a_507_bc wrote: Find all strictly increasing sequences of positive integers $a_1, a_2, \ldots$ with $a_1=1$, satisfying $$3(a_1+a_2+\ldots+a_n)=a_{n+1}+\ldots+a_{2n}$$for all positive integers $n$. We proof by induction that $a_n=2n-1$. We have $a_1=1$ and $a_2=3a_1$ so the claim holds for $1,2$. Assume the claim holds for $1,2,\ldots,2n$. By subtracting $3(a_1+a_2+\ldots+a_n)=a_{n+1}+\ldots+a_{2n}$ from $3(a_1+a_2+\ldots+a_{n+1})=a_{n+2}+\ldots+a_{2n+2}$ we get $a_{2n+2}+a_{2n+1}=4a_{n+1}=8n+4$. Since $4n-1=a_{2n}<a_{2n+1}<a_{2n+2}$ we have $a_{2n+1}\in\{4n,4n+1\}$. Assume $a_{2n+1}=4n$. Then we have $a_{4n-1}+a_{4n}=4a_{2n}=8n-4$ and $a_{4n+1}+a_{4n+2}=4a_{2n+1}=8n$ which contadicts $a_{4n-1}<a_{4n}<a_{4n+1}<a_{4n+2}$. So $a_{2n+1}=4n+1$ and $a_{2n+2}=4n+3$. This completes the induction. $a_n=2n-1$ is a solution since it satisfies $\sum_{i=1}^na_n=n^2$ and thus \[ 3(a_1+a_2+\ldots+a_n)=3n^2=(2n)^2-n^2=(a_1+a_2+\ldots+a_{2n})-(a_1+a_2+\ldots+a_n)=a_{n+1}+\ldots+a_{2n} \]

pi_quadrat_sechstel wrote: We proof by induction that $a_n=2n-1$. This doesn’t show that 2n-1 is the only solution