%I #125 Jul 25 2023 22:43:51
%S 1,3,3,5,3,9,3,7,5,9,3,15,3,9,9,9,3,15,3,15,9,9,3,21,5,9,7,15,3,27,3,
%T 11,9,9,9,25,3,9,9,21,3,27,3,15,15,9,3,27,5,15,9,15,3,21,9,21,9,9,3,
%U 45,3,9,15,13,9,27,3,15,9,27,3,35,3,9,15,15,9,27,3,27
%N a(n) = d(n^2), where d(k) = A000005(k) is the number of divisors of k.
%C Inverse Moebius transform of A034444: Sum_{d|n} 2^omega(d), where omega(n) = A001221(n) is the number of distinct primes dividing n.
%C Number of elements in the set {(x,y): x|n, y|n, gcd(x,y)=1}.
%C Number of elements in the set {(x,y): lcm(x,y)=n}.
%C Also gives total number of positive integral solutions (x,y), order being taken into account, to the optical or parallel resistor equation 1/x + 1/y = 1/n. Indeed, writing the latter as X*Y=N, with X=x-n, Y=y-n, N=n^2, the one-to-one correspondence between solutions (X, Y) and (x, y) is obvious, so that clearly, the solution pairs (x, y) are tau(N)=tau(n^2) in number. - _Lekraj Beedassy_, May 31 2002
%C Number of ordered pairs of positive integers (a,c) such that n^2 - ac = 0. Therefore number of quadratic equations of the form ax^2 + 2nx + c = 0 where a,n,c are positive integers and each equation has two equal (rational) roots, -n/a. (If a and c are positive integers, but, instead, the coefficient of x is odd, it is impossible for the equation to have equal roots.) - _Rick L. Shepherd_, Jun 19 2005
%C Problem A1 on the 21st Putnam competition in 1960 (see John Scholes link) asked for the number of pairs of positive integers (x,y) such that xy/(x+y) = n: the answer is a(n); for n = 4, the a(4) = 5 solutions (x,y) are (5,20), (6,12), (8,8), (12,6), (20,5). - _Bernard Schott_, Feb 12 2023
%C Numbers k such that a(k)/d(k) is an integer are in A217584 and the corresponding quotients are in A339055. - _Bernard Schott_, Feb 15 2023
%D A. M. Gleason et al., The William Lowell Putnam Mathematical Competitions, Problems & Solutions:1938-1960 Soln. to Prob. 1 1960, p. 516, MAA, 1980.
%D Ross Honsberger, More Mathematical Morsels, Morsel 43, pp. 232-3, DMA No. 10 MAA, 1991.
%D Loren C. Larson, Problem-Solving Through Problems, Prob. 3.3.7, p. 102, Springer 1983.
%D Alfred S. Posamentier and Charles T. Salkind, Challenging Problems in Algebra, Prob. 9-9 pp. 143 Dover NY, 1988.
%D D. O. Shklarsky et al., The USSR Olympiad Problem Book, Soln. to Prob. 123, pp. 28, 217-8, Dover NY.
%D Wacław Sierpiński, Elementary Theory of Numbers, pp. 71-2, Elsevier, North Holland, 1988.
%D Charles W. Trigg, Mathematical Quickies, Question 194, pp. 53, 168, Dover, 1985.
%H Enrique Pérez Herrero, <a href="/A048691/b048691.txt">Table of n, a(n) for n = 1..10000</a>
%H Ovidiu Bagdasar, <a href="/A048691/a048691.pdf">On Some Functions Involving the lcm and gcd of Integer Tuples</a>.
%H Umberto Cerruti, <a href="/A146564/a146564.pdf">Percorsi tra i numeri</a> (in Italian), pages 3-4.
%H Daniele A. Gewurz and Francesca Merola, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL6/Gewurz/gewurz5.html">Sequences realized as Parker vectors of oligomorphic permutation groups</a>, J. Integer Seq., Vol. 6 (2003), Article 03.1.6.
%H John Scholes, <a href="https://web.archive.org/web/20071103150812/http://www.kalva.demon.co.uk/putnam/psoln/psol601.html">Problem A1 of 21st Putnam Competition 1960</a>, 2002. [Wayback Machine link]
%H Wacław Sierpiński, <a href="http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.dl-catalog-556369c7-b6cc-4a5b-be36-bfc8e0ca7cfa">Elementary Theory of Numbers</a>, Warszawa 1964.
%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads and other Mathematical competitions</a>.
%F a(n) = A000005(A000290(n)).
%F tau(n^2) = Sum_{d|n} mu(n/d)*tau(d)^2, where mu(n) = A008683(n), cf. A061391.
%F Multiplicative with a(p^e) = 2e+1. - _Vladeta Jovovic_, Jul 23 2001
%F Also a(n) = Sum_{d|n} (tau(d)*moebius(n/d)^2), Dirichlet convolution of A000005 and A008966. - _Benoit Cloitre_, Sep 08 2002
%F a(n) = A055205(n) + A000005(n). - _Reinhard Zumkeller_, Dec 08 2009
%F Dirichlet g.f.: (zeta(s))^3/zeta(2s). - _R. J. Mathar_, Feb 11 2011
%F a(n) = Sum_{d|n} 2^omega(d). Inverse Mobius transform of A034444. - _Enrique Pérez Herrero_, Apr 14 2012
%F G.f.: Sum_{k>=1} 2^omega(k)*x^k/(1 - x^k). - _Ilya Gutkovskiy_, Mar 10 2018
%F Sum_{k=1..n} a(k) ~ n*(6/Pi^2)*(log(n)^2/2 + log(n)*(3*gamma - 1) + 1 - 3*gamma + 3*gamma^2 - 3*gamma_1 + (2 - 6*gamma - 2*log(n))*zeta'(2)/zeta(2) + (2*zeta'(2)/zeta(2))^2 - 2*zeta''(2)/zeta(2)), where gamma is Euler's constant (A001620) and gamma_1 is the first Stieltjes constant (A082633). - _Amiram Eldar_, Jan 26 2023
%t A048691[n_]:=DivisorSigma[0,n^2] (* _Enrique Pérez Herrero_, May 30 2010 *)
%t DivisorSigma[0,Range[80]^2] (* _Harvey P. Dale_, Apr 08 2015 *)
%o (PARI) A048691(n)=prod(i=1,#n=factor(n)[,2],n[i]*2+1) /* or, of course, a(n)=numdiv(n^2) */ \\ _M. F. Hasler_, Dec 30 2007
%o (MuPAD) numlib::tau (n^2)$ n=1..90 // _Zerinvary Lajos_, May 13 2008
%o (Haskell)
%o a048691 = product . map (a005408 . fromIntegral) . a124010_row
%o -- _Reinhard Zumkeller_, Jul 12 2012
%o (PARI) for(n=1, 100, print1(direuler(p=2, n, (1 - X^2)/(1 - X)^3)[n], ", ")) \\ _Vaclav Kotesovec_, Aug 21 2021
%Y Cf. A000005, A034444, Mobius transf. of A035116, A048785, A061391, A018805, A002088, A015614, A018892, A063647, A182139.
%Y Partial sums give A061503.
%Y For similar LCM sequences, see A070919, A070920, A070921.
%Y Cf. A005408, A124010.
%Y For the earliest occurrence of 2n-1 see A016017.
%Y Cf. A001620, A082633, A217584, A339055.
%K nonn,easy,mult
%O 1,2
%A _David Johnson-Davies_
%E Additional comments from _Vladeta Jovovic_, Apr 29 2001