REVIEW 2 major objections 5 minor 13 references
A Recreational Application of Two Integer Sequences and the Generalized Repetitious Number Puzzle
T0 review · 2 major / 5 minor · reviewed 2026-08-14 · deepseek-v4-flash
Pith's one-line read Repeating a k-digit block r times always produces the block multiplied by one fixed co-divisor, so dividing by its prime factors recovers the block.
desk verdict A correct but very small recreational extension of Gardner's repetitious-number trick; Theorem 3.5 has a fixable but real flaw when the co-divisor has repeated prime factors. 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 central object is the $(k,r)$ co-divisor: the integer $\left(1(0)^{k-1}\right)_{r-1}1$ that multiplies a $k$-digit generator to produce its $r$-fold repetition. It carries the argument because the whole proof reduces to recognizing that the repetitious number is exactly $g$ times this co-divisor; afterward, the puzzle becomes the problem of factoring the co-divisor. The Fundamental Theorem of Arithmetic supplies the finite divisor sequence, and ordinary positional block notation supplies the factorization that starts the argument.
What would settle it
Test any single instance by hand: for $g=45$, $k=2$, $r=3$, the repetitious number is $454545$ and the co-divisor is $10101$; $454545$ divided by $10101$ must be exactly $45$. Any one case where the co-divisor does not evenly divide the repetitious number would disprove the claim.
Extended reading notes
Core claim
The central claim is Theorem 3.5: given the number formed by writing a $k$-digit block $g$ exactly $r$ times, that number factors as $g$ times the co-divisor $s(k,r)=\left(1(0)^{k-1}\right)_{r-1}1$, a numeral built from the block $1$ followed by $k-1$ zeros, repeated $r-1$ times, and ending in $1$. The proof writes the repetitious number as a sum of $r$ shifted copies of $g$, factors $g$ out, and then applies the Fundamental Theorem of Arithmetic to obtain a finite sequence of prime divisors whose successive division returns $g$. The two catalogued sequences under study appear as special cases: $s(k,2)$ for duplicating any $k$-digit number and $s(3,r)$ for replicating any three-digit number $r$ times.
Load-bearing premise
The proof assumes ordinary base-ten positional notation, in which writing a $k$-digit block after itself means the block multiplied by $10^k$ plus the block; under any other digit-block convention the co-divisor formula would not follow.
Editorial extensions
If this is right
- The classic three-digit, two-repetition puzzle extends to every block length and every replication count: any $k$-digit block repeated $r$ times is undone by dividing through the prime factors of the co-divisor $\left(1(0)^{k-1}\right)_{r-1}1$.
- The order of the divisions is irrelevant, because the divisors are just prime factors of one product; this extends the paper's observation about $7$, $11$, and $13$ to the general case.
- If the co-divisor is prime, a single division suffices; if it is composite, the identity yields a ready-made division relay with one division per prime factor, counting multiplicity.
- Each pair $(k,r)$ defines a member of an infinite family of co-divisor sequences, and the paper's two starting sequences are the special cases $s(k,2)$ and $s(3,r)$.
- The generalized puzzle gives a direct classroom check for a long chain of divisions: whoever performs the final step must end exactly at the original generator.
Reading between the lines
- The identity is essentially base-independent: in any base $b$, the same argument gives $n = g(1 + b^k + \cdots + b^{k(r-1)})$, so the puzzle could be posed in other bases once the corresponding repunit-like co-divisor is factored.
- The practical bottleneck for designing new versions of the puzzle is factoring $s(k,r)$; the paper's worked examples rely on computer-generated factorizations for even moderate parameters, so a usable puzzle needs a known factorization of its co-divisor.
- Because $s(k,r)$ has the form $1 + 10^k + \cdots + 10^{k(r-1)}$, the co-divisor sequences sit naturally inside the classical study of repunits and their prime divisors, where known divisibility results could suggest or rule out puzzle parameters.
- A natural testable extension would be to ask which pairs $(k,r)$ make the co-divisor prime, since those pairs yield the simplest one-step version of the trick.
Signed reviews
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper revisits Martin Gardner's 'Repetitious Number' puzzle, in which a three-digit number is duplicated and then divided successively by 7, 11, and 13. It observes that the divisor 1001 is term a(3) of OEIS A000533 and that, for a k-digit generator duplicated once, the co-divisor is 10^k+1 (Theorem 3.1). It then treats a fixed 3-digit generator replicated r times, obtaining the co-divisor b(r-1) from OEIS A261544 (Theorem 3.3), and states a full generalization: for a k-digit generator g replicated r times, n = g*S with S = (1(0)^{k-1})_{r-1}1 (Theorem 3.5, Eq. (2)). The paper concludes by defining the (l,r) co-divisor number and a family of (l,r) co-divisor sequences that contains A000533 and A261544 as special cases.
Significance. The central factorization identity n = g*S is elementary, correct, and proved by direct algebra from the base-10 positional representation; it cleanly explains the classical trick and produces a family of division puzzles. The connection to OEIS sequences A000533 and A261544 is a pleasant observation. The proofs are self-contained, the worked examples match the given factorizations, and the paper is appropriately modest in scope, presenting a recreational application rather than a deep new theory. The main weakness is a missing multiplicity specification in the recovery procedure, with a concrete data error in Table 1; once this is corrected, the generalized instruction is valid for all k and r.
major comments (2)
- [§3.3, Theorem 3.5 and Eq. (2)] The theorem's recovery instruction must specify that the prime factors are taken with multiplicity. If 'prime divisors' is read in the standard sense of distinct primes, the theorem is false: for k=3, r=9 the co-divisor is 1001001001001001001001001 = 3^2 * 757 * 333667 * 440334654777631, as Table 2 itself indicates, so dividing n=gS by the four distinct prime divisors once divides only by S/3 and leaves 3g. The same issue affects Theorem 3.1: for k=11, 10^11+1 = 11^2 * 23 * 4093 * 8779, so the duplicated number requires division by 11 twice. The fix is to state throughout the statements, proofs, and puzzle instructions that one divides by the prime-power factors p^e, equivalently by the prime factors counted with multiplicity; the current wording 'prime divisors' is ambiguous and, under the standard reading, incorrect.
- [Table 1, row k=11] The factorization 100000000001 = 11 * 23 * 4093 * 8779 is arithmetically incorrect. The product 11*23*4093*8779 equals 9090909091, which is 100000000001/11, and 9090909091 is again divisible by 11; the correct factorization is 11^2 * 23 * 4093 * 8779. Because this row supplies the divisor set for the duplicated 11-digit generator in Theorem 3.1, the table entry propagates the multiplicity error into the proposed puzzle instructions and should be corrected.
minor comments (5)
- [§1.2, definition of A000533] The definition 'a(n) = 10n + 1' appears to be a typo for 'a(n) = 10^n + 1'; as written, a(2) would be 21 rather than the printed value 101.
- [§3.2] The sentence 'The proof of Theorem 3.2 gives us a method' should refer to Theorem 3.3, which is the theorem proved in that subsection.
- [§3.3, proof of Theorem 3.5] The proof says 'express it as a sum of r positive integers both divisible by g'; since there are r summands, the phrase should read 'all divisible by g' to avoid confusion.
- [Table 2, row r=9] The entry '33' in the factorization column is not a prime factor and appears to be a formatting error for 3^2; the row should read 3^2 * 757 * 333667 * 440334654777631.
- [§4, Conclusion] The conclusion says the generalization first takes 'k ≥ 3', but Section 3.1 treats arbitrary k for replication number 2; this should be phrased as 'arbitrary k with r = 2'.
Circularity Check
No significant circularity: the core identity n = g((1(0)^{k-1})_{r-1}1) is derived directly from positional notation, and the co-divisor name is assigned after the fact.
full rationale
The paper's central claim, Theorem 3.5, derives the factorization n = g((1(0)^{k-1})_{r-1}1) by expanding the repetitious number as a sum of shifted generator blocks and factoring out g. This is direct algebra from the standard base-10 positional representation, not an assumption of the conclusion. The quantity (1(0)^{k-1})_{r-1}1 is named the (l,r) co-divisor only after the identity is proved, so the naming is not load-bearing. The OEIS sequences A000533 and A261544 are recognized as special cases of this formula, but the paper does not fit any parameter to data or import an unverified self-citation: the cited sources are Gardner's puzzle, standard number theory, OEIS descriptions, and Wolfram Alpha factorization tables. The factorization tables are external computational checks, not fitted inputs. The proof is self-contained apart from the Fundamental Theorem of Arithmetic and elementary divisibility facts, which are standard and not the paper's own results. There is no prediction that reduces by construction to an input, no self-citation chain supporting the main theorem, and no ansatz smuggled in via citation. A possible correctness concern is that the recovery procedure says to divide by 'prime divisors' without clarifying multiplicity; for example, when the co-divisor contains a repeated prime factor, dividing by distinct prime factors once would not fully recover g. That is a precision or correctness issue, not a circularity, because the underlying identity n = gS remains independently derived. Accordingly, the circularity score is 0.
Assumptions & free parameters
assumptions (3)
- standard math Unique base-10 positional representation of positive integers
- standard math Fundamental Theorem of Arithmetic
- standard math Divisibility properties, including Corollary 2.12 and the Division Algorithm
Cite this review
Pith. "Pith review of A Recreational Application of Two Integer Sequences and the Generalized Repetitious Number Puzzle." pith.science (2026). https://pith.science/paper/7355D3JH
@misc{pith2026190806014,
author = {Pith},
title = {Pith review of: A Recreational Application of Two Integer Sequences and the Generalized Repetitious Number Puzzle},
year = {2026},
howpublished = {\url{https://pith.science/paper/7355D3JH}},
note = {Machine review of arXiv:1908.06014}
}
read the original abstract
In this article, we give a particular recreational application of the sequence A000533 and A261544 in "The On-line Encyclopedia of Integer Sequences" (OEIS). The recreational application provides a direct extension to "The Repetitious Number" puzzle of Martin Gardner contained in The Second Scientific American Book of Mathematical Puzzles and Diversions published in 1961. We then provide a generalization to the repetitious number puzzle and give a related puzzle as an illustrative example. Finally, as a consequence of the generalization, we define a family of sequence in which the sequences A000533 and A261544 belong.
Reference graph
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Reviewed August 14, 2026 · model on record in the stance chip above.
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