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Math-1016 · mathematics · Combinatorics

Dividing up the unrestricted partitions

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Statement

Definitions and mathematical statement

Work in the ring $\mathbb Z[[x]]$ of formal power series. For each pair of positive integers $j,m$, choose a sign $\epsilon_{j,m}\in\{-1,1\}$, and set $\epsilon_{j,0}=1$. Form the product \[ F(x)=\prod_{j=1}^{\infty}\left(\sum_{m=0}^{\infty} \epsilon_{j,m}x^{jm}\right) =\sum_{N=0}^{\infty}c_Nx^N. \] Each coefficient is well defined: only factors $j\le N$ and terms with $jm\le N$ can contribute to $c_N$. No analytic convergence condition is involved, and signs for different pairs $(j,m)$ may be chosen independently.

The question is whether one infinite choice of these signs can make $c_N\in\{-1,0,1\}$ for every $N\ge0$. More explicitly, if a partition of $N$ has multiplicities $(m_j)_{j\ge1}$, give it the sign $\prod_j\epsilon_{j,m_j}$. Then \[ c_N=\sum_{\substack{m_j\ge0\text{ finitely supported}\\ \sum_jjm_j=N}}\prod_j\epsilon_{j,m_j}. \] Thus the same rule for a part size and its multiplicity must balance the positive and negative partitions at every integer simultaneously. The coefficient $c_0$ is automatically $1$. Modulo $2$, all signs are $1$, so $c_N$ has the same parity as the ordinary partition number $p(N)$. The signs cannot be assigned independently to individual partitions: their multiplicative factorization through $(j,m)$ is essential. Satisfying only a finite initial range of coefficients does not itself furnish the required infinite assignment.

Short English statement

Can signs in the unrestricted-partition product be chosen so every resulting coefficient is zero, one, or minus one?

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Math-1016-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

gpt 6 astra ultra attempt on Math-1016

Dividing up the unrestricted partitions

Definitions and mathematical statement

Work in the ring $\mathbb Z[[x]]$ of formal power series. For each pair of positive integers $j,m$, choose a sign $\epsilon_{j,m}\in\{-1,1\}$, and set $\epsilon_{j,0}=1$. Form the product \[ F(x)=\prod_{j=1}^{\infty}\left(\sum_{m=0}^{\infty} \epsilon_{j,m}x^{jm}\right) =\sum_{N=0}^{\infty}c_Nx^N. \] Each coefficient is well defined: only factors $j\le N$ and terms with $jm\le N$ can contribute to $c_N$. No analytic convergence condition is involved, and signs for different pairs $(j,m)$ may be chosen independently.

The question is whether one infinite choice of these signs can make $c_N\in\{-1,0,1\}$ for every $N\ge0$. More explicitly, if a partition of $N$ has multiplicities $(m_j)_{j\ge1}$, give it the sign $\prod_j\epsilon_{j,m_j}$. Then \[ c_N=\sum_{\substack{m_j\ge0\text{ finitely supported}\\ \sum_jjm_j=N}}\prod_j\epsilon_{j,m_j}. \] Thus the same rule for a part size and its multiplicity must balance the positive and negative partitions at every integer simultaneously. The coefficient $c_0$ is automatically $1$. Modulo $2$, all signs are $1$, so $c_N$ has the same parity as the ordinary partition number $p(N)$. The signs cannot be assigned independently to individual partitions: their multiplicative factorization through $(j,m)$ is essential. Satisfying only a finite initial range of coefficients does not itself furnish the required infinite assignment.

Short English statement

Can signs in the unrestricted-partition product be chosen so every resulting coefficient is zero, one, or minus one?

Research attempt

Scope and process.
This is a GPT-6 Astra Ultra research attempt dated 1 October 2026, using primary-source browsing and elementary mathematical checks. The canonical definition above is retained. Results below are restricted results or reductions, not a claim of a new solution to the entire catalogue question. No formal proof assistant or external mathematical referee checked this text.
Literature and strategy.
Newman's original posting [S2, R1] describes finite searches, but those do not by themselves produce an infinite sign choice. We analyze an extension one degree at a time, obtain an exact extension criterion, and exhibit a valid finite prefix which has no extension. This makes the obstacle to a naive greedy proof explicit.
The new variables at degree $N$.
Assume all $\epsilon_{j,m}$ with $jm<N$ have been chosen. There are exactly $\tau(N)$ new variables of weight $N$, namely $\epsilon_{j,N/j}$ for divisors $j$ of $N$. A term of weight $N$ can contribute to $c_N$ only on its own, because multiplication by any other positive-degree term exceeds degree $N$. Thus \[ c_N=B_N+\sum_{j\mid N}\epsilon_{j,N/j}, \] where $B_N$ is determined entirely by earlier choices. If $t=\tau(N)$, the new sum can be any member of $\{-t,-t+2,\ldots,t\}$. Hence an extension with $|c_N|\le1$ exists exactly when \[ |B_N|\le t+1. \] Indeed, outside this interval every available sum has absolute value at least two after adding $B_N$. Inside it, if zero has the required parity it is available whenever $|B_N|\le t$; otherwise one of $1,-1$ is available. In the boundary case $|B_N|=t+1$ the parity is the odd one and the nearest endpoint gives absolute value one. The parity agrees with $p(N)$ automatically, since every signed partition still contributes one modulo two.
A checked obstruction to arbitrary greedy extension.
The following rows list $(\epsilon_{j,1},\ldots, \epsilon_{j,\lfloor9/j\rfloor})$: \[ \begin{array}{c|l} j&\text{signs}\\ \hline 1&(1,-1,1,1,-1,1,-1,1,1)\\ 2&(1,1,1,-1)\\ 3&(-1,1,1)\\ 4&(1,1)\\ 5&(-1)\\6&(-1)\\7&(1)\\8&(-1)\\9&(1) \end{array} \] Truncated multiplication gives \[ (c_0,c_1,\ldots,c_9)=(1,1,0,1,1,1,-1,1,0,0). \] All are allowed. With the weight-ten coefficients omitted, the coefficient of $x^{10}$ is $B_{10}=6$. The four new signs are $\epsilon_{1,10},\epsilon_{2,5},\epsilon_{5,2},\epsilon_{10,1}$, whose sum is at least $-4$. Consequently $c_{10}\ge2$ for every extension of this prefix. This is a counterexample to the assertion that every valid prefix can be extended, not to Newman's existential infinite-assignment conjecture.
Verification of the numerical certificate.
The computation uses exact integers. Begin with coefficient vector $(1,0,\ldots,0)$ through degree ten. For each row $j$, replace it by its convolution with the factor having constant one and the listed signed coefficients in degrees $j,2j,\ldots$. Unlisted coefficients are zero only for this computation of $B_{10}$. The resulting vector is $(1,1,0,1,1,1,-1,1,0,0,6)$. The independent verification script scripts/check_attempt_batch_algebra_2026_10_01.py recomputes this vector and checks all $2^4$ extensions at degree ten. Unlisted signs in the original problem must eventually be chosen as $\pm1$; the zero convention is just a device for isolating the old contribution before adding the new signs explicitly.
What compactness would and would not give.
At level $N$, keep all choices of signs with $jm\le N$ satisfying the coefficient constraints through degree $N$. This is a finite set, with finitely many extensions per node. If every level is nonempty, there is an infinite branch: choose a child admitting arbitrarily deep descendants at each step, which is possible by finite branching. Every coefficient of the resulting infinite assignment is fixed at its corresponding level, so all constraints hold. Conversely an infinite assignment makes every level nonempty. Thus proving nonemptiness for every degree would suffice, even without an explicit formula for the signs. Verifying a fixed large degree, or finding many nodes there, does not prove this premise.
Final status and first remaining gap.
UNRESOLVED. The exact one-step criterion and a genuine dead-end prefix are established. The missing invariant must select prefixes that keep $|B_N|\le\tau(N)+1$ for all future degrees, or otherwise prove arbitrarily deep feasible levels. This attempt has no such invariant and no obstruction affecting every infinite choice.

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