

A048691


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


66



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

A048691 is the 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
a(n) = A055205(n) + A000005(n).  Reinhard Zumkeller, Dec 08 2009


REFERENCES

A. M. Gleason et al., The William Lowell Putnam Mathematical Competitions, Problems & Solutions:19381960 Soln. to Prob. 1 1960 p. 516 MAA
R. Honsberger, More Mathematical Morsels, Morsel 43 pp. 2323, DMA No. 10 MAA 1991
L. C. Larson, ProblemSolving Through Problems, Prob. 3.3.7, p. 102 Springer 1983
A. S. Posamentier & C. 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
W. Sierpiński, Elementary Theory of Numbers, pp. 712 Elsevier, North Holland 1988
Charles W. Trigg, Mathematical Quickies, Question 194 pp. 53;168, Dover 1985


LINKS

Enrique Pérez Herrero, Table of n, a(n) for n = 1..10000
Ovidiu Bagdasar, On Some Functions Involving the lcm and gcd of Integer Tuples
Umberto Cerruti, Percorsi tra i numeri (in Italian), pages 34.
Daniele A. Gewurz and Francesca Merola, Sequences realized as Parker vectors of oligomorphic permutation groups, J. Integer Seqs., Vol. 6, 2003.
J. Scholes, Problem A1 of 21st Putnam Competition 1960
W. Sierpiński, Elementary Theory of Numbers, Warszawa 1964.


FORMULA

tau(n^2) = Sum_{dn} mu(n/d)*tau(d)^2, where mu(n) = A008683(n), cf. A061391.
Multiplicative with a(p^e) = 2e+1.  Vladeta Jovovic, Jul 23 2001
Also a(n) = sum(dn, tau(d)*moebius(n/d)^2).  Benoit Cloitre, Sep 08 2002
Dirichlet g.f. (zeta(s))^3/zeta(2s).  R. J. Mathar, Feb 11 2011
a(n) = Sum_{dn} 2^omega(d).  Enrique Pérez Herrero, Apr 14 2012
a(n) = A000005(A000290(n)).  Omar E. Pol, Oct 24 2013


MATHEMATICA

A048691[n_]:=DivisorSigma[0, n^2] (* Enrique Pérez Herrero, May 30 2010 *)
DivisorSigma[0, Range[80]^2] (* Harvey P. Dale, Apr 08 2015 *)


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
(MuPAD) numlib::tau (n^2)$ n=1..90 // Zerinvary Lajos, May 13 2008
(Haskell)
a048691 = product . map (a005408 . fromIntegral) . a124010_row
 Reinhard Zumkeller, Jul 12 2012


CROSSREFS

Cf. A000005, A035116, A048785, A061391, A018805, A002088, A015614, A018892, A063647, A182139.
Partial sums give A061503.
For similar LCM sequences, see A070919, A070920, A070921.
Cf. A005408, A124010.
Sequence in context: A015909 A029620 A204100 * A248955 A071053 A094439
Adjacent sequences: A048688 A048689 A048690 * A048692 A048693 A048694


KEYWORD

nonn,mult


AUTHOR

David JohnsonDavies


EXTENSIONS

Additional comments from Vladeta Jovovic, Apr 29 2001


STATUS

approved



