MathLabs

Problem 5

Consider an infinite sequence a1,a2,…a_1,a_2,\ldots of positive integers such that 100!(am+am+1+⋯+an)100!(a_m+a_{m+1}+\cdots+a_n) is a multiple of an−m+1an+ma_{n-m+1}a_{n+m} for all positive integers m,nm,n with m≤nm\le n. Prove that the sequence is either bounded or linear. A sequence is bounded if there is a constant NN such that an<Na_n<N for every positive integer nn, and it is linear if an=n⋅a1a_n=n\cdot a_1 for every positive integer nn.
Step 7 of 10: A high-lcm pair forces the whole sequence to be linear
lcm⁡(apd,akd−1)>c2(apd+akd−1),gcd⁡(pd,kd−1)=1⟹aj=ja1\operatorname{lcm}(a_{pd},a_{kd-1})>c^2(a_{pd}+a_{kd-1}),\qquad \gcd(pd,kd-1)=1\Longrightarrow a_j=ja_1
Detailed analysis

Choose kk so large that b=akd−1>c2tb=a_{kd-1}>c^2t, then choose a prime p>bp>b with p∤(kd−1)p\nmid(kd-1) and sufficiently large that pb>c2(pt+b)pb>c^2(pt+b). Since apd=pta_{pd}=pt, we have lcm⁡(apd,b)≥pb\operatorname{lcm}(a_{pd},b)\ge pb, so Step 2 applies. The pair (pd,kd−1)(pd,kd-1) has gcd 11, because gcd⁡(d,kd−1)=1\gcd(d,kd-1)=1 and p∤(kd−1)p\nmid(kd-1). Steps 3–5 therefore give aj=ja1a_j=j a_1 for all jj (in the normalized sequence).