{"id":"c15e1a99-411c-4ddf-bc6f-09a512b161f1","arxiv_id":"1908.00713","paper_version":1,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":5.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"Even-length palindromes are weak additive Ramanujan-Hardy numbers, squares of palindromes with at least two digits are weak multiplicative Ramanujan-Hardy numbers, and fixed extra terms admit only finitely many such numbers.","lead":"This paper introduces two new families of integers, the weak additive and weak multiplicative Ramanujan-Hardy numbers, defined using digit sums, extra terms, and reversals. It proves that all even-length palindromes belong to the first family and all squares of two-or-more-digit palindromes belong to the second.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Proposition 17(a) Case 3 relies on a false inequality (13); for a=b−1 it is positive for all b>3, so the central inclusion 'squares of palindromes are wMRH' is not proved as written.","rationale":"The strongest claim requires s_b(P^2) ≤ P for all two-digit palindromes. The reader's weakest assumption correctly points at this inequality and the finite checks in Section 6, but the more serious defect is that Case 3's global estimate is invalid for a whole family, not just a finite subcase. The displayed evaluation f(b−1)=6−7b is arithmetically wrong (correct value 3(b−1)(b−3)), and the inequality (13) fails already at b=4,a=3. Since this is the keystone of the proof of Proposition 17(a), the central multiplicative inclusion is unproven as written. The claim may still be true; all small examples we checked satisfy the inequality. Therefore the reader's CONDITIONAL verdict remains appropriate, but the requested revision is more substantial than correcting a few finite checks and arithmetic typos: the Case 3 argument must be replaced by a correct digit-sum bound. I did not find any indication that the theorem itself is false.","tokens_in":15209,"tokens_out":9601,"duration_ms":85642,"concrete_test":"Evaluate Section 6, Case 3 at b=4, a=3: inequality (13) becomes 9 ≤ 0, immediately falsifying the printed bound; then, to decide whether the underlying claim is true despite the proof gap, directly compute s_b([aa]_b^2) for all 2 ≤ b ≤ 10^6 and 1 ≤ a ≤ b−1 and compare with [aa]_b. If every case satisfies s ≤ [aa]_b, the claim survives and only a corrected Case 3 proof is needed; if any case fails, Proposition 17(a) is false.","verdict_should_be":"UNCHANGED","load_bearing_attack":"The multiplicative half of the abstract's claim rests on Proposition 17(a): for every palindrome P with at least two digits, A = P − s_b(P^2) must be nonnegative, i.e. s_b(P^2) ≤ P. Prop 5(c) covers P with at least three digits, so the two-digit case P=[aa]_b is handled in Section 6. In Case 3 (a^2 ≥ b), the proof asserts that each carry reduces the digit sum by b and raises it by 1, that there are at least four carries, and hence s_b(P^2) ≤ 4a^2 − 4b + 4. This bound is then used to derive inequality (13): 4a^2 − a(b+1) + 4(1−b) ≤ 0. But at a = b−1 the left-hand side is 3(b−1)(b−3), not the printed '6 − 7b'; it is positive for every b > 3. For example, (b,a) = (4,3) gives (13) as 9 ≤ 0, which is false. Although the underlying inequality s_b(P^2) ≤ P still holds in that example (P=[33]_4, P^2=[3201]_4, s=6), the proof's estimate is far too weak, so the argument in Case 3 fails for infinitely many two-digit palindromes, not only for the finite list marked 'can be checked numerically.' Thus Proposition 17(a), and with it the central claim that all squares of palindromes with at least two digits are wMRH numbers, is not established by the printed proof.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper introduces two new classes of integers, b-wARH and b-wMRH numbers, which relax the earlier b-ARH and b-MRH definitions by replacing the multiplier with an extra term A. The main claims are that every palindrome with an even number of digits, or with an odd number of digits and even middle digit, is b-wARH, and that every square of a palindrome with at least two digits is b-wMRH. The proofs rest on digit-sum inequalities in Proposition 5, a case analysis for two-digit palindromes, and several corollaries about infinitude, divisibility, and digit-length bounds. The paper also contains computational tables and OEIS references.","tokens_in":15562,"tokens_out":20656,"duration_ms":191081,"significance":"If established, the two inclusion theorems are attractive and give a large, easily described supply of numbers in the new classes. The lemma s_b(P^2) <= P for palindromic P is a useful observation, and Corollary 20, stating that every integer divides some b-wMRH number, is striking. The computational tables and OEIS entries are valuable additions. However, the manuscript as written does not prove the central inclusions: the main two-digit case for Proposition 17(a) contains a false inequality, Proposition 5(a) is false as stated, and the proof of Proposition 5(c) has a digit-count gap. These issues are load-bearing for the abstract's claims, so the paper needs substantial revision before the results can be accepted.","major_comments":[{"comment":"The proof of Case 3 in Proposition 17(a) is invalid. The quadratic in (13), f(a)=4a^2-a(b+1)+4(1-b), evaluates at a=b-1 to 3(b-1)(b-3), not the printed \"6-7b\"; this is positive for every b>3. Hence the claimed inequality fails at a=b-1, so the upper bound s_b(P^2) <= 4a^2-4b+4 is not shown to be at most P. Consequently the argument does not establish s_b(P^2) <= P for the infinitely many two-digit palindromes [b-1,b-1]_b. The underlying statement may still be true, but the proof as written does not prove it.","section":"§6, Case 3, Eq. (13)"},{"comment":"The proof of Proposition 5(c) asserts that N^2 has 2n-1 digits, but for N=b^n-1 one has N^2=b^{2n}-2b^n+1, which has 2n digits; for example, 999^2=998001 in base 10. The digit-sum bound should therefore be (b-1)*2n, and the printed inequality (12), with 2n-1, does not suffice. The text also states that (12) holds for n=3 and b>=2, but for b=2,n=3 it reads 5<=4, forcing the subsequent numerical check. Thus the load-bearing inequality s_b(N^2) <= N for N with at least three digits is not established by the printed proof.","section":"§3, Prop. 5(c), Eqs. (11)-(12)"},{"comment":"The sentence \"Clearly b) implies a)\" is incorrect, since the right-hand side of b) contains N*b rather than N; in fact Proposition 5(a) is false as stated. A counterexample is N=[11]_2=3, for which s_b(N)=2 and 2s_b(N)=4>3. This lemma is cited in the proof of Proposition 7(a) to justify A>=0. The construction in Proposition 7 can be repaired by noting directly that the digit string defining N' has value at least its digit sum, but the manuscript's current proof relies on a false statement.","section":"§3, Prop. 5(a)-(b)"}],"minor_comments":[{"comment":"The cases [22]_5, [22]_6, [22]_7, and [22]_8 are dismissed with \"can be checked numerically\" but the computations are not shown; a proof should either give the values of s_b(P^2) and P for these cases or replace the case split with a uniform argument.","section":"§6, Case 2"},{"comment":"The displayed extra terms in Proposition 16 do not satisfy the defining equation. For N_k=[1(0)^k(0)^k1]_b and A_k=b^{2k}-2, one has s_b(N_k)=2, so A_k+s_b(N_k)=b^{2k}; its reversal is 1, giving b^{2k}+1, which is not N_k=b^{2k+1}+1 for k>=1. The intended sequence is likely [1(0)^k1]_b with A=b^{k+1}-2.","section":"§5, Prop. 16"},{"comment":"In Proposition 23, for N_k=([1(0)^{k-1}1]_b)^2 and A_k=b^k-1, one has s_b(N_k)=4, so A_k+s_b(N_k)=b^k+3; multiplying by its reversal does not equal (b^k+1)^2 for general k. The correct extra term appears to be b^k-3, giving A_k+s_b(N_k)=b^k+1, whose square is N_k.","section":"§5, Prop. 23"},{"comment":"The claim that a b-MRH number must be divisible by s_b(N)^2 is not generally true. For example, 90 is a 10-wMRH number with s_10(90)=9, and 9^2=81 does not divide 90. The final divisibility step in Proposition 14 therefore needs a different justification.","section":"§5, Prop. 14"},{"comment":"In the odd-length case of Proposition 7(a), the first displayed equality omits the middle digit in the summand; the printed equation should read [a_1...a_n(a_{n+1}/2)(0)^n]_b + [(a_{n+1}/2)a_n...a_1]_b = N. This is likely a typographical error.","section":"§4, Prop. 7(a)"}],"recommendation":"major_revision","confidential_remarks":"The central claims appear likely to be salvageable, but the manuscript needs substantive mathematical revision, not just copy-editing. The Case 3 bound in Proposition 17 must be replaced by a correct digit-sum estimate, the proof of Proposition 5(c) needs a valid argument for the 2n-digit case, and Proposition 5(a) should be corrected or removed. The false extra-term formulas in Propositions 16 and 23 and the unjustified divisibility step in Proposition 14 should also be fixed before publication."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"What you should know: this paper introduces two new integer classes, b-wARH and b-wMRH, and claims all suitable palindromes are in the first and all squares of palindromes with at least two digits are in the second. The definitions are natural variants of the authors' earlier b-ARH and b-MRH classes, and the additive half is clean. The multiplicative half, however, is not proved as written.\n\nThe additive result, Proposition 7(a), is genuinely nice. The construction chooses A as a shifted palindrome minus the digit sum, and Proposition 5(a) ensures nonnegativity. That part works. The paper also has useful corollaries about divisors and non-Niven examples, and the digit-length bounds in Theorems 38–43 look plausible.\n\nThe trouble is Proposition 17(a), the square-of-palindromes result. For two-digit P=[aa]_b with a^2 ≥ b, the proof bounds s_b(P^2) by 4a^2 − 4b + 4 and then claims inequality (13) holds at the endpoints. At a=b−1 the left side is 3(b−1)(b−3), not the printed 6−7b, and it is positive for every b>3. So the proof fails for infinitely many two-digit palindromes, not just the finite cases left as numeric checks. The underlying claim may still be true — [33]_4 works, for example — but the printed argument does not establish it. This is a load-bearing gap because the abstract's second headline depends on exactly this proposition.\n\nThere are also mechanical errors that look like typos but as printed are false. Proposition 16 gives A_k = b^{2k}−2 for N_k = [1 0^{2k} 1]_b, but the defining equation needs A_k = b^{2k+1}−2. Proposition 23 gives A_k = b^k−1, which is off by two; the calculation needs b^k−3. The table counts for the two classes are swapped (77 vs 365), and some small two-digit subcases are dismissed as \"can be checked numerically.\"\n\nThe additive half and the general framework are solid enough that this paper deserves a serious referee, but the referee needs to redo the two-digit computation. If the case-3 gap is fixed with a sharper carry estimate, the paper is publishable as an elementary number theory note. As it stands, I would not cite it yet. My recommendation: send it to peer review, but flag the specific hole in Proposition 17(a) so the referee actually checks the arithmetic.","headline":"New digit-sum classes with a clean additive half, but the headline square-of-palindromes theorem is not proved as printed; the two-digit case has a false inequality at a=b−1.","tokens_in":16070,"tokens_out":3873,"would_cite":false,"duration_ms":40269,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11B83","11B99"],"pacs":[],"model":"deepseek-v4-flash","headline":"This paper defines two new classes of integers and proves that all palindromes with an even number of digits or an even middle digit are in the first class, while all squares of palindromes with at least two digits are in the second.","keywords":["b-wARH numbers","b-wMRH numbers","palindrome","digit sum","reversal","numeration base","b-Niven numbers","Ramanujan-Hardy numbers"],"falsifier":"For the two-digit palindromes the paper leaves to numerical checking, $P=[22]_6$, $P=[22]_7$, and $P=[22]_8$, find all integers $X$ with $X\\cdot X^R=P^2$ and check whether any of them satisfies $X\\ge s_b(P^2)$. If none does, Proposition 17(a) fails for that base; a single such case would refute the claim that all squares of palindromes with at least two digits are b-wMRH numbers.","tokens_in":15021,"feed_emoji":"🔁","tokens_out":13314,"duration_ms":111954,"temperature":0.7,"pith_summary":"The paper introduces two families of integers, b-wARH and b-wMRH numbers, defined by asking that a number $N$ can be reproduced from the sum of its own base-b digits plus an extra term $A$, either by adding that sum's reversal (wARH) or by multiplying by it (wMRH). Its main results are inclusion theorems: every palindrome with an even number of digits, and every palindrome with an odd number of digits and an even middle digit, is b-wARH, and every square of a palindrome with at least two digits is b-wMRH, for every base $b\\ge 2$. These families strictly contain the previously studied b-ARH and b-MRH numbers. The paper also proves there are infinitely many b-wARH and b-wMRH numbers in every base, gives bounds on how many digits such numbers can have for a fixed extra term, and relates the new classes to b-Niven numbers.","feed_headline":"Palindromes and their squares fill two new number classes","feed_subtitle":"A digit-sum trick with reversals puts even-length palindromes and squares of palindromes into new families, in any base.","key_machinery":"The load-bearing object is the extra term $A$, a nonnegative integer satisfying $N = A+s_b(N)+(A+s_b(N))^R$ for b-wARH numbers and $N=(A+s_b(N))((A+s_b(N))^R)$ for b-wMRH numbers. The proofs work by choosing $A$ so that $A+s_b(N)$ becomes a prescribed number whose reversal is already known: the zero-padded half of the palindrome for the additive class, and the palindrome itself for the multiplicative class. The analytic ingredient that keeps $A$ nonnegative is Proposition 5, especially the digit-sum inequality $s_b(P^2)\\le P$ for integers with at least three base-b digits.","core_discovery":"The central discovery is that a single extra-term construction turns digit sums into reversals that rebuild a number. For an even-length palindrome $N$ whose digits are $a_1\\ldots a_n\\,a_n\\ldots a_1$, the choice $A=[a_1\\ldots a_n\\,0^n]_b - s_b(N)$ makes $A+s_b(N)$ equal to the zero-padded left half of $N$, whose reversal is the right half; adding them reproduces $N$. For a palindrome $P$ with at least two digits, the choice $A=P-s_b(P^2)$ makes $A+s_b(P^2)=P$, so $P^2$ is the product of $P$ with its own reversal. The same equations give the inclusions of the old classes: any b-ARH number is b-wARH and any b-MRH number is b-wMRH, using $A=s_b(N)(M-1)$. The paper claims these two inclusion statements hold in every base $b\\ge 2$.","pith_inferences":["The construction behind Proposition 17(a) actually works for any product $P\\cdot P^R$ when $P$ has at least three base-b digits, not only for palindromic squares; the paper's statement singles out palindromes, but the same choice of $A$ proves the wider claim for this product-with-reversal family.","Because the digit-sum inequality is the only analytic input, the inclusion results may be sensitive to base only through the size of $s_b$; bases with slack in the inequality are likely to admit many alternative extra terms, which connects to the paper's open multiplicity questions.","A systematic scan of $s_b(P^2)-P$ over all two-digit palindromes would settle the finite cases the paper leaves to numerical checking and could reveal whether a uniform proof exists for all bases."],"forward_implications":["Every base-b palindrome of even length, and every base-b palindrome of odd length with an even middle digit, is a b-wARH number for every base $b\\ge 2$.","Every base-b square of a palindrome with at least two digits is a b-wMRH number; more generally, if $P$ has at least three base-b digits, the same construction shows $P\\cdot P^R$ is b-wMRH.","Every b-ARH number is b-wARH and every b-MRH number is b-wMRH, with extra term $A=s_b(N)(M-1)$, so the new classes strictly contain the old ones.","For every base $b\\ge 2$ there are infinitely many b-wARH numbers, infinitely many b-wMRH numbers, and infinitely many corresponding extra terms.","There are infinitely many b-wARH numbers that are not b-MRH numbers and infinitely many b-wMRH numbers that are not b-MRH numbers."],"supporting_citations":[{"why":"Defines the b-ARH and b-MRH classes whose inclusion into the new classes is the main extension; its Theorem 35 is used in the classification of wARH numbers with extra term 0.","marker":"[6]"},{"why":"Supplies the infinite families of b-ARH and b-MRH numbers and the palindrome-multiplication result used in the corollaries about divisors and infinitude of the new classes.","marker":"[8]"},{"why":"Provides the online integer-sequence data used for the computed lists of wARH and wMRH numbers below 10000.","marker":"[9]"},{"why":"Carries the taxicab-number motivation behind the Ramanujan-Hardy terminology.","marker":"[5]"}],"fun_headline_variants":["Even-length palindromes join new class via digit-sum reversal","Squares of palindromes join a new number class","Two larger classes defined by reversing digit-sums","Palindromes and squares give two new number classes","Digit-sum reversal builds two new integer families"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The proof that every square of a palindrome with at least two digits is a b-wMRH number rests on the digit-sum inequality $s_b(P^2)\\le P$; if this fails for some palindrome, the chosen extra term $A=P-s_b(P^2)$ would be negative and the construction would not give a witness.","fun_headline_variants_meta":{"raw":{"variants":["Even-length palindromes join new class via digit-sum reversal","Squares of palindromes join a new number class","Two larger classes defined by reversing digit-sums","Palindromes and squares give two new number classes","Digit-sum reversal builds two new integer families"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000779,"raw_usage":{"total_tokens":3433,"prompt_tokens":926,"completion_tokens":2507,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":542,"completion_tokens_details":{"reasoning_tokens":2431}},"tokens_in":542,"tokens_out":2507,"duration_ms":19836,"temperature":1.0,"reasoning_tokens":2431,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-14T15:36:25.121736+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"For the two-digit palindromes the paper leaves to numerical checking, $P=[22]_6$, $P=[22]_7$, and $P=[22]_8$, find all integers $X$ with $X\\cdot X^R=P^2$ and check whether any of them satisfies $X\\ge s_b(P^2)$. If none does, Proposition 17(a) fails for that base; a single such case would refute the claim that all squares of palindromes with at least two digits are b-wMRH numbers.","supporting_citations":[{"cited_title":"Nit ¸ic˘ a, About some relatives of the taxicab number,J","cited_arxiv_id":null,"evidence_quote":"Defines the b-ARH and b-MRH classes whose inclusion into the new classes is the main extension; its Theorem 35 is used in the classification of wARH numbers with extra term 0."},{"cited_title":"Nit ¸ic˘ a, Infinite sets ofb-additive and b-multiplicative Ramanujan-Hardy numbers, J","cited_arxiv_id":null,"evidence_quote":"Supplies the infinite families of b-ARH and b-MRH numbers and the palindrome-multiplication result used in the corollaries about divisors and infinitude of the new classes."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"Provides the online integer-sequence data used for the computed lists of wARH and wMRH numbers below 10000."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"Carries the taxicab-number motivation behind the Ramanujan-Hardy terminology."}],"review_version":1}