{"id":"06830111-41e6-491f-8690-487291714cb9","arxiv_id":"1908.03855","paper_version":2,"verdict":"CONDITIONAL","confidence":"HIGH","novelty_score":6.0,"correctness_risk":"low","formal_verification":"none","parameter_count":0,"one_line_summary":"The infinite decimal formed by the digit right before the last non-zero digit of n^n is transcendental.","lead":"A short number theory paper defines a decimal from the digit just before the last non-zero digit of n^n and proves this decimal is transcendental. It extends a known technique by Dresden to a new digit position.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Lemma 2.4 Case 2's 'Equivalently' step hides an essential modular identity; without it, the periodicity of R_0 and R_1 is unsupported.","rationale":"The reader identified the unstated modular fact in Lemma 2.4 Case 2 as the weakest assumption, and this stress-test agrees: that step is load-bearing because Lemma 2.4 underpins the rationality of R_0 and R_1, which are needed to form the rational approximants p_n/q_n for the Roth argument. The concern is not that the central claim is false; the modular identity appears to be true and provable by a case split, and the approximation argument for P_n against 7/(10^{2·10^n}-1) is sound. However, as written, the proof leaves a nontrivial step unjustified, and it also contains a false intermediate divisibility claim in Case 1. These are fixable, but a strict referee should require the missing justifications before accepting the paper unconditionally. The reader's verdict of CONDITIONAL is therefore appropriate, and this stress-test does not move the verdict.","tokens_in":5556,"tokens_out":39786,"duration_ms":353767,"concrete_test":"Run a small script to verify, for every integer a with 1 ≤ a ≤ 1000 and 10 ∤ a, the congruence (a+10)^{10a+100} ≡ a^{10a} (mod 100). If a counterexample appears, Lemma 2.4 Case 2 is false and the periodicity argument collapses. If the congruence holds for the tested range and the case analysis confirms it in general, the gap is an omitted justification rather than a fatal flaw.","verdict_should_be":"UNCHANGED","load_bearing_attack":"The proof of Lemma 2.4 Case 2 jumps from the congruence n'^n ≡ (n'+10)^{n+100} (mod 100) to the congruence n'^n ≡ n'^n n'^{100} (mod 100) with the word 'Equivalently'. This is not self-evident because n'+10 is not congruent to n' modulo 100. Via the binomial theorem the step reduces to proving (n'+10)^{10n'+100} ≡ n'^{10n'+100} mod 100, which in turn requires the nontrivial identity n'^{10n'}(n'^{100}-1) ≡ 0 mod 100 for every n' not divisible by 10. The paper supplies no proof of this identity, and a case split on the last digit of n' (even, 5, or coprime to 10) is necessary. Since Lemma 2.4 is used to prove that R_0 and R_1 are rational, this gap directly affects the construction of the infinitely many rational approximants that drive the transcendence argument. Separately, Lemma 2.4 Case 1 contains a false divisibility claim: it states that 75 | (n^{100}-1) because n^{100} ends in 76, but 75 does not divide every number ending in 75. For example, 12^{100}-1 is not divisible by 75. The needed conclusion still follows from the valid facts 4 | n^n and 25 | n^{100}-1, so this is a local, repairable error.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper defines P = 0.d1d2d3..., where dn is the digit immediately preceding the last non-zero digit of n^n (with a convention when n^n has exactly one non-zero digit), and proves that P is transcendental (Theorem 1.3). The proof follows Dresden's method: it establishes periodicities of the rbln function (Lemmas 2.1 and 2.4), then forms rational approximants p_n/q_n with |P - p_n/q_n| < 1/q_n^{2.4} for infinitely many n, which contradicts Roth's theorem if P were algebraic.","tokens_in":5881,"tokens_out":26189,"duration_ms":210473,"significance":"If the proof is completed, this is a genuine new example of a transcendental number defined by a natural digit sequence, extending Dresden's earlier results on last non-zero digits of n^n and n!. The construction of the approximants is explicit and the argument is self-contained modulo the standard Roth theorem. The paper is clearly written and accessible, and it gives appropriate credit to prior work, especially Dresden's technique. The significance is moderate: it is a new instance in an established framework rather than a new method, but the rbln digit is a natural and previously unstudied variant, so the result is worth publishing if the proof gaps are fixed. The paper also includes a useful historical summary of transcendental numbers, though this part is longer than the novel mathematical content.","major_comments":[{"comment":"The 'Equivalently' step is not justified. From n'^{10n'} ≡ (n'+10)^{10n'+100} mod 100, the text immediately replaces (n'+10)^{10n'+100} by n'^{10n'+100} and reduces the desired congruence to n'^{10n'}(n'^{100}-1) ≡ 0 mod 100. This is not immediate because n'+10 is not congruent to n' modulo 100. The congruence (n'+10)^{10n'+100} ≡ n'^{10n'+100} mod 100 is in fact true (it follows from the binomial theorem, since (10n'+100)·10 ≡ 0 mod 100), but this argument is not supplied. Moreover, the further reduction to n'^{10n'}(n'^{100}-1) ≡ 0 mod 100 requires a case split on the last digit of n' (for example, when n' is coprime to 10 one needs n'^{100} ≡ 1 mod 100, and when n' ends in 5 one needs 25 | n'^{10n'} and 4 | n'^{100}-1). Because Lemma 2.4 is used to show that R_0 and R_1 are rational, and these rationals are part of the approximants p_n/q_n, this gap is load-bearing and should be filled with a explicit proof.","section":"Section 2.1, Lemma 2.4, Case 2"}],"minor_comments":[{"comment":"The claim '75 | (n^{100} - 1)' is false; for example, 12^{100} ends in 76, but 12^{100} - 1 is not divisible by 3, hence not by 75. The desired conclusion 100 | n^n(n^{100}-1) still follows from 4 | n^n and 25 | n^{100}-1, so the proof is repairable, but the false statement should be corrected.","section":"Section 2.1, Lemma 2.4, Case 1"},{"comment":"The statement 'Because the denominator of each R_i divides t_n' is false for n=1, since R_0 has denominator 10^{100}-1 while t_1 = 10^{20}-1. The construction works for all n≥2, which still supplies infinitely many approximants and proves transcendence, so the claim should be restricted to n≥2.","section":"Section 2.2, construction of p_n/q_n"},{"comment":"In the proof of item (1), the dichotomy 'If rbln(n')=0' is irrelevant; the argument only uses that n' ends in 2, and the phrase should be removed or clarified.","section":"Section 2.1, Lemma 2.1, proof of item (1)"},{"comment":"There are several minor typographical and formatting issues: the Latin word in the introduction is given as 'transcend˘ere' but should be 'transcendere'; in the displayed rational representation of N there is an errant space in both numerator and denominator; and in the future work section, 'is the decimals' should be 'are the decimals'.","section":"General presentation"}],"recommendation":"major_revision","confidential_remarks":"The paper is a short, well-motivated contribution that fits the journal's scope. The main theorem is likely correct and the proof strategy is sound, but the current version has a genuine gap in Lemma 2.4 Case 2 that must be repaired before the proof is complete. The other issues are local and easily fixable. I recommend major revision rather than rejection, as the gap is within the scope of a normal revision."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Short version: this is a modest, correct extension of Dresden's technique to the digit before the last non-zero digit of n^n. The main theorem—that the decimal formed from those digits is transcendental—is proved by a sound Roth argument. The paper has two local blemishes, both repairable, and neither threatens the result.\n\nWhat's new: Dresden (and later Ikeda-Matsuoka) looked at the last non-zero digit. Chu moves one digit to the left, defines rbln(n^n), and shows the decimal is transcendental. The key lemmas establish that for multiples of 100 the rbln digit depends only on the last non-zero digit of n, and that rbln is invariant under n → n+100 when 100∤n. The proof then proceeds exactly as Dresden's: build rational approximants from periodic blocks, and use Roth's theorem. The construction is clean and the approximant denominator analysis checks out.\n\nSoft spots:\n- In Lemma 2.4 Case 1, the line '75 | (n^100 - 1)' is false. For n ending in 2 (e.g., 12), n^100 ends in 76, so n^100-1 ends in 75, but it need not be divisible by 75: 12^100 ≡ 0 mod 3, so n^100-1 ≡ 2 mod 3. The desired conclusion 100 | n^n(n^100-1) still follows from 4 | n^n and 25 | n^100-1. This is a one-line repair.\n- Lemma 2.4 Case 2: the 'Equivalently' step is under-explained. It's valid, but only after noting that (n'+10)^{n+100} ≡ n'^{n+100} mod 100 because the cross term 10(n+100)n'^{n+99} vanishes mod 100 when n=10n'. Then the required congruence follows by applying Case 1 to n'. The paper skips these details.\n- Lemma 2.1 uses a computer check that (ℓd)^100 ends in 01 for 40 values. That's fine, but a short modular proof would make the paper self-contained.\n\nNone of this is load-bearing. The central argument holds up. The paper is honest about being a direct adaptation of Dresden's method, and it cites the relevant work.\n\nWho it's for: people interested in digit-based transcendence. It's a small step, not a breakthrough, but it's a correct new result. I'd send it to peer review and accept after minor revision.","headline":"A modest but correct extension of Dresden's technique to the rbln digit; two local repairable errors, main theorem stands.","tokens_in":6376,"tokens_out":13176,"would_cite":false,"duration_ms":104250,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11J81","11J82","11A63"],"pacs":[],"model":"deepseek-v4-flash","headline":"The decimal formed from the digit immediately before the last non-zero digit of $n^n$ is transcendental.","keywords":["transcendental number","rbln digit","last non-zero digit","n^n","decimal expansion","rational approximation","transcendence criterion","digit sequence"],"falsifier":"Compute $d^{100} \\bmod 100$ for $d = 1,2,\\ldots,9$ directly; the proof requires exactly 01 for odd digits other than 5, 76 for even digits, and 25 for 5, and any violation would invalidate the periodicity lemma $\\mathrm{rbln}(n^n) = \\mathrm{rbln}((n+100)^{n+100})$ on which the theorem rests.","tokens_in":107,"feed_emoji":"🔢","tokens_out":14949,"duration_ms":200099,"temperature":0.7,"pith_summary":"This paper proves that the infinite decimal obtained by writing, for each positive integer $n$, the digit immediately before the last non-zero digit of $n^n$, is transcendental. Earlier results had shown that the last digit of $n^n$ forms a rational decimal and that the last non-zero digit forms a transcendental decimal; this paper moves one step further to the left. The proof adapts a known rational-approximation technique: it exhibits infinitely many rational numbers with denominators $10^{2\\cdot 10^n}-1$ that approximate the decimal with error below $1/q^{2.4}$, a rate that the classical theorem on algebraic numbers forbids. If correct, the paper adds a new explicit transcendental decimal built from a simple digit rule.","feed_headline":"A new transcendental number from n^n's penultimate nonzero digit","feed_subtitle":"By approximating the decimal more closely than any algebraic number can be, the proof shows it is transcendental.","key_machinery":"The key object is the function $\\mathrm{rbln}(m)$, the digit just before the last non-zero digit of $m$, with $\\mathrm{rbln}(m)=0$ when $m$ has a single non-zero digit. The argument rests on two periodicity facts: for $n$ not divisible by $100$, $\\mathrm{rbln}(n^n) = \\mathrm{rbln}((n+100)^{n+100})$, and for $n$ divisible by $100$ the value is forced by the last non-zero digit of $n$ to be $7$, $2$, or $0$ according as that digit is even, $5$, or odd and not $5$. These facts make the sub-sequence of digits at positions $10^n, 2\\cdot 10^n, \\ldots$ equal to $0,7,0,7,2,7,0,7,0,0$ for every $n\\ge 2$, and the denominator $10^{2\\cdot 10^n}-1$ comes from recognizing that repeated block as the decimal expansion of $7/(10^{2\\cdot 10^n}-1)$.","core_discovery":"The main result is Theorem 1.3: let $P = 0.d_1d_2d_3\\ldots$ with $d_n = \\mathrm{rbln}(n^n)$, where $\\mathrm{rbln}(m)$ is the digit immediately to the left of the last non-zero digit of $m$ (and is $0$ when $m$ has only one non-zero digit). Then $P$ is transcendental. The proof shows that the digits of $P$ at positions that are multiples of $10^n$ stabilize to the repeating pattern $0,7,0,7,2,7,0,7,0,0$ as $n$ grows, so $P$ can be written as a sum of periodic rational blocks plus a tail very close to $7/(10^{2\\cdot 10^n}-1)$. Truncating this decomposition gives rationals $p_n/q_n$ with $q_n = 10^{2\\cdot 10^n}-1$ and $|P - p_n/q_n| < 1/q_n^{2.4}$. By the classical theorem on rational approximations to algebraic numbers, such good approximations can happen infinitely often only if $P$ is not algebraic.","pith_inferences":["Editorial inference: the same residue facts underlying the proof (100th powers ending in 01, 76, or 25 depending on the last digit) would likely control $\\mathrm{rbln}(n!)$ as well, so the same rational-approximation technique may produce a transcendental decimal from factorials; the paper does not examine that case.","Editorial inference: because the approximation exponent $2.4$ is well above the threshold $2$, the argument could tolerate some irregularity in the repeating pattern; even a sparse infinite set of block lengths satisfying the pattern would still force transcendence.","Editorial inference: the periodicity facts imply a fixed 10-adic limit for the subsequence of digits at positions $10^n$, and searching for analogous 10-adic regularities in other digit functions could yield further transcendental decimals."],"forward_implications":["$P$ is transcendental, so in particular it is not rational and its decimal expansion is not eventually periodic.","The construction gives an explicit infinite family of rational approximations with error exponent $2.4$, a concrete case where the rational-approximation criterion forces non-algebraicity.","The same digit-shift moves from a rational construction (the last digit of $n^n$) to a transcendental one, showing that the position just left of the last non-zero digit carries genuinely new information.","The paper enlarges the family of digit-generated decimals known to be transcendental by one explicit example whose digit rule is easy to state."],"supporting_citations":[{"why":"Supplies the rational-approximation technique used to prove transcendence.","marker":"[4]"},{"why":"Introduces the last non-zero digit construction and proves the resulting decimal is irrational, the starting point extended here.","marker":"[3]"},{"why":"Shows the ordinary last-digit decimal is rational, motivating the shift one digit to the left.","marker":"[6]"}],"fun_headline_variants":["Penultimate digits of n^n yield a new transcendental number","New transcendental number from n^n's penultimate digit","n^n's penultimate nonzero digit yields a transcendental","A transcendental from the digit before n^n's last nonzero digit","The penultimate digit of n^n creates a transcendental"],"cache_read_input_tokens":8448,"weakest_assumption_plain":"The proof's core step rests on the unproved modular claim that $d^{100}$ modulo 100 depends only on the last digit $d$, taking values 01, 76, or 25; if that fact fails, the periodic pattern that produces the rational approximations is not established.","fun_headline_variants_meta":{"raw":{"variants":["Penultimate digits of n^n yield a new transcendental number","New transcendental number from n^n's penultimate digit","n^n's penultimate nonzero digit yields a transcendental","A transcendental from the digit before n^n's last nonzero digit","The penultimate digit of n^n creates a transcendental"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000498,"raw_usage":{"total_tokens":2381,"prompt_tokens":831,"completion_tokens":1550,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":447,"completion_tokens_details":{"reasoning_tokens":1471}},"tokens_in":447,"tokens_out":1550,"duration_ms":11337,"temperature":1.0,"reasoning_tokens":1471,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-14T14:01:34.006610+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Compute $d^{100} \\bmod 100$ for $d = 1,2,\\ldots,9$ directly; the proof requires exactly 01 for odd digits other than 5, 76 for even digits, and 25 for 5, and any violation would invalidate the periodicity lemma $\\mathrm{rbln}(n^n) = \\mathrm{rbln}((n+100)^{n+100})$ on which the theorem rests.","supporting_citations":[{"cited_title":"Dresden, Three transcendental numbers from the last non-zero digits of nn, F n, and n!, Math","cited_arxiv_id":null,"evidence_quote":"Supplies the rational-approximation technique used to prove transcendence."},{"cited_title":"Dresden, Two irrational numbers from the last non-zero digits ofnn and n!, Math","cited_arxiv_id":null,"evidence_quote":"Introduces the last non-zero digit construction and proves the resulting decimal is irrational, the starting point extended here."},{"cited_title":"Euler and J","cited_arxiv_id":null,"evidence_quote":"Shows the ordinary last-digit decimal is rational, motivating the shift one digit to the left."}],"review_version":1}