GPT 6 Astra Ultra attempt on Math-5574

Erdős Problem #315: strict growth loss away from Sylvester's sequence

Reconstruction of Kamio's comparison proof, 23 September 2026.

1) Statement and normalization

Let $u_1=1$, $u_{j+1}=u_j(u_j+1)$, and $c_0=\lim u_j^{2^{-j}}$. The reference denominator sequence is $u_j+1=2,3,7,43,\ldots$, whose reciprocals sum to one. We prove that every different increasing sequence of positive integers $(a_j)$ with $\sum_j1/a_j=1$ satisfies \[ \liminf_{j\to\infty}a_j^{2^{-j}}<c_0. \] We use the recurrence to define the reference sequence, avoiding any ambiguous shift in a floor formula. The proof follows Kamio's argument, with its finite comparison lemma proved in full. The strict inequality is the essential point.

2) Generalized reference sequences

For an integer $r\ge1$, put $s_1(r)=r+1$ and $s_{j+1}(r)=s_j(r)^2-s_j(r)+1$. Telescoping gives \[ \sum_{i<j}\frac1{s_i(r)}+\frac1{s_j(r)-1}=\frac1r, \qquad s_j(r)-1=r\prod_{i<j}s_i(r).\tag{1} \] In particular $\sum_i1/s_i(r)=1/r$. The quantities $s_j(r)^{2^{-j}}$ strictly decrease, and $(s_j(r)-1)^{2^{-j}}$ strictly increase. Their logarithmic difference tends to zero. Their common limit $c_r$ satisfies \[ \sqrt r<c_r<\sqrt{r+1}. \] Thus $c_r<c_t$ for integers $r<t$. The recurrence also gives \[ s_{i+j-1}(r)=s_i(s_j(r)-1),\qquad c_r^{2^{j-1}}=c_{s_j(r)-1}.\tag{2} \] Here $s_j(1)=u_j+1$, so $c_1=c_0$ in the question's notation.

3) Two finite comparison facts

First suppose $x_1\ge\cdots\ge x_n>0$ and $y_1\ge\cdots\ge y_n>0$ satisfy $\prod_{i\le j}x_i\ge\prod_{i\le j}y_i$ for every $j$. Put $S_j=\sum_{i\le j}(\log x_i-\log y_i)\ge0$. Convexity of the exponential and summation by parts give \[ \sum_i(x_i-y_i)\ge\sum_i y_i(\log x_i-\log y_i) =y_nS_n+\sum_{j<n}(y_j-y_{j+1})S_j\ge0.\tag{3} \] Thus prefix product dominance implies total sum dominance.

Second, let $U>0$ and let $(v_i),(w_i)$ be positive nonincreasing sequences satisfying, for every $n\ge t$, \[ \sum_{i\le n}w_i+U\prod_{i\le n}w_i=U, \qquad \sum_{i\le n}v_i+U\prod_{i\le n}v_i\le U,\tag{4} \] and suppose $v_i\le w_i$ for $i<t$. Then $\sum_{i\le n}v_i\le\sum_{i\le n}w_i$ for every $n$. We prove this by induction. If $v_n\le w_n$, use the preceding prefix inequality. If $\prod_{i\le n}v_i\ge\prod_{i\le n}w_i$, use $4$ directly. In the remaining case $v_n>w_n$ and the full product of $v$ is smaller. Let $m$ be the largest index with $\prod_{i=m}^n v_i<\prod_{i=m}^n w_i$. Then $m<n$, and for every $j\in[m,n]$, \[ \prod_{i=m}^j v_i<\prod_{i=m}^j w_i, \] because for $j<n$ the complementary suffix ratio is at least one by maximality of $m$. Apply $3$ to this block and add the induction hypothesis up to $m-1$. This proves the second comparison fact, including the case $m=1$ with an empty preceding prefix.

4) A first deviation when the target sum is $1/r$

Consider a positive nondecreasing integer sequence $(a_i)$ with sum of reciprocals $1/r$ and $a_1\ne r+1$. The infinite positive tail implies $a_1>r$, hence $a_1\ge r+2$. Set $v_i=1/a_i$ and $U=1/r$. Each remainder is positive and has denominator dividing $r\prod_{i\le n}a_i$, so \[ U-\sum_{i\le n}v_i\ge\frac1{r\prod_{i\le n}a_i} =U\prod_{i\le n}v_i.\tag{5} \]

For $r\ge2$, let $L=r(r+1)^2(r+2)/2$ and define \[ w_1=\frac1{r+2},\quad w_2=\frac2{(r+1)^2},\quad w_i=\frac1{s_{i-2}(L)}\quad(i\ge3). \] This sequence is nonincreasing: the first inequality is $(r+1)^2\ge2(r+2)$, valid for $r\ge2$, and the later ones follow from the large value of $L$ and its increasing reference sequence. Direct calculation gives \[ U-w_1-w_2=Uw_1w_2=1/L. \] Together with $1$, this verifies equality in $4$ for every $n\ge2$. Also $v_1\le w_1$. Apply the second comparison fact with $t=2$ and $5$.

Both infinite sums equal $U$. If $v_i<w_i$ for all sufficiently large $i$, their tail sums would have that strict inequality, while prefix dominance forces the opposite tail inequality. Therefore $v_i\ge w_i$ infinitely often. Along those indices $a_i\le1/w_i$, yielding \[ \liminf_i a_i^{2^{-i}}\le c_L^{1/4}<c_r.\tag{6} \] For the strict step, $2$ gives $c_r^4=c_M$ with $M=s_3(r)-1=r(r+1)(r^2+r+1)$, and \[ M-L=\frac{r^2(r+1)(r-1)}2>0. \] Integer monotonicity of $c_r$ proves $6$.

If $r=1$, use $w_1=w_2=1/3$, $w_3=3/10$ and $w_i=1/s_{i-3}(30)$ for $i\ge4$. This is nonincreasing, its first three entries have sum $29/30$ and product $1/30$, and $1$ proves $4$ for $t=3$. Since $a_1\ge3$ and $a_2\ge a_1$, the two initial comparisons hold. The same reasoning gives \[ \liminf_i a_i^{2^{-i}}\le c_{30}^{1/8}<c_{42}^{1/8}=c_1, \] where $42=s_4(1)-1$. This handles every first-term deviation.

5) Moving a later deviation to the first position

For a sequence different from $s_i(1)$, let $j$ be its first differing index and put $r=s_j(1)-1$, $b_i=a_{i+j-1}$. By $1$, the remaining reciprocal sum is $1/r$, and $b_1\ne r+1$. Section 4 gives $\liminf_i b_i^{2^{-i}}<c_r$. Raising this inequality to $2^{-(j-1)}$ and using $2$ yields \[ \liminf_n a_n^{2^{-n}} <c_r^{2^{-(j-1)}}=c_1=c_0. \] This is the required strict loss, even when the first deviation occurs arbitrarily late.

6) Outcome and sources

SOLVED by reconstruction. All comparison and strictness steps are supplied. The argument even allows nondecreasing denominators, a stronger setting than the strictly increasing sequence asked for. It is attributed to Kamio, whose method builds on Soundararajan; the independent resolution by Li and Tang is also recorded on the problem page. No novelty or independent Lean verification is claimed.

Sources: https://www.erdosproblems.com/315, checked 23 September 2026; Y. Kamio, Asymptotic analysis of infinite decompositions of a unit fraction into unit fractions, Lemmas 6--7 and Theorem 8, https://arxiv.org/pdf/2503.02317; Z. Li and Q. Tang, https://arxiv.org/abs/2503.12277. The inequality directions and reciprocal/liminf conversion above are written explicitly rather than relying on typographical errors in the source exposition.

The estimate measures complete coverage of the strict inequality, not a correctness probability.

7) COMPLETION ESTIMATE

COMPLETION: 100%