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Statement

The question as the note states it (printed p. 6199): "Let $1\le a_1<a_2<\cdots<a_k\le n$ be integers such that no $a_i$ is the sum of (any 2 or more) consecutive $a_j$-s. Is it possible for $k$ to be significantly larger than $n/2$?" The note first records Pomerance's example $k=12$, $n=20$, the set $\{4,5,6,7,8,10,12,14,16,17,19,20\}$, and the family for $n=4m$ with $m$ odd, $\{m-1,m,m+1,\tfrac{3m-1}2,\tfrac{3m+1}2,2m,2m+2,2m+3,\ldots,4m\}\setminus\{\tfrac{5m+1}2,3m,\tfrac{7m+1}2\}$, "a set having $\tfrac n2+2$ elements, which is one more than the trivial set $\{2m,2m+1,2m+2,\ldots4m\}$".

The construction (pp. 6199--6201). "The following construction shows that $k=19n/36+O(1)$ may be attained." For positive integers $x,y$ take

• (A) the $4y+1$ consecutive integers $2x-2y,\ldots,2x+2y$;
• (B) the $4y$ integers in $[3x-3y+1,3x+3y-1]$ not divisible by $3$;
• (C) the $4y-1$ even integers in $[4x-4y+2,4x+4y-2]$;
• (D) all the integers from $4x+4y+2$ to $8x+8y+4$,

$4x+16y+3$ numbers in all, and delete from (D) the elements equal to a sum of consecutive elements: the sums of three consecutive elements of (A) (I), of two consecutive elements of (B) (II), of four consecutive elements of (A) (III), of two consecutive elements of (C) (IV), and of two or three consecutive elements straddling a junction (A)--(B), (B)--(C) or (C)--(D) (V). The conditions (i) $2x+2y\le3x-3y$ and $3x+3y\le4x-4y+2$, (ii) $4x-4y+1>3x+3y-1$, (iii) $9x-9y+7>8x+8y+4$ and (iv) $10x-10y+10>8x+8y+4$ make the blocks increasing and keep the other consecutive sums out of the set (p. 6200). The note then states (p. 6201) that classes I and II coincide and III and IV coincide, so that $(4y-1)+(4y-2)+5=8y+2$ elements are deleted; that the conditions (i)--(iv), in effect (iii), require $x\ge17y-2$; and that with $x=17y-2$ the remaining set has $76y-7$ members up to $n=8x+8y+4=144y-12$, the proportion $19/36$.

The infinite version (pp. 6201--6202). With $A(n)$ the number of members at most $n$ of an infinite sequence: "We can achieve $\limsup A(n)/n=19/36$ using the previous construction" (p. 6201). The construction is repeated with very rapidly growing $y$: with $T$ the sum of all members so far, take $y=T^2$, form the next finite segment, and delete the $z$ in four ranges of length about $T$ ($2x-2y+1\le z\le2x-2y+T$, $4x-4y+T\le z\le4x-4y+2T+1$, $6x-6y+2T+3\le z\le6x-6y+3T+3$, $8x-8y+3T+1\le z\le8x-8y+4T+6$). The note asserts that no remaining member is then a sum of consecutive others, and that the loss of about $4T=4\sqrt y$ members is negligible against $n=144y-12$, so the proportion $19/36$ is kept (p. 6202).

Source. R. Freud, Adding numbers, James Cook Mathematical Notes 6 (1993), issue 60 (January 1993), 6199--6202; the scan of the whole issue read for this page has two printed pages per PDF page (printed pp. 6198--6199 on PDF p. 10, 6200--6201 on PDF p. 11, 6202--6203 on PDF p. 12), read on the page images at 150 dpi.

Read depth. Claims checked: the question, Pomerance's examples, the four blocks with their counts, the conditions (i)--(iv), the deletion list and its count, the closing sentence with $x\ge17y-2$, $76y-7$ and $144y-12$, and the infinite version were read clause by clause on the page images. Two arithmetic checks were made here from the printed figures: $4x+16y+3-(8y+2)=76y-7$ at $x=17y-2$, and $8x+8y+4=144y-12$. The verification that no remaining member is a sum of consecutive members is the note's (the conditions (i)--(iv) and the coincidences I = II, III = IV are asserted with brief reasons) and is not checked here; an external Lean file behind the catalog's label proves it for its encoding of the same blocks (Problem 867's page), and nothing is independently reviewed here.

Proof pointer

pp. 6200--6201 as summarized above: the block sizes and spacings are chosen so that the only consecutive sums landing in the set are the listed sums of two, three or four consecutive members of (A), (B) or (C) and at their borders, all of which fall in (D) and are deleted; the coincidences I = II and III = IV keep the deletion count at $8y+2$.

Dependencies

None.

Bears on

• Problem 867: for $N=144y-12$ the set has $76y-7=\tfrac{19}{36}(N+12)-7=\tfrac{19}{36}N-\tfrac23$ members, so $|A|-N/2=N/36-\tfrac23$ is unbounded along these $N$, against the problem's bound $|A|\le N/2+O(1)$; the verification that the set has no member a sum of consecutive members is the note's (the site: "Freud [Fr93] constructed a sequence with density $\ge19/36$").
• Problem 839: the infinite version gives a sequence with no member a sum of two or more consecutive members and $\limsup A(n)/n=19/36$; that problem asks whether every such sequence has $\limsup a_n/n=\infty$, and whether its logarithmic density is $0$, and this sequence decides neither.