Wiki
Wiki

Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

Updated


Source. Theorem 4.3 of Section 4, p. 137, with Lemma 4.1 (pp. 137--138), the proof (pp. 138--139) and Remark 4.1 (p. 139), of A. Sárközy and V. T. Sós, On additive representation functions, in R. L. Graham et al. (eds.), The Mathematics of Paul Erdős I, Springer, 1997, 129--150, doi:10.1007/978-3-642-60408-9_11, as identified on the source card.

Statement

Setting (pp. 130 and 137). For A⊂N0\mathcal A\subset\mathbb N_0 and n∈N0n\in\mathbb N_0, r2(A,n)r_2(\mathcal A,n) is the number of solutions of a+a′=na+a'=n with a,a′∈Aa,a'\in\mathcal A and a≤a′a\le a'. For u∈Nu\in\mathbb N, Su(A)\mathcal S_u(\mathcal A) is the set of n∈Nn\in\mathbb N with r2(A,n)=ur_2(\mathcal A,n)=u, and Su(A,N)S_u(\mathcal A,N) is its counting function. The counting function of a set B\mathcal B is B(N)=∣{b:0<b≤N, b∈B}∣B(N)=|\{b:0<b\le N,\ b\in\mathcal B\}|.

Theorem 4.3 (p. 137, quoted). "Let k∈Nk\in\mathbb N and let u1<u2<…<uku_1<u_2<\ldots<u_k be positive integers. Then there is an infinite set A⊂N0\mathcal A\subset\mathbb N_0 such that writing"

B=N∖(⋃i=1kSui(A))\mathcal B=\mathbb N\setminus\Bigl(\textstyle\bigcup_{i=1}^k\mathcal S_{u_i}(\mathcal A)\Bigr)

"we have"

Sui(A,N)=Nk+O(Nα)S_{u_i}(\mathcal A,N)=\frac Nk+O(N^\alpha)

"and"

B(N)=O(Nα)B(N)=O(N^\alpha)

"where α=log⁡3log⁡4\alpha=\frac{\log 3}{\log 4}."

The first estimate holds for each i=1,…,ki=1,\ldots,k. The paper notes the case k=1k=1, u1=2u_1=2: there is a set with r2(A,n)=2r_2(\mathcal A,n)=2 for all but O(Nα)O(N^\alpha) of the n≤Nn\le N (p. 137). The theorem answers, in the negative, the authors' earlier expectation that the conclusion of their Problem 4.1 survives when r2(A,n)r_2(\mathcal A,n) is bounded only outside a thin set of nn (p. 137). Here α=0.7924…\alpha=0.7924\ldots.

Remark 4.1 (p. 139). For positive rationals r1,…,rkr_1,\ldots,r_k with sum 11, the authors say the same idea gives an infinite A⊂N0\mathcal A\subset\mathbb N_0 with Sui(A,N)=riN+O(Nα)S_{u_i}(\mathcal A,N)=r_iN+O(N^\alpha) for some 0<α<10<\alpha<1; the print gives the range of ii as 1≤i≤11\le i\le1, read as 1≤i≤k1\le i\le k, and refers to the theorem as Theorem 4. No proof is given. They expect the analogue with arbitrary densities λi\lambda_i to hold, with a harder proof.

Read depth. Claims checked: the setting, the statement, Lemma 4.1 and Remark 4.1 were read clause by clause on the printed pages, and the proof was read for its structure. Nothing here is independently reviewed.

Proof pointer

Pages 137--139. Lemma 4.1 (pp. 137--138) takes F\mathcal F, the integers whose base-44 digits are all 00 or 11, and G=2×F\mathcal G=2\times\mathcal F, the integers whose base-44 digits are all 00 or 22. Its four parts are: every n∈Nn\in\mathbb N is f+gf+g with f∈Ff\in\mathcal F, g∈Gg\in\mathcal G in exactly one way; the counting functions of F+F\mathcal F+\mathcal F and of G+G\mathcal G+\mathcal G are O(Nα)O(N^\alpha), since the sums in F+F\mathcal F+\mathcal F have no base-44 digit 33; and, for H=F∪G\mathcal H=\mathcal F\cup\mathcal G, the n≤Nn\le N with r2(H,n)>1r_2(\mathcal H,n)>1 number O(Nα)O(N^\alpha) (the print's display of this last part omits the restriction n≤Nn\le N). With 0=g1<g2<⋯0=g_1<g_2<\cdots the elements of G\mathcal G and Gi={g1,…,gui}\mathcal G_i=\{g_1,\ldots,g_{u_i}\}, the set is

A=(⋃i=1k(k×(F+Gi)+{i}))∪(k×G).\mathcal A=\Bigl(\textstyle\bigcup_{i=1}^k\bigl(k\times(\mathcal F+\mathcal G_i)+\{i\}\bigr)\Bigr)\cup(k\times\mathcal G).

A large n≡i(modk)n\equiv i\pmod k then has exactly uiu_i representations as a point of the ii-th block plus a point of k×Gk\times\mathcal G, by the unique representation in Lemma 4.1 applied once for each gtg_t, t≤uit\le u_i; the sums of two block elements and of two elements of k×Gk\times\mathcal G are O(Nα)O(N^\alpha) in number by Lemma 4.1. The print's index ⋃j=kk\bigcup_{j=k}^{k} in the first step (p. 138) is read as ⋃j=1k\bigcup_{j=1}^{k}, which is how the step's display (4.21) writes it.

Dependencies

Lemma 4.1 (pp. 137--138) of the same paper, proved there in a few lines.

Bears on

  • Problem 14: the case k=1k=1, u1=1u_1=1 gives an infinite A⊂N0\mathcal A\subset\mathbb N_0 for which all but O(Nlog⁡3/log⁡4)O(N^{\log3/\log4}) integers in {1,…,N}\{1,\ldots,N\} have exactly one representation a+a′a+a' with a≤a′a\le a'. The set contains 00; translating it by 11 (an observation of this page, not of the paper) gives a set of positive integers with the same bound, since it moves each count of representations from nn to n+2n+2. The exponent log⁡3/log⁡4\log3/\log4 exceeds 1/21/2, so this neither contradicts the lower bound the problem asks about nor answers whether o(N1/2)o(N^{1/2}) exceptions are possible.