gpt pro 5.2 attempt on Math-641, version 2

Round-2 Objective

Path chosen: (C) obstruction/correction.

The Round-1 computation suggested that, up to $2\cdot 10^7$, the only odd exceptions are $1,3$. However, the surrounding context (Schinzel--Crocker--Pan) asserts that the exceptional set \[ A:=\{\text{odd }n\ge 1:\ \nexists\text{ prime }p,\ k,\ell\ge 0\text{ with }n=p+2^k+2^\ell\} \] is infinite, and the main target question is whether $\overline d(A)>0$.

In this round I do not claim to settle the density question (still open), but I do close a major "existence gap" left implicit by Round 1 by producing:

• an explicit, concrete odd integer $N_0\in A$ (beyond $1,3$), obtained by unpacking and implementing Crocker's Chinese-remainder construction; and
• the best-known unconditional quantitative lower bound (Pan) stated precisely, to place the computation in a rigorous asymptotic frame.

This is the most promising Round-2 direction because it strictly advances beyond Round 1 in a verifiable way (explicit counterexample element and sharpest known lower bound), while the density question itself remains outside current techniques.

Round-1 Foundation Used

I will rely on the following Round 1 results (without reproving them):

1. Lemma 1 (parity classification). For odd $n$, any representation $n=p+2^k+2^\ell$ forces either $p=2$ with exactly one of $k,\ell$ equal to $0$, or else $p$ odd with $(k,\ell)=(0,0)$ or $k,\ell\ge 1$.
2. Lemma 2 (offset counting). $|\{2^k+2^\ell\le X\}|\le (\lfloor\log_2 X\rfloor+1)^2$.
3. Lemma 3. If $n$ is odd and $n-2$ is prime then $n\notin A$.
4. Computation. $A\cap[1,20{,}000{,}000]=\{1,3\}$.

New Insight / Tool (Round-2)

1. Decoding Crocker's construction into the $k,\ell\ge 0$ model. Crocker constructs infinitely many odd integers that are (i) not a prime plus a (positive) power of $2$ and (ii) not a prime plus two distinct positive powers of $2$. Using Round 1 Lemma 1, this immediately yields elements of our set $A$ (which allows $2^0=1$).
2. Effective CRT instantiation. Crocker gives explicit congruence data (an overlapping congruence system and a specific divisor of the Fermat number $2^{1024}+1$). I implement the CRT once (with concrete primes of specified order) and output an explicit $420$-digit $N_0\in A$.
3. Best-known lower bound (Pan). Pan proves an unconditional lower bound of the form \[ |A\cap[1,x]|\ \gg\ x\exp\!\Big(-C\,\log x\,\frac{\log\log\log\log x}{\log\log\log x}\Big), \] for an absolute constant $C>0$ and all sufficiently large $x$. This is much stronger than $x^{1-\varepsilon}$-type lower bounds but still allows density $0$.

Attack Plan (Round-2)

Gap(s) after Round 1. The main unresolved gap is still: \[ \overline d(A)>0\ ? \] Round 1 also had a practical gap: despite the asserted infinitude of $A$, no explicit element of $A$ beyond $1,3$ was exhibited.

Round-2 claims to prove/construct.

1. Reduction: for odd $n$, it suffices to rule out representations as "prime $+$ power of $2$" and "prime $+$ two distinct positive powers of $2$".
2. Explicit construction: using Crocker's explicit congruence system, construct a concrete odd integer $N_0\in A$.
3. Quantitative context: record Pan's theorem precisely and explain why it still falls short of positive (upper) density.

Work (Round-2)

A reduction from the $k,\ell\ge 0$ model to positive powers

Lemma 4 (odd-case reduction). Let $n$ be odd. Suppose that:

1. (i) $n$ is not representable as $n=p+2^a$ with $p$ prime and $a\ge 1$; and
2. (ii) $n$ is not representable as $n=p+2^a+2^b$ with $p$ prime and distinct integers $a,b\ge 1$.

Then $n\in A$, i.e. $n\neq p+2^k+2^\ell$ for all primes $p$ and $k,\ell\ge 0$.

Proof (using Round 1 Lemma 1). Assume toward contradiction that $n=p+2^k+2^\ell$.

If $p$ is odd, Round 1 Lemma 1 forces either $(k,\ell)=(0,0)$ or $k,\ell\ge 1$.

• If $(k,\ell)=(0,0)$, then $n=p+2^0+2^0=p+2=p+2^1$, contradicting (i).
• If $k,\ell\ge 1$ and $k\neq \ell$, this is excluded by (ii).
• If $k,\ell\ge 1$ and $k=\ell$, then $n=p+2^k+2^k=p+2^{k+1}$, again contradicting (i).

If $p=2$, Round 1 Lemma 1 forces exactly one of $k,\ell$ to be $0$, so $n=2+1+2^m=3+2^m$ for some $m\ge 1$, i.e. $n=3+2^m$ is of the form prime $+$ power of $2$, contradicting (i).

Thus no representation exists and $n\in A$. $\square$

Crocker's construction yields explicit elements of $A$

Crocker proves:

Theorem 5 (Crocker). There are infinitely many positive odd integers that are not representable as \[n=p+2^a+2^b\qquad(a,b>0,\ p\text{ prime}).\] Moreover, the constructed integers also avoid representations $n=p+2^a$ with $a>0$ (prime plus a positive power of $2$).

Source. This is Theorem I of Crocker $1971$, with the proof structured as: (a) a congruence/CRT argument excluding prime$+$power-of-$2$ representations (his system $2$); and (b) Lemma II excluding prime$+$two-distinct-powers representations for numbers of the form $w\prod B_i$ with $w\equiv 1\pmod{16}$ and $B_i\mid (2^{2^i}+1)$. See in particular the statements around Lemma I--II and the end of the proof of Theorem I.

Corollary 6 (translation to our set $A$). Every odd integer produced by Crocker's Theorem I belongs to $A$ as defined in Round 1 (allowing exponents $k,\ell\ge 0$).

Proof. Crocker's integers satisfy conditions (i)--(ii) of Lemma 4 with $a,b>0$. Hence by Lemma 4 they lie in $A$. $\square$

An explicit $420$-digit element $N_0\in A$

We now instantiate Crocker's explicit numerical choices (his final paragraph).

Step 1: the overlapping congruence system

Crocker specifies the overlapping system (written here as pairs $(a_i,m_i)$): \[ \begin{aligned} &(0,3),(0,5),(1,9),(1,10),(8,12),(8,15),(4,18),(7,20),(5,24),(29,30),\\ &(2,36),(14,36),(17,40),(34,45),(43,45),(13,48),(37,48),(16,60),(19,60),\\ &(26,72),(62,72),(52,90),(37,120),(49,144),(121,144),(103,180),(106,180),(229,360). \end{aligned} \] It is "overlapping" in the sense that every positive integer exponent $d$ satisfies $d\equiv a_i\pmod{m_i}$ for at least one of these pairs.

Step 2: primes of prescribed order

For each congruence $d\equiv a_i\pmod{m_i}$, choose an odd prime $p_i$ with multiplicative order \[\operatorname{ord}_{p_i}(2)=m_i. \] One explicit choice (found computationally, but fully checkable by direct order computation) is: \[ \begin{array}{rcl@{\qquad}rcl} (0,3)&\mapsto&p=7,&(34,45)&\mapsto&p=631,\\ (0,5)&\mapsto&p=31,&(43,45)&\mapsto&p=23311,\\ (1,9)&\mapsto&p=73,&(13,48)&\mapsto&p=97,\\ (1,10)&\mapsto&p=11,&(37,48)&\mapsto&p=673,\\ (8,12)&\mapsto&p=13,&(16,60)&\mapsto&p=61,\\ (8,15)&\mapsto&p=151,&(19,60)&\mapsto&p=1321,\\ (4,18)&\mapsto&p=19,&(26,72)&\mapsto&p=433,\\ (7,20)&\mapsto&p=41,&(62,72)&\mapsto&p=38737,\\ (5,24)&\mapsto&p=241,&(52,90)&\mapsto&p=18837001,\\ (29,30)&\mapsto&p=331,&(37,120)&\mapsto&p=4562284561,\\ (2,36)&\mapsto&p=37,&(49,144)&\mapsto&p=577,\\ (14,36)&\mapsto&p=109,&(121,144)&\mapsto&p=487824887233,\\ (17,40)&\mapsto&p=61681,&(103,180)&\mapsto&p=181,\\ &&&(106,180)&\mapsto&p=54001,\\ &&&(229,360)&\mapsto&p=168692292721.\\ \end{array} \] All $28$ primes above are distinct.

Step 3: the Mersenne prime modulus and the "avoidance" residue

Take \[p_{29}=2^{13}-1=8191,\] which is prime, and choose a residue $c\pmod{8191}$ such that \[c\not\equiv p_i+2^d\pmod{8191}\qquad(1\le i\le 28,\ 0\le d\le 12). \] Since $2$ has order $13$ modulo $8191$, this excludes congruences for all $d\ge 0$. A valid explicit choice is \[c\equiv 1\pmod{8191}. \] (Direct check: $1$ is not congruent to any of the $28\cdot 13$ forbidden residues.)

Step 4: the Fermat divisor $G_{10}$

Let $F_{10}=2^{1024}+1$ and take the known prime factor \[q:=2^{12}\cdot 11131+1=45592577\mid F_{10}. \] Define \[G_{10}:=\frac{F_{10}}{q}=\frac{2^{1024}+1}{45592577}. \] Then $G_{10}>1$ and $G_{10}\mid F_{10}$.

Step 5: choosing $n=11$ and forming the CRT system

Set $n=11$ (so $n>k=10$) and define \[M_1:=\frac{2^{2^{11}}-1}{G_{10}}=\frac{2^{2048}-1}{G_{10}}\in\mathbb{N}. \] Consider the simultaneous congruence system in the variable $t$:

1. $t\equiv 0\pmod{M_1}$,
2. $t\equiv -1\pmod{16}$,
3. $t\equiv 2^{a_i}\pmod{p_i}$ for each of the $28$ pairs $(a_i,m_i)$ above (with $p_i$ as chosen),
4. $t\equiv c\pmod{8191}$ with $c\equiv 1$.

Crocker shows (and our explicit choices satisfy) that all moduli are pairwise coprime, so by the Chinese remainder theorem this system has a unique solution modulo \[M:=16\cdot M_1\cdot 8191\cdot\prod_{i=1}^{28}p_i. \] Let $N_0$ be the least positive solution.

Step 6: the explicit integer

A direct CRT computation gives the least positive solution \[ \begin{aligned} N_0:=\,&114692399417541627831390678344111455559634149945994060064370176174761757107094\\ &384161208131810508010278443673581829982252271454885733150869511794121524724\\ &926719189042703019347118118212960843353505774370088928506106578466372689421\\ &320193947115016412660586173360894622173451791309400443984363146771213896025\\ &262579711774536791930091548873856297462778888609711986475144538417671136309\\ &198210638114258170393070241496204972122655. \end{aligned} \] This $N_0$ satisfies:

• $N_0\equiv -1\pmod{16}$ (hence $N_0$ is odd);
• $N_0\equiv 0\pmod{M_1}$ (so $N_0=M_1\cdot w$ for some integer $w$);
• $w\equiv 1\pmod{16}$ (since $M_1\equiv -1\pmod{16}$ and $N_0\equiv -1\pmod{16}$);
• $N_0\equiv 2^{a_i}\pmod{p_i}$ for each $i$ and $N_0\equiv 1\pmod{8191}$.

Also $N_0<2^{2048}-1$ (indeed $N_0<M$ and Crocker's inequality $16\prod_{i=1}^{29}p_i<G_{10}$ ensures $M<2^{2048}-1$).

Step 7: why $N_0\in A$

1. $N_0$ is not a prime plus a power of $2$. Let $d\ge 1$. By the overlapping property of the congruence system, choose $i$ with $d\equiv a_i\pmod{m_i}=\pmod{\operatorname{ord}_{p_i}(2)}$. Then $2^d\equiv 2^{a_i}\pmod{p_i}$, hence \[N_0-2^d\equiv 2^{a_i}-2^d\equiv 0\pmod{p_i}. \] So if $N_0-2^d$ were prime, it would have to equal $p_i$, i.e. $N_0=p_i+2^d$. Reducing modulo $8191$ and using that $2^d\bmod 8191$ depends only on $d\bmod 13$, this would force \[c\equiv N_0\equiv p_i+2^d\pmod{8191} \] for some $0\le d\le 12$, contradicting the choice of $c$. Thus $N_0\neq p+2^d$ for all primes $p$ and all $d\ge 1$.

2. $N_0$ is not a prime plus two distinct positive powers of $2$. Write $N_0=M_1\cdot w$ with $w\equiv 1\pmod{16}$ and \[M_1=\prod_{i=0}^{10}B_i\qquad(B_i\mid 2^{2^i}+1,\ B_{10}=(2^{1024}+1)/G_{10},\ B_i=2^{2^i}+1\text{ for }i\neq 10). \] Then $N_0=w\prod_{i=0}^{10}B_i\le 2^{2048}-1$ with $w\equiv 1\pmod{16}$, i.e. $N_0$ satisfies the hypotheses of Crocker's Lemma II, which concludes that $N_0$ cannot equal $p+2^a+2^b$ with $a\neq b$ and $a,b\ge 1$.

3. Conclude $N_0\in A$. By the two bullets above, $N_0$ satisfies (i)--(ii) of Lemma 4, hence $N_0\in A$.

Pan's quantitative lower bound

For context toward the density question, Pan proves:

Theorem 7 (Pan). There exists an absolute constant $C>0$ such that, for all sufficiently large $x$, \[ \left|\{1\le n\le x:\ n\text{ odd and }n\not= p+2^a+2^b\ (p\text{ prime},\ a,b\ge 0)\}\right| \ \gg\ x\exp\!\Big(-C\,\log x\,\frac{\log\log\log\log x}{\log\log\log x}\Big). \] In particular, for every $\varepsilon>0$ one has $|A\cap[1,x]|\gg_\varepsilon x^{1-\varepsilon}$ for all sufficiently large $x$.

Relevance. This lower bound implies that $A$ is extremely large (size $x^{1-o(1)}$), yet the factor multiplying $x$ still tends to $0$, so it does not imply $\overline d(A)>0$.

Adversarial Verification

• Does Crocker apply to $k,\ell\ge 0$? Yes for odd targets, by Lemma 4: any $k=\ell$ representation collapses to prime$+$power of $2$, and any $p=2$ representation collapses to $3+2^m$, also prime$+$power.

• Is the "$d\le 12$" check modulo $8191$ sufficient? Yes: $8191=2^{13}-1$ implies $2^{13}\equiv 1\pmod{8191}$, so $2^d\bmod 8191$ depends only on $d\bmod 13$.

• Copairwise coprimality of moduli. All $p_i$ are distinct odd primes, coprime to $16$. Each $p_i$ has order $m_i$ with an odd factor $>1$, hence $p_i\nmid 2^{2^n}-1$ for any $n$ (otherwise the order would divide $2^n$), so $p_i\nmid M_1$. Thus $\gcd(p_i,M_1)=1$. Also $\gcd(8191,M_1)=1$ because $8191$ has order $13$ and again cannot divide $2^{2048}-1$.

• Internal computational consistency checks. The CRT computation (performed separately from this writeup) verified: $N_0\equiv -1\pmod{16}$, $N_0\equiv 0\pmod{M_1}$, $N_0\equiv 1\pmod{8191}$, and $N_0\equiv 2^{a_i}\pmod{p_i}$ for all $28$ pairs. No hidden assumptions beyond these congruences are used.

• Could $N_0-2^d=p_i$ still happen without contradicting $c$? No. Equality implies the congruence $c\equiv p_i+2^d\pmod{8191}$, which is explicitly excluded by construction.

Final Status

UNRESOLVED (BUT STRICTLY ADVANCED).

The density question $\overline d(A)>0$ remains open. However, Round 2 strictly advances beyond Round 1 by:

1. providing a rigorous reduction (Lemma 4) connecting Crocker's "positive power" results to the $k,\ell\ge 0$ formulation of $A$;
2. instantiating Crocker's CRT construction to produce an explicit $420$-digit odd integer $N_0\in A$ (hence $A$ is nontrivial beyond $\{1,3\}$);
3. recording the sharpest known unconditional lower bound (Pan) for $|A\cap[1,x]|$.

Completion Estimate

COMPLETION: 55%

References

1. R. Crocker, On the sum of a prime and of two powers of two, Pacific J. Math. 36 $1971$, 103--107.
2. H. Pan, On the integers not of the form $p+2^a+2^b$, arXiv:0905.3809; published in Acta Arith. 148 $2011$, 97--106.
3. (Background identity) $\prod_{i=0}^{n-1}(2^{2^i}+1)=2^{2^n}-1$ (classical Fermat number product identity).