GPT 6 Astra Ultra attempt on Math-6153

Erdős Problem 1099: power sums of consecutive divisor ratios

Statement and scope

For the ordered divisors $1=d_1<\cdots<d_{\tau(n)}=n$, put \[ h_\alpha(n)=\sum_{i=1}^{\tau(n)-1}\left(\frac{d_{i+1}}{d_i}-1\right)^\alpha. \] The question is whether $\liminf_{n\to\infty}h_\alpha(n)<\infty$ for every $\alpha>1$. Vose proved this in Integers with consecutive divisors in small ratio, J. Number Theory 19 $1984$, 233--238, https://doi.org/10.1016/0022-314X(84)90107-0. We do not reconstruct his sufficient construction. We prove a universal lower bound, clarify the critical exponent, and derive the auxiliary sum-of-ratios consequence. The existence assertion remains unresolved in this attempt.

Refining a multiplicative partition decreases the cost

If $a<b<c$ are positive reals and $\alpha\ge1$, write $u=b/a-1$, $v=c/b-1$. Then \[ c/a-1=u+v+uv\ge u+v,\qquad (u+v)^\alpha\ge u^\alpha+v^\alpha. \] Consequently inserting $b$ between $a,c$ cannot increase the sum of powered relative gaps. Repeated insertion proves this for any finite refinement.

Let $m=\lfloor\sqrt n\rfloor$. Refine the divisor list by the positive real numbers \[ 1,2,\ldots,m,\quad n/m,n/(m-1),\ldots,n. \] This list already contains every original divisor: one of $d,n/d$ is at most $\sqrt n$ and is an integer. Delete the repeated central point if $n=m^2$. The ratios in its two halves are $(j+1)/j$, in reverse orders, and its middle ratio is $n/m^2$. Thus \[ \boxed{h_\alpha(n)\ge 2\sum_{j=1}^{m-1}\frac1{j^\alpha} +\left(\frac n{m^2}-1\right)^\alpha.} \tag{1} \] This is a comparison with a refinement by real numbers, not an assertion that all the inserted numbers divide $n$.

In particular, for $\alpha>1$, \[ \liminf_{n\to\infty}h_\alpha(n)\ge2\zeta(\alpha). \] For $\alpha=1$, $1$ grows like $\log n$. For $0<\alpha<1$, the elementary inequality $(e^t-1)^\alpha\ge t$ for all sufficiently small positive $t$, together with $(e^t-1)^\alpha/t$ having a positive infimum on $(0,\infty)$, gives $h_\alpha(n)\gg_\alpha\sum_i\log(d_{i+1}/d_i)=\log n$. Alternatively, the following exact identity immediately handles the endpoint $\alpha=1$: \[ \sum_i\log(d_{i+1}/d_i)=\log n. \] Thus the restriction $\alpha>1$ is necessary. The positive infimum just used follows from the ratio tending to infinity both as $t\downarrow0$ and as $t\to\infty$, and continuity in between.

A necessary abundance of divisors

For $\alpha>1$, let $u_i=d_{i+1}/d_i-1$. Since $\log(1+u_i)\le u_i$, Hölder's inequality gives \[ \log n\le\sum_i u_i \le(\tau(n)-1)^{1-1/\alpha}h_\alpha(n)^{1/\alpha}. \] Any sequence with $h_\alpha(n)\le H$ must therefore satisfy \[ \tau(n)-1\ge H^{-1/(\alpha-1)} (\log n)^{\alpha/(\alpha-1)}. \tag{2} \] Prime powers alone cannot work: $h_\alpha(p^a)=a(p-1)^\alpha\to\infty$. Nor does a uniformly bounded largest divisor ratio by itself suffice; it controls each summand but not their number.

The auxiliary sum of ratios

Carefully retaining the $\tau(n)-1$ gaps, define \[ E(n)=\sum_{i=1}^{\tau(n)-1}\frac{d_{i+1}}{d_i}-\tau(n)-\log n. \] Then \[ E(n)=\sum_i\left(u_i-\log(1+u_i)\right)-1. \] For $u\ge0$, \[ 0\le u-\log(1+u)=\int_0^u\frac t{1+t}\,dt\le\frac{u^2}{2}. \] It follows that \[ -1\le E(n)\le\frac12h_2(n)-1. \] So a bounded sequence for the main question at $\alpha=2$ indeed gives the finite-liminf assertion about $E(n)$. The formula also prevents an off-by-one error in the unpowered sum.

Assessment

The lower bound $1$, the necessity of $\alpha>1$, the divisor-count obstruction $2$, and the auxiliary implication are proved. What is missing is a sequence whose entire logarithmic divisor mesh has uniformly bounded powered cost, including its long central region. Neither numerical examples nor the refinement lower bound supplies such a sequence. Vose's theorem is correctly attributed as the known affirmative answer, not inserted as a proof of our missing construction. Source statement: https://www.erdosproblems.com/1099.

COMPLETION ESTIMATE

COMPLETION: 40%