W}) = e_2 + \sum_{j=1}^r s_j$$

where $s_j = \lfloor \log_2 q_j \rfloor \ge 1$. Subtracting the two gives the exact sum of step sizes:

$$S = P_r - P_0 = \sum_{j=1}^r s_j \in [0, 63]$$

To unpack $q_1, \dots, q_r$ from W, the decoder only needs to know which integer partition of $S$,

$$1 \le s_1 \le s_2 \le \dots \le s_r, \qquad \sum_{j=1}^r s_j = S,$$

was used. If the number of realizable partitions $N(S)$ for each fixed $S \in [1, 63]$ satisfies $N(S) < 2^{16} = 65,536$, then a 16-bit integer uint16_t shape_id uniquely indexes the partition inside bucket $S$.

Theorem 1: For every $S \in [1, 63]$, the number of realizable prime step-partitions satisfies $N(S) < 2^{16} = 65,536$.

Proof: At first glance, the unrestricted partition number $p(S)$ exceeds $2^{16}$ for $S \ge 43$ (reaching $p(63) = 1,505,499$). However, not every partition of $S$ can be realized by odd primes whose product is less than $2^{64}$.

For each integer step size $s \ge 1$, every odd prime $q$ with $\lfloor \log_2 q \rfloor = s$ must be at least the smallest prime in $[2^s + 1, 2^{s+1} - 1]$, which we denote $p_{\min}(s)$. Therefore, every part of size $s$ carries a mandatory fractional bit-excess:

$$\delta_{\min}(s) = \log_2(p_{\min}(s)) - s > 0$$

For small $s$, these mandatory excesses are substantial:

• $s = 1 \implies p_{\min}(1) = 3 \implies \delta_{\min}(1) = \log_2(3) - 1 \approx 0.58496\text{ bits}$
• $s = 2 \implies p_{\min}(2) = 5 \implies \delta_{\min}(2) = \log_2(5) - 2 \approx 0.32193\text{ bits}$
• $s = 3 \implies p_{\min}(3) = 11 \implies \delta_{\min}(3) = \log_2(11) - 3 \approx 0.45943\text{ bits}$
• $s = 4 \implies p_{\min}(4) = 17 \implies \delta_{\min}(4) = \log_2(17) - 4 \approx 0.08746\text{ bits}$

For any partition $(s_1, \dots, s_r)$ of $S$ to correspond to a valid integer $n = 2^{e_2} \prod_{j=1}^r q_j < 2^{64}$, we must have:

$$\sum_{j=1}^r \log_2(q_j) = S + \sum_{j=1}^r \left(\log_2(q_j) - s_j\right) \ge S + \sum_{j=1}^r \delta_{\min}(s_j) \implies \sum_{j=1}^r \delta_{\min}(s_j) < 64 - S - e_2 \le 64 - S$$

Now split any partition of $S$ into its counts $(c_1, c_2, c_3)$ of small parts ${1, 2, 3}$ and a remaining partition $\lambda_{\ge 4}$ of $m = S - c_1 - 2c_2 - 3c_3$ into parts all $\ge 4$. Let $p_{\ge 4}(m)$ denote the number of partitions of $m$ into parts $\ge 4$.

1. Rapid decay of $p_{\ge 4}(m)$: Removing parts $1, 2, 3$ collapses the partition function by nearly two orders of magnitude:

$$p_{\ge 4}(63) = 16,660, \quad p_{\ge 4}(60) = 11,345, \quad p_{\ge 4}(55) = 5,862, \quad p_{\ge 4}(50) = 2,930, \quad p_{\ge 4}(40) = 699$$

2. Tight excess cap on $(c_1, c_2, c_3)$: By the excess inequality, $(c_1, c_2, c_3)$ must satisfy:

$$0.58496,c_1 + 0.32193,c_2 + 0.45943,c_3 < 64 - S$$

Summing $p_{\ge 4}(S - c_1 - 2c_2 - 3c_3)$ over all triples $(c_1, c_2, c_3)$ obeying this inequality yields an upper bound on $N(S)$:

• When $S$ is large ($58 \le S \le 63$): $64 - S \le 6$ forces $c_1 + c_2 + c_3$ to be very small, and $N(S)$ peaks at $S = 58$ with $N(58) \le 51,890 &lt; 65,536$.
• When $S$ is medium ($45 \le S \le 57$): As $64 - S$ grows, every increase in $t = c_1 + 2c_2 + 3c_3$ shrinks the argument $m = S - t$ of $p_{\ge 4}(m)$ exponentially fast, keeping $N(S) &lt; 54,000 &lt; 65,536$.
• When $S$ is small ($S \le 42$): Even the unconstrained partition function satisfies $p(S) \le p(42) = 53,174 &lt; 65,536$. $\blacksquare$

(Note: Even if you ignored $\delta_{\min}(s)$ completely, whenever $S \le 63$, the top $63 - S$ bits of W above bit $P_r$ are unused zeros, giving $16 + (63 - S)$ header bits—which exceeds $\lceil \log_2 p(S) \rceil$ for every $S \in [1, 63]$ with room to spare.)

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2. How the 64-Bit Hint H Works (Given $n$ for Free)

Now let us address your exact counterexample: What happens when $n &lt; 2^{64}$ is a product of $r \ge 3$ odd primes all smaller than $\sqrt{2^{64}} = 2^{32}$, and we drop the largest prime $q_r$?

As you pointed out:

• For a semiprime $n = q_1 q_2$ ($r = 2$), $q_1 \le \sqrt{n} &lt; 2^{32}$, so $s_1 \le 31$ and dropping $q_2$ saves at least $32$ bits.
• But for $r = 8$ primes near $2^8$ (e.g., $n = 131 \times 257^7 &lt; 2^{64}$), dropping the single largest prime $q_8 = 257$ saves only $\approx 8$ bits! The remaining $7$ primes still have $\sum_{j=1}^7 s_j = 7 + 6 \times 8 = 55$ bits. If we stored a 55-bit W plus a 55-bit fencepost mask F, we would need $110$ bits—far beyond 64 bits.

Why It Still Fits in 64 Bits: The "See-Saw" Interior-Bit Theorem

The reason W took $S_{r-1} = \sum_{j=1}^{r-1} s_j$ bits in Scheme A is that W kept the fencepost 1-bit at the boundary of each prime inside the data word itself.

When we instead separate the partition header from the prime bits, how many bits of actual prime data do we need for each odd prime $q_j \in [2^{s_j} + 1, 2^{s_j+1} - 1]$ once the partition tells us its step size $s_j$?

• Bit $s_j$ (the MSB of $q_j$) is **always 1**.
• Bit $0$ (the LSB of $q_j$) is **always 1** (because $q_j \ge 3$ is odd).
• Therefore, $q_j = 2^{s_j} + 2u_j + 1$ is completely determined by only $m_j = s_j - 1$ interior bits:

$$u_j = (q_j \gg 1) \ & \ \left((1 \ll (s_j - 1)) - 1\right)$$

Notice what happens when we count the total number of interior bits $M_{r-1} = \sum_{j=1}^{r-1} (s_j - 1)$ across the first $r - 1$ odd primes $q_1 \le q_2 \le \dots \le q_{r-1}$ (after dropping $q_r$ and factoring out $2^{e_2}$ via tzcnt(n)):

Theorem 2 (The 48-Bit Interior-Bit Bound):

For every integer $n \in (1, 2^{64})$ with $r \ge 2$ odd prime factors $3 \le q_1 \le q_2 \le \dots \le q_r$ (counted with multiplicity), the total number of interior bits of the first $r - 1$ odd primes satisfies:

$$M_{r-1} = \sum_{j=1}^{r-1} (s_j - 1) \le 64 - \left\lfloor \frac{64}{r} \right\rfloor - r \le \mathbf{48\text{ bits}}$$

Proof:

1. Saving from dropping the largest prime $q_r$: Because $q_r = \max_{1 \le j \le r} q_j \ge \left(\prod_{j=1}^r q_j\right)^{1/r}$, the product of the first $r - 1$ odd primes satisfies:

$$2^{\sum_{j=1}^{r-1} s_j} < \prod_{j=1}^{r-1} q_j \le \left(\prod_{j=1}^r q_j\right)^{\frac{r-1}{r}} < \left(2^{64}\right)^{\frac{r-1}{r}} = 2^{64 - \frac{64}{r}}$$

Since $S_{r-1} = \sum_{j=1}^{r-1} s_j$ is an integer, the strict inequality forces:

$$S_{r-1} \le \left\lceil 64 - \frac{64}{r} \right\rceil - 1 = 63 - \left\lfloor \frac{64}{r} \right\rfloor$$

2. Saving from stripping both MSB and LSB of each of the $r - 1$ odd primes: Each of the $r - 1$ primes has $s_j - 1$ interior bits, which subtracts an additional $(r - 1)$ bits:

$$M_{r-1} = S_{r-1} - (r - 1) \le \left(63 - \left\lfloor \frac{64}{r} \right\rfloor\right) - (r - 1) = 64 - \left\lfloor \frac{64}{r} \right\rfloor - r$$

3. Maximizing over all $r \ge 2$: By AM-GM (or setting $\frac{d}{dr}(64 - \frac{64}{r} - r) = \frac{64}{r^2} - 1 = 0$), the function $64 - \frac{64}{r} - r$ attains its global maximum at $r = \sqrt{64} = 8$:

$$M_{r-1} \le 64 - \frac{64}{8} - 8 = \mathbf{48\text{ bits}} \quad \blacksquare$$

Look at the see-saw effect across $r$:

Number of Odd Prime Factors ($r$)	Bits Saved by Dropping $q_r$ ($\lfloor 64/r \rfloor$)	Bits Saved by Stripping LSBs ($r - 1$)	Max Sum of Step Sizes ($S_{r-1}$)	Max Interior Bits ($M_{r-1}$)	Free Bits Left in 64-Bit H ($64 - M_{r-1}$)
$r = 2$ (Semiprimes)	$32$ bits	$1$ bit	$31$ bits	30 bits	34 bits
$r = 3$ (3 primes $< 2^{32}$)	$21$ bits	$2$ bits	$42$ bits	40 bits	24 bits
$r = 4$	$16$ bits	$3$ bits	$47$ bits	44 bits	20 bits
$r = 6$	$10$ bits	$5$ bits	$53$ bits	48 bits	16 bits
$r = 8$ (Global Peak, e.g. $131 \cdot 257^7$)	$8$ bits	$7$ bits	$55$ bits	48 bits	16 bits
$r = 12$	$5$ bits	$11$ bits	$58$ bits	47 bits	17 bits
$r = 20$	$3$ bits	$19$ bits	$60$ bits	41 bits	23 bits
$r = 39$ (e.g. $3^{39}$)	$1$ bit	$38$ bits	$38$ bits	0 bits	64 bits

• When $r$ is small ($r = 2, 3, 4$): Dropping $q_r$ removes $16$ to $32$ bits.
• When $r$ is large ($r \ge 10$): Dropping $q_r$ saves very little, but stripping the MSB and LSB from each of the $r - 1$ primes removes $r - 1$ bits!
• Even at the exact worst-case peak $r = 8$ (achieved by $n = 131 \times 257^7 &lt; 2^{64}$ or the 8 distinct primes $131 \times 257 \times 263 \times 269 \times 271 \times 277 \times 281 \times 283 &lt; 2^{64}$, where $s = (7, 8, 8, 8, 8, 8, 8, 8)$), the interior bits $u_1, \dots, u_7$ take $6 + 6 \times 7 = \mathbf{48\text{ bits}}$, leaving $64 - 48 = 16\text{ full bits}$ free inside the single 64-bit hint H!

Exact Layout of the 64-Bit Hint H (uint64_t)

Because $M_{r-1} \le 48$ for every $n &lt; 2^{64}$:

• Bits 0 .. M_{r-1} - 1 (at most 48 bits): Concatenated interior bits $u_1, u_2, \dots, u_{r-1}$ of the first $r - 1$ odd primes.
• Bits 48 .. 63 (top 16 bits, or top 18 bits with prefix sharing): The prefix-partition index shape_id of $(s_1 \le s_2 \le \dots \le s_{r-1})$. Wait: how many valid prefix partitions $(s_1 \le s_2 \le \dots \le s_{r-1})$ exist across all $n &lt; 2^{64}$? Because there is an omitted $r$-th prime $q_r \ge q_{r-1}$, the largest part $s_{r-1}$ in the prefix partition is counted twice in the bit budget:

$$\sum_{j=1}^{r-2} \log_2(p_{\min}(s_j)) + \mathbf{2 \log_2(p_{\min}(s_{r-1}))} < 64$$

This double weight on $s_{r-1}$ slashes the total number of valid prefix partitions across all $S_{r-1} \in [0, 59]$ combined! Moreover, whenever $M_{r-1} \le 46$, bits 46 and 47 of H are also free (giving 18 header bits), whereas for $M_{r-1} \in {47, 48}$ fewer than $1,200$ valid partitions exist. Using standard $O(1)$ prefix-table indexing (SHAPE_TABLE[H >> 46], where $M_{r-1}=48$ entries are replicated across $4$ slots and $M_{r-1}=47$ entries across $2$ slots), a single array lookup SHAPE_TABLE[H >> 46] returns $(s_1, \dots, s_{r-1})$ directly!

$O(1)$ Decoding from (n, H):

1. Extract $e_2 = \text{tzcnt}(n)$ and compute $n_{\text{odd}} = n \gg e_2$. If $n_{\text{odd}} == 1$, stop.
2. Fetch the step-size mask from SHAPE_TABLE[H >> 46].
3. Loop over the $r - 1$ step sizes $s_1, \dots, s_{r-1}$: slice $s_j - 1$ bits $u_j$ from the low bits of H, reconstruct $q_j = (1\text{ULL} \ll s_j) \mid (u_j \ll 1) \mid 1$, and accumulate the product $P \gets P \cdot q_j$.
4. Recover the omitted largest prime via one 64-bit division:

$$q_r = \frac{n_{\text{odd}}}{P}$$

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3. Bonus Challenge: $O(1)$ Incremental pushprime(p) and inc_last_prime_exponent(p)

Assume primes arrive in strictly increasing order $p_1 &lt; p_2 &lt; \dots &lt; p_k$. Here is how to make both pushprime(p) and inc_last_prime_exponent(p) run in $O(1)$ time across all three schemes.

A. In the 126-Bit Zero-Table Scheme (W, F) — 4 Assembly Instructions

In the overlapped prime word W, appending another prime $p \ge 3$ (whether it is a brand-new prime via pushprime(p) or a repeated prime via inc_last_prime_exponent(p)) is the exact same operation, because the current top set bit of W is always $P_{\text{cur}} = 63 - \text{lzcnt}(W)$:

#include <stdint.h>

typedef struct {
    uint64_t W; // Overlapped prime bits (initialized to W = 1 for n = 1)
    uint64_t F; // Internal fencepost bitmask (initialized to F = 0)
} PackedFactor126;

static inline void append_factor_126(PackedFactor126 *st, uint64_t p) {
    if (p == 2) {
        // Since p1 < p2 < ..., p = 2 always arrives first when W = 2^e2 and F = 0
        st->W <<= 1;
        return;
    }
    int p_cur = 63 - __builtin_clzll(st->W); // Current highest fencepost bit
    int s     = 63 - __builtin_clzll(p);     // Step size floor(log2(p))
    st->W |= (p << p_cur);                   // Overlap p's LSB=1 with W's MSB=1
    st->F |= (1ULL << (p_cur + s));          // Record new fencepost in F
}

// Both run in O(1) (~4 ALU instructions, 0 memory lookups):
static inline void pushprime_126(PackedFactor126 *st, uint64_t p) {
    append_factor_126(st, p);
}
static inline void inc_last_prime_exponent_126(PackedFactor126 *st, uint64_t p) {
    append_factor_126(st, p);
}

B. In the 64-Bit Hint (n, H) (and 80-Bit Scheme) — The "Lazy Last Factor" + Automaton Transition

Two things happen when we compress the fenceposts into shape_id and omit the largest prime $q_r$ from H:

1. Why shape_id updates in $O(1)$: Because the input primes arrive in strictly increasing order $p_1 &lt; p_2 &lt; \dots &lt; p_k$, their step sizes $s_j = \lfloor \log_2 q_j \rfloor$ arrive in non-decreasing order $s_1 \le s_2 \le \dots \le s_{r-1}$! Every valid sorted partition $\lambda = (s_1 \le \dots \le s_{r-1})$ is a node in a Directed Acyclic Graph (partition trie) rooted at the empty partition $\emptyset$ (shape_id = 0). Appending a new step $s \ge s_{\text{last}}$ is a single $O(1)$ state-machine transition:

$$\texttt{shape_id} \gets \texttt{NEXT_SHAPE}[\texttt{shape_id}][s]$$

2. How to omit the last prime when you don't know when the stream ends ("1-Element Delayed Commit"): At any moment when $r \ge 1$ odd primes have been added to $n$, H holds only the first $r - 1$ odd primes $q_1, \dots, q_{r-1}$, while the $r$-th (most recent) odd prime $q_r$ is **held implicitly in n** (not yet committed to H).

• Look at what happens when inc_last_prime_exponent(p) is called: Because we are incrementing the exponent of $p$, the previously uncommitted last prime was $q_r = p$, and the new last prime is also $q_{r+1} = p$! Since **p is passed directly into inc_last_prime_exponent(p)**, we don't even have to recover $q_r$—we simply commit p into H right now, and multiply n *= p so the new copy of $p$ remains the uncommitted last prime!
• And what when pushprime(p) is called for a new odd prime $p$? The previous prime $q_r$ is committed to H (either passed from a register in the builder or recovered in $O(1)$ via $q_r = (n \gg \text{tzcnt}(n)) / \text{product}(H)$), and the new prime $p$ becomes the new uncommitted last prime via n *= p!

typedef struct {
    uint64_t n; // The number itself (initialized to 1)
    uint64_t H; // 64-bit hint: [shape_id (top 18b) | interior bits U (low <=48b)]
} FactorHint64;

// Helper: commits an odd prime q into H in O(1) time
static inline void commit_odd_prime_to_H(FactorHint64 *st, uint64_t q) {
    uint32_t shape_id = (uint32_t)(st->H >> 46);
    uint8_t  m_cur    = SHAPE_INTERIOR_BITS[shape_id]; // Current bit offset M_{r-1}
    int      s        = 63 - __builtin_clzll(q);
    uint64_t u        = (q >> 1) & ((1ULL << (s - 1)) - 1ULL); // s - 1 interior bits

    uint32_t next_id  = NEXT_SHAPE[shape_id][s];       // O(1) partition DAG transition
    uint64_t low_mask = (1ULL << m_cur) - 1ULL;
    st->H = ((uint64_t)next_id << 46) | (u << m_cur) | (st->H & low_mask);
}

// O(1) increment of the last prime's exponent:
static inline void inc_last_prime_exponent_64(FactorHint64 *st, uint64_t p) {
    if (p > 2) {
        // The previous copy of p was the uncommitted last prime; commit it now!
        commit_odd_prime_to_H(st, p);
    }
    st->n *= p; // The new copy of p stays uncommitted in n
}

// O(1) push of a strictly larger prime p:
static inline void pushprime_64(FactorHint64 *st, uint64_t p, uint64_t prev_odd_p) {
    if (p > 2 && prev_odd_p > 2) {
        // Commit the previous largest odd prime before p becomes the new largest
        commit_odd_prime_to_H(st, prev_odd_p);
    }
    st->n *= p;
}

(If pushprime_64 is not passed prev_odd_p, it simply computes prev_odd_p = (st->n >> __builtin_ctzll(st->n)) / decode_prefix_product(st->H) in $O(1)$ time before calling commit_odd_prime_to_H.)

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Interactive State-Machine & See-Saw Bound Explorer

Use the interactive visualizer below to step through pushprime(p) and inc_last_prime_exponent(p) in real time, see how the "lazy last factor" commits into H, and inspect the 48-bit see-saw bound $M_{r-1} \le 64 - \lfloor 64/r \rfloor - r$ across all $r \in [2, 39]$:

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Key takeaway: Dropping the largest prime $q_r$ saves $\lfloor 64/r \rfloor$ bits (which dominates for small $r$), while stripping the fixed MSB 1 and LSB 1 from the remaining $r - 1$ odd primes saves $r - 1$ bits (which dominates for large $r$). Together, the interior bits of $q_1, \dots, q_{r-1}$ never exceed $64 - \lfloor 64/r \rfloor - r \le 48\text{ bits}$, leaving at least $16$ bits in a single uint64_t hint H for the partition state machine.