Abstract

Let DD be a subset of a finite commutative ring RR with identity. Let f(x)∈R[x]f(x)\in R[x] be a polynomial of positive degree dd. For integer 0≤k≤∣D∣0\leq k \leq |D|, we study the number Nf(D,k,b)N_f(D,k,b) of kk-subsets S⊆DS\subseteq D such that \begin{align*} \sum_{x\in S} f(x)=b. \end{align*} In this paper, we establish several asymptotic formulas for Nf(D,k,b)N_f(D,k, b), depending on the nature of the ring RR and ff. For R=ZnR=\mathbb{Z}_n, let p=p(n)p=p(n) be the smallest prime divisor of nn, ∣D∣=n−c≥Cdnp−1d+c|D|=n-c \geq C_dn p^{-\frac 1d }+c and f(x)=adxd+⋯+a0∈Z[x]f(x)=a_dx^d +\cdots +a_0\in \mathbb{Z}[x] with (ad,…,a1,n)=1(a_d, \dots, a_1, n)=1. Then ∣Nf(D,k,b)−1n(n−ck)∣≤(δ(n)(n−c)+(1−δ(n))(Cdnp−1d+c)+k−1k),\left| N_f(D, k, b)-\frac{1}{n}{n-c \choose k}\right|\leq {δ(n)(n-c)+(1-δ(n))(C_dnp^{-\frac 1d}+c)+k-1\choose k}, partially answering an open question raised by Stanley \cite{St}, where δ(n)=∑i∣n,μ(i)=−11iδ(n)=\sum_{i\mid n, μ(i)=-1}\frac 1 i and Cd=e1.85dC_d=e^{1.85d}. Furthermore, if nn is a prime power, then δ(n)=1/pδ(n) =1/p and one can take Cd=4.41C_d=4.41. For R=FqR=\mathbb{F}_q of characteristic pp, let f(x)∈Fq[x]f(x)\in \mathbb{F}_q[x] be a polynomial of degree dd not divisible by pp and D⊆FqD\subseteq \mathbb{F}_q with ∣D∣=q−c≥(d−1)q+c|D|=q-c\geq (d-1)\sqrt{q}+c. Then ∣Nf(D,k,b)−1q(q−ck)∣≤(q−cp+p−1p((d−1)q12+c)+k−1k).\left| N_f(D, k, b)-\frac{1}{q}{q-c \choose k}\right|\leq {\frac{q-c}{p}+\frac {p-1}{p}((d-1)q^{\frac 12}+c)+k-1 \choose k}. If f(x)=ax+bf(x)=ax+b, then this problem is precisely the well-known subset sum problem over a finite abelian group. Let GG be a finite abelian group and let D⊆GD\subseteq G with ∣D∣=∣G∣−c≥c|D|=|G|-c\geq c. Then ∣Nx(D,k,b)−1∣G∣(∣G∣−ck)∣≤(c+(∣G∣−2c)δ(e(G))+k−1k),\left| N_x(D, k, b)-\frac{1}{|G|}{|G|-c \choose k}\right|\leq {c + (|G|-2c)δ(e(G))+k-1 \choose k}, where e(G)e(G) is the exponent of GG and δ(n)=∑i∣n,μ(i)=−11iδ(n)=\sum_{i\mid n, μ(i)=-1}\frac 1 i. In particular, we give a new short proof for the explicit counting formula for the case D=GD=G.

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