REVIEW 1 major objections 4 minor 10 references
A New Transcendental Number from $N^N$
T0 review · 1 major / 4 minor · reviewed 2026-08-14 · deepseek-v4-flash
Pith's one-line read The decimal formed from the digit immediately before the last non-zero digit of $n^n$ is transcendental.
desk verdict A modest but correct extension of Dresden's technique to the rbln digit; two local repairable errors, main theorem stands. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
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 $7/(10^{2\cdot 10^n}-1)$.
What would settle it
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.
Extended reading notes
Core claim
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.
Load-bearing premise
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.
Editorial extensions
If this is right
- $P$ 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.
Reading between the lines
- 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.
Signed reviews
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
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.
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 (1)
- [Section 2.1, Lemma 2.4, Case 2] 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.
minor comments (4)
- [Section 2.1, Lemma 2.4, Case 1] 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 2.2, construction of p_n/q_n] 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 2.1, Lemma 2.1, proof of item (1)] 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.
- [General presentation] 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'.
Circularity Check
No circularity: the transcendental number P is defined directly from the rbln digits, and the approximating rationals are constructed from P itself, with no fitted parameter or self-citation chain.
full rationale
The proof of Theorem 1.3 is not circular. P = 0.d1d2... is defined from the rbln digits of n^n, and the approximating rationals p_n/q_n are built explicitly from the decimal structure of P (the periodic rationals R_i and the comparison rational s_n/t_n). The approximation inequality is computed directly from that construction, and Roth's theorem is then applied as an external, independent criterion. No parameter is fitted to the target quantity, and no load-bearing assertion is taken from the author's own prior work; the only cited technique is Dresden's method [4], which is used as a proof strategy rather than as a premise equivalent to the result. The local difficulties in Lemma 2.4 — the unproved modular step "n'^n ≡ n'^n n'^100 mod 100" in Case 2 and the false statement that 75 | (n^100 − 1) because n^100 ends in 76 — are correctness or rigor gaps in the proof of a supporting lemma, not instances of circularity: the modular facts are asserted as claims to be proved, and the target transcendence statement is not assumed. Therefore the derivation chain is self-contained in the sense relevant to circularity analysis.
Assumptions & free parameters
assumptions (2)
- standard math Roth's theorem (Thue-Siegel-Roth): for algebraic α and any ε > 0, there are only finitely many rationals p/q with |α - p/q| < 1/q^{2+ε}.
- standard math Modular arithmetic facts about powers: numbers ending in 76, 25, or 01 have stable powers modulo 100 (Remark 2.2), and a^k mod 100 for k a multiple of 10 depends only on the last digit of a.
Cite this review
Pith. "Pith review of A New Transcendental Number from $N^N$." pith.science (2026). https://pith.science/paper/JKITCJDJ
@misc{pith2026190803855,
author = {Pith},
title = {Pith review of: A New Transcendental Number from $N^N$},
year = {2026},
howpublished = {\url{https://pith.science/paper/JKITCJDJ}},
note = {Machine review of arXiv:1908.03855}
}
abstract
We first give a summary of the history of transcendental numbers then use a nice technique by G. Dresden to prove a new transcendental number. In particular, while previous work looked at the last non-zero digit of $n^n$, we consider the digit right before its last non-zero digit and show that the infinite decimal built from these digits is transcendental.
Reference graph
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Reviewed August 14, 2026 · model on record in the stance chip above.
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