MathLabs

Problem 6

Find all positive integers nn for which there exist nonnegative integers a1,…,ana_1,\dots,a_n such that ∑i=1n2−ai=∑i=1ni3−ai=1\sum_{i=1}^n2^{-a_i}=\sum_{i=1}^n i3^{-a_i}=1.
Step 2 of 4: Pass from an odd length to the next even length
In plain words

One term can be replaced by two equal binary halves whose ternary indices add to three times the old index.

am+1′=am+1+1,a2m+2′=am+1+1a'_{m+1}=a_{m+1}+1,\quad a'_{2m+2}=a_{m+1}+1
Detailed analysis

Suppose (a1,…,a2m+1)(a_1,\ldots,a_{2m+1}) is feasible. Define aj′=aja'_j=a_j except am+1′=am+1+1a'_{m+1}=a_{m+1}+1 and a2m+2′=am+1+1a'_{2m+2}=a_{m+1}+1. The binary sum is unchanged because one term 2−am+12^{-a_{m+1}} becomes two copies of 2−(am+1+1)2^{-(a_{m+1}+1)}. The ternary change is −(m+1)3−am+1+((m+1)+(2m+2))3−(am+1+1)=0-(m+1)3^{-a_{m+1}}+((m+1)+(2m+2))3^{-(a_{m+1}+1)}=0. Thus feasibility at 2m+12m+1 implies feasibility at 2m+22m+2.