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A092089 Number of odd-length palindromes among the k-tuples of partial quotients of the continued fraction expansions of n/r, r = 1, ..., n.

%I A092089 #53 Feb 16 2025 08:32:52
%S A092089 1,2,3,4,3,6,3,8,5,6,3,12,3,6,9,12,3,10,3,12,9,6,3,24,5,6,7,12,3,18,3,
%T A092089 16,9,6,9,20,3,6,9,24,3,18,3,12,15,6,3,36,5,10,9,12,3,14,9,24,9,6,3,
%U A092089 36,3,6,15,20,9,18,3,12,9,18,3,40,3,6,15,12,9,18,3,36,9,6,3,36,9,6,9,24,3
%N A092089 Number of odd-length palindromes among the k-tuples of partial quotients of the continued fraction expansions of n/r, r = 1, ..., n.
%C A092089 Suggested by _R. K. Guy_, Mar 26 2004
%C A092089 From _Jianing Song_, Mar 24 2019: (Start)
%C A092089 a(n) is also the number of inequivalent residue classes modulo n where the equivalence relation is defined as [a] ~ [b] (mod n) if and only if there exists some k such that gcd(k, n) = 1 and that a*k^2 == b (mod n). For example, for n = 16, the inequivalent residue classes are {[0], [1], [2], [3], [4], [5], [6], [7], [8], [10], [12], [14]}, so a(16) = 14.
%C A092089 Proof: let S(n) be the set of inequivalent residue classes modulo n, so our goal is to show that |S(n)| = a(n) for all n. By the Chinese Remainder Theorem, if gcd(s, t) = 1, then [a] ~ [b] (mod s*t) if and only if [a] ~ [b] (mod s) and [a] ~ [b] (mod t), so there is a one-to-one correspondence between S(s*t) and S(s) X S(t), that is, |S(n)| is multiplicative. It is obvious that |S(p^e)| = a(p^e), so |S(n)| = a(n) for all n. (End)
%H A092089 Antti Karttunen, <a href="/A092089/b092089.txt">Table of n, a(n) for n = 1..10000</a>
%H A092089 Agustina Czenky, <a href="https://arxiv.org/abs/2404.08084">Diagramatics for cyclic pointed fusion categories</a>, arXiv:2404.08084 [math.QA], 2024. See p. 15.
%H A092089 László Tóth, <a href="http://www.seminariomatematico.polito.it/rendiconti/69-1/97.pdf">Menon's identity and arithmetical sums representing functions of several variables</a>, Rend. Sem. Mat. Univ. Politec. Torino, 69 (2011), 97-110 (see (37) in Corollary 15, p. 108).
%H A092089 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/PartialQuotient.html">Partial quotient</a>.
%F A092089 Conjecture: Let n = (2^k0)*(p1^k1)*(p2^k2)*...*(pm^km) be the prime factorization of n where p1, p2, ..., pm are distinct primes. Then a(n) is multiplicative and is given by a(n) = f(k0)*g(k1)*g(k2)*...*g(km), where f(0) = 1, f(1) = 2, f(k) = 4(k-1) if k>1 and g(k) = 2k+1 (This has been verified for n = 1-10000.) [Corrected by _Jianing Song_, Mar 24 2019]
%F A092089 Multiplicative with a(p^e) = 2e+1 if p is odd; a(2) = 2, a(2^e)= 4*(e-1), if e > 1. - _Michel Marcus_, Jun 26 2014
%F A092089 Dirichlet g.f.: zeta(s)^3/zeta(2*s) * (1 - 1/2^s + 1/2^(2*s-1)). - _Jianing Song_, Mar 25 2019 [corrected by _Amiram Eldar_, Dec 18 2023]
%F A092089 Sum_{k=1..n} a(k) ~ (3/Pi^2) * n * (log(n^2) + c_1 * log(n) + c_2), where c_1 = 6 * gamma - 2 - log(2) - 4*zeta'(2)/zeta(2) = 3.04999078122..., gamma is Euler's constant (A001620), c_2 = 2 - 6 * gamma + 6 * gamma^2 + log(2) - 3 * gamma * log(2) + 3*log(2)^2/2 - 6 *gamma_1 + 4*zeta'(2)/zeta(2) + (2 * log(2) - 12 * gamma) * zeta'(2)/zeta(2) + 8 * (zeta'(2)/zeta(2))^2 - 4 * zeta''(2)/zeta(2) = -0.1743888255..., and gamma_1 is the 1st Stieltjes constant (A082633). - _Amiram Eldar_, Dec 18 2023
%e A092089 [1, 2, 1, 2, 1] <-> 1+1/(2+1/(1+1/(2+1/1))) = 15/11 is one of the nine palindromes {[15], [5], [3, 1, 3], [3], [1, 1, 1], [1, 2, 1, 2, 1], [1, 3, 1], [1, 13, 1], [1]} among the continued fraction expansions of 15/r for r = 1..15. Thus a(15)=9.
%t A092089 Table[Apply[Times, FactorInteger[n] /. {p_, e_} /; p > 0 :> Which[p == 1, 1, OddQ@ p, 2 e + 1, And[p == 2, e == 1], 2, True, 4 (e - 1)]], {n, 89}] (* _Michael De Vlieger_, Sep 11 2017 *)
%o A092089 (PARI) a(n) = if (n % 2, numdiv(n^2), if (n/2 % 2, 2*numdiv((n/2)^2), val = valuation(n, 2); 4*(val-1)*numdiv((n/2^val)^2))); \\ _Michel Marcus_, Jun 26 2014
%o A092089 (Scheme) (define (A092089 n) (cond ((= 1 n) n) ((zero? (modulo n 4)) (* 4 (+ -1 (A067029 n)) (A092089 (A000265 n)))) ((even? n) (* 2 (A092089 (/ n 2)))) (else (* (+ 1 (* 2 (A067029 n))) (A092089 (A028234 n)))))) ;; _Antti Karttunen_, Sep 11 2017
%Y A092089 Cf. A001620, A201994, A306016.
%K A092089 nonn,easy,mult
%O A092089 1,2
%A A092089 _John W. Layman_, Mar 29 2004