

A048691


a(n) = d(n^2), where d(k) = A000005(k) is the number of divisors of k.


98



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, 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, 45, 3, 9, 15, 13, 9, 27, 3, 15, 9, 27, 3, 35, 3, 9, 15, 15, 9, 27, 3, 27
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OFFSET

1,2


COMMENTS

Inverse Moebius transform of A034444: Sum_{dn} 2^omega(d), where omega(n) = A001221(n) is the number of distinct primes dividing n.
Number of elements in the set {(x,y): xn, yn, gcd(x,y)=1}.
Number of elements in the set {(x,y): lcm(x,y)=n}.
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=xn, Y=yn, N=n^2, the onetoone 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
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
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


REFERENCES

A. M. Gleason et al., The William Lowell Putnam Mathematical Competitions, Problems & Solutions:19381960 Soln. to Prob. 1 1960, p. 516, MAA, 1980.
Ross Honsberger, More Mathematical Morsels, Morsel 43, pp. 2323, DMA No. 10 MAA, 1991.
Loren C. Larson, ProblemSolving Through Problems, Prob. 3.3.7, p. 102, Springer 1983.
Alfred S. Posamentier and Charles T. Salkind, Challenging Problems in Algebra, Prob. 99 pp. 143 Dover NY, 1988.
D. O. Shklarsky et al., The USSR Olympiad Problem Book, Soln. to Prob. 123, pp. 28, 2178, Dover NY.
Wacław Sierpiński, Elementary Theory of Numbers, pp. 712, Elsevier, North Holland, 1988.
Charles W. Trigg, Mathematical Quickies, Question 194, pp. 53, 168, Dover, 1985.


LINKS



FORMULA

tau(n^2) = Sum_{dn} mu(n/d)*tau(d)^2, where mu(n) = A008683(n), cf. A061391.
Dirichlet g.f.: (zeta(s))^3/zeta(2s).  R. J. Mathar, Feb 11 2011
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


MATHEMATICA



PROG

(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
(Haskell)
a048691 = product . map (a005408 . fromIntegral) . a124010_row
(PARI) for(n=1, 100, print1(direuler(p=2, n, (1  X^2)/(1  X)^3)[n], ", ")) \\ Vaclav Kotesovec, Aug 21 2021


CROSSREFS

Cf. A000005, A034444, Mobius transf. of A035116, A048785, A061391, A018805, A002088, A015614, A018892, A063647, A182139.
For the earliest occurrence of 2n1 see A016017.


KEYWORD

nonn,easy,mult


AUTHOR



EXTENSIONS



STATUS

approved



