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Statement

Setting (pp. 1--2). Td(n)T_d(n) is the largest number of equilateral triangles, of all side lengths together, determined by nn points of Rd\mathbb R^d. 1P\mathbb 1_P is 11 when the condition PP holds and 00 otherwise, and [r]={1,…,r}[r]=\{1,\ldots,r\}.

Theorem 3 (p. 2). Let r≥3r\ge3 be a fixed integer, let nn be sufficiently large, and let pp be the remainder of nn on division by 2r2r. Then

T2r(n)=∑1≤i<j<k≤rninjnk+∑i∈[r]((ni−1ni∉4Z)(n−ni)+ni−pi3+1pi>8(pi−8)),T_{2r}(n)=\sum_{1\le i<j<k\le r}n_in_jn_k +\sum_{i\in[r]}\Bigl((n_i-\mathbb 1_{n_i\notin4\mathbb Z})(n-n_i) +\frac{n_i-p_i}3+\mathbb 1_{p_i>8}(p_i-8)\Bigr),

where pip_i is the remainder of nin_i on division by 1212, and the parts n1,…,nrn_1,\ldots,n_r, with n1+⋯+nr=nn_1+\cdots+n_r=n, are chosen as follows; write q=⌊n/r⌋q=\lfloor n/r\rfloor.

  • If p∈[0,r)p\in[0,r) and pp is even: r−p/2r-p/2 parts equal to qq and p/2p/2 parts equal to q+2q+2.
  • If p∈[0,r)p\in[0,r) and pp is odd: r−(p+1)/2r-(p+1)/2 parts equal to qq, one part equal to q+1q+1, and (p−1)/2(p-1)/2 parts equal to q+2q+2.
  • If p∈[r,2r)p\in[r,2r) and pp is even: r−p/2r-p/2 parts equal to q−1q-1 and p/2p/2 parts equal to q+1q+1.
  • If p∈[r,2r)p\in[r,2r) and pp is odd: r−(p+1)/2r-(p+1)/2 parts equal to q−1q-1, one part equal to qq, and (p−1)/2(p-1)/2 parts equal to q+1q+1.

In words: when nn is even all parts are even and any two differ by at most 22; when nn is odd exactly one part is odd and it differs by 11 from every other part (p. 16, end of the proof).

Corollary 4 (p. 2). Let r≥3r\ge3 be a fixed integer. If nn is sufficiently large and divisible by 12r12r, then

T2r(n)=(r3)(nr)3+(r−1)n2r+n3.T_{2r}(n)=\binom r3\Bigl(\frac nr\Bigr)^3+\frac{(r-1)n^2}r+\frac n3.

Here every part is n/rn/r, a multiple of 1212, so each indicator and each pip_i in Theorem 3 vanishes. For r=3r=3 the corollary reads T6(n)=n3/27+2n2/3+n/3T_6(n)=n^3/27+2n^2/3+n/3 when 36∣n36\mid n and nn is large.

Proof pointer

§ 6, pp. 12--16. The lower bound is the even-dimensional Lenz construction of § 2.2 (pp. 4--5): rr pairwise orthogonal unit circles with a common center in R2r\mathbb R^{2r}, the nin_i points on the ii-th circle placed in copies of a regular dodecagon, so that triangles of side 2\sqrt2 come from points on three different circles or from a pair at distance 2\sqrt2 on one circle and a point on another, and triangles of side 3\sqrt3 lie on one circle. For the upper bound, the stability result Theorem 7 puts all but o(n)o(n) points of an extremal set on rr such circles; Claim 20 (p. 12) shows an extremal set has no point off the circles, and Claim 21 (p. 13) bounds the count by the maximum of the construction's count over all splittings n1+⋯+nr=nn_1+\cdots+n_r=n, which is display (6) (p. 14). The proof of Theorem 3 (pp. 14--16) then finds the maximizing splitting: Claims 22--24 show that two parts differ by at most 22, that parts differing by 22 are even, and that at most one part is odd.

Read depth

Claims checked: Theorem 3 and Corollary 4 were read clause by clause on the page image of p. 2 (arXiv version 4), and Corollary 4 was checked against Theorem 3 by substitution. The proofs (§ 2.2 and § 6, pp. 4--5 and 12--16) were read for structure only; no estimate was checked. Nothing here is independently reviewed.

Dependencies

Theorem 7 and, through it, Theorem 2; the proof of Theorem 3 also uses Lemma 13 (p. 7).

Source. F. C. Clemen, A. Dumitrescu and D. Liu, The number of regular simplices in higher dimensions, arXiv:2507.19841 (2025), read in version 4 (28 July 2026); see the source card.

Bears on

  • Problem 755: with r=3r=3 the theorem gives the exact maximum number of equilateral triangles of all sizes together spanned by nn points of R6\mathbb R^6 for large nn, which is n3/27+O(n2)n^3/27+O(n^2); this contains the problem's bound for triangles of side 11. The exact count for side 11 alone is Proposition 25.