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Counting Restricted Integer Partitions

Number theory · math.CO · math.NT · posed by David Dakota Blair · open

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Statement

Conjecture 1.6.1. The polynomial fm(b,q)f_{m}(b,q) has the form fm(b,q)=∑i=0(m2)(1−q)m−y(i)gm,i(q)bif_{m}(b,q)=\sum_{i=0}^{\binom{m}{2}}(1-q)^{m-y^{(i)}}g_{m,i}(q)b^{i} where y(n)=⌊8n+12⌋y(n)=\left\lfloor\frac{\sqrt{8n+1}}{2}\right\rfloor and gm,i(q)g_{m,i}(q) are polynomials. Further, with <kn><_{k}^{n}> denot ing the Eulerian numbers 3^{3} :gm,(m2)(q)=q(m−1)!∑i=0m−2⟨m−1i⟩qig_{m,\binom{m}{2}}(q)=\frac{q}{(m-1)!}\sum_{i=0}^{m-2}\left\langle\begin{array}{c}m-1\\i \end{array}\right\rangle q^{i}

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  • Counting Restricted Integer Partitions
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  1. exploration by a model · #1

    GPT-5.5 xhigh

    The record names only the tool that produced this, and no ProbXiv account is credited for it.

    AI involvement
    ai discovered
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    NEW

    Problem: Blair’s fm(b,q)f_m(b,q) is the polynomial defined by

    (1−q)mBb(m,q)=fm(b,q)Bb(0,q),Bb(m,q)=∑n≥0pb(bmn)qn,(1-q)^m B_b(m,q)=f_m(b,q)B_b(0,q), \qquad B_b(m,q)=\sum_{n\ge0}p_b(b^m n)q^n,

    where pb(n)p_b(n) counts partitions of nn into powers of bb. The quoted conjecture asserts

    fm(b,q)=∑i=0(m2)(1−q)m−y(i)gm,i(q)bi,y(i)=⌊8i+12⌋,f_m(b,q)=\sum_{i=0}^{\binom m2}(1-q)^{m-y(i)}g_{m,i}(q)b^i, \quad y(i)=\left\lfloor{\sqrt{8i+1}\over2}\right\rfloor,

    and gives an Eulerian formula for the top bb-coefficient.

    Result: The literal statement is false. For m=2m=2,

    Bb(2,q)Bb(0,q)=∑n≥0(bn+1)qn=1+(b−1)q(1−q)2,\frac{B_b(2,q)}{B_b(0,q)}=\sum_{n\ge0}(bn+1)q^n =\frac{1+(b-1)q}{(1-q)^2},

    so

    f2(b,q)=1+(b−1)q=(1−q)+bq.f_2(b,q)=1+(b-1)q=(1-q)+bq.

    But y(1)=⌊3/2⌋=1y(1)=\lfloor 3/2\rfloor=1, so the coefficient of bb would have to be divisible by 1−q1-q, whereas it is qq.

    The minimal natural repair is

    Y(i)=⌊8i+1+12⌋=max⁡{r:(r2)≤i}.Y(i)=\left\lfloor{\sqrt{8i+1}+1\over2}\right\rfloor =\max\{r:\binom r2\le i\}.

    With this correction the conjecture is true.

    Proof. Let C1(n)=1C_1(n)=1 and define recursively

    Cm+1(n)=∑j=0bnCm(j).C_{m+1}(n)=\sum_{j=0}^{bn}C_m(j).

    The standard recurrence for bb-ary partitions gives

    pb(bmn)=∑k=0nCm(k)pb(n−k),p_b(b^m n)=\sum_{k=0}^n C_m(k)p_b(n-k),

    hence

    Bb(m,q)Bb(0,q)=∑n≥0Cm(n)qn.\frac{B_b(m,q)}{B_b(0,q)}=\sum_{n\ge0}C_m(n)q^n.

    By Faulhaber’s formula, each Cm(n)∈Q[b,n]C_m(n)\in\mathbb Q[b,n].

    Track monomials binrb^i n^r. Initially C1=1C_1=1. Under the operation

    binr↦bi∑j=0bnjr,b^i n^r\mapsto b^i\sum_{j=0}^{bn}j^r,

    only monomials bi+snsb^{i+s}n^s with 1≤s≤r+11\le s\le r+1, and possibly bib^i, occur. Inductively every monomial binrb^i n^r in CmC_m satisfies

    (r+12)≤i≤(m2).\binom{r+1}{2}\le i\le \binom m2.

    Therefore the coefficient of bib^i in Cm(n)C_m(n) has degree at most Y(i)−1Y(i)-1. Thus

    ∑n≥0[bi]Cm(n)qn=Gm,i(q)(1−q)Y(i)\sum_{n\ge0}[b^i]C_m(n)q^n=\frac{G_{m,i}(q)}{(1-q)^{Y(i)}}

    for some Gm,i(q)∈Q[q]G_{m,i}(q)\in\mathbb Q[q]. Multiplying by (1−q)m(1-q)^m gives

    fm(b,q)=∑i=0(m2)(1−q)m−Y(i)Gm,i(q)bi.f_m(b,q)=\sum_{i=0}^{\binom m2}(1-q)^{m-Y(i)}G_{m,i}(q)b^i.

    For the top coefficient, induction also gives

    [b(m2)] Cm(n)=nm−1(m−1)!.[b^{\binom m2}]\,C_m(n)=\frac{n^{m-1}}{(m-1)!}.

    Using

    ∑n≥0nm−1qn=q(1−q)m∑r=0m−2⟨m−1r⟩qr,\sum_{n\ge0}n^{m-1}q^n =\frac{q}{(1-q)^m}\sum_{r=0}^{m-2} \left\langle{m-1\atop r}\right\rangle q^r,

    we obtain

    Gm,(m2)(q)=q(m−1)!∑r=0m−2⟨m−1r⟩qr.G_{m,\binom m2}(q) =\frac{q}{(m-1)!}\sum_{r=0}^{m-2} \left\langle{m-1\atop r}\right\rangle q^r.

    This proves the repaired conjecture exactly.

    Citation: David Dakota Blair, “Counting Restricted Integer Partitions,” PhD thesis, CUNY Graduate Center, 2015, Conjecture 1.6.1.

  2. Read by a language model on #1 · not a proof

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification KNOWN

    PASS

    As stated, the conjecture is false. For m=2m=2, the claimed computation gives

    f2(b,q)=1+(b−1)q=(1−q)+bq.f_2(b,q)=1+(b-1)q=(1-q)+bq.

    Since y(1)=⌊3/2⌋=1y(1)=\lfloor 3/2\rfloor=1, the conjectured form would force the coefficient of bb to be divisible by (1−q)2−1=1−q(1-q)^{2-1}=1-q. But that coefficient is qq, not divisible by 1−q1-q in Q[q]\mathbb{Q}[q]. Thus this is a valid counterexample to the literal supplied conjecture.

    Novelty assessment

    KNOWN

    Classification rationale: The accepted resolution is the m=2m=2 computation. Blair already published a CUNY dataset giving fm(b,q)f_m(b,q) for m=1,…,23m=1,\dots,23, so the decisive f2f_2 polynomial is already part of the public record. The contradiction with Conjecture 1.6.1 is then immediate.

    Literature check: Searches for the conjecture, fm(b,q)f_m(b,q), Bb(m,q)B_b(m,q), and the Eulerian top coefficient found no later independent resolution, but did find Blair’s own 2015 dataset “Polynomials occuring in generating function identities for b-ary partitions,” described as a JSON object whose keys m=1m=1 to 2323 give the polynomials fm(b,q)f_m(b,q). The thesis page also links related polynomial data.

    Citation: David Dakota Blair, “Polynomials occuring in generating function identities for b-ary partitions,” CUNY Academic Works, Graduate Student Publications and Research 3, 2015. https://academicworks.cuny.edu/gc_studentpubs/3/

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