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A048691
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tau(n^2), where tau = A000005.
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51
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| A048691 is the inverse Moebius transform of A034444: Sum_{d|n} 2^omega(d), where omega(n) = A001221(n) is the number of distinct primes dividing n.
Number of elements in the set {(x,y): x|n, y|n, 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=x-n, Y=y-n, N=n^2, the one-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 (blekraj(AT)yahoo.com), 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 (rshepherd2(AT)hotmail.com), Jun 19 2005
a(n) = A055205(n) + A000005(n). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Dec 08 2009]
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REFERENCES
| A. M. Gleason et al., The William Lowell Putnam Mathematical Competitions, Problems & Solutions:1938-1960 Soln. to Prob. 1 1960 p. 516 MAA
R. Honsberger, More Mathematical Morsels, Morsel 43 pp. 232-3, DMA No. 10 MAA 1991
L. C. Larson, Problem-Solving Through Problems, Prob. 3.3.7, p. 102 Springer 1983
A. S. Posamentier & C. T. Salkind, Challenging Problems in Algebra, Prob. 9-9 pp. 143 Dover NY 1988
D. O. Shklarsky et al., The USSR Olympiad Problem Book, Soln. to Prob. 123 pp. 28;217-8 Dover NY
W. Sierpinski, Elementary Theory of Numbers, pp. 71-2 Elsevier, North Holland 1988
Charles W. Trigg, Mathematical Quickies, Question 194 pp. 53;168, Dover 1985
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LINKS
| Daniele A. Gewurz and Francesca Merola, Sequences realized as Parker vectors ..., J. Integer Seqs., Vol. 6, 2003.
J. Scholes, Problem A1 of 21-st Putnam Competition 1960
W. Sierpi\'{n}ski, Elementary Theory of Numbers, Warszawa 1964.
Enrique Pérez Herrero, Table of n, a(n) for n = 1..10000
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FORMULA
| tau(n^2) = Sum_{d|n} mu(n/d)*tau(d)^2, where mu(n) = A008683(n), cf. A061391.
Multiplicative with a(p^e) = 2e+1. - Vladeta Jovovic (vladeta(AT)eunet.rs), Jul 23 2001
Also a(n) = sum(d|n, tau(d)*moebius(n/d)^2 ) - Benoit Cloitre (benoit7848c(AT)orange.fr), Sep 08 2002
Dirichlet g.f. (zeta(s))^3/zeta(2s). - R. J. Mathar, Feb 11 2011
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MATHEMATICA
| A048691[n_]:=DivisorSigma[0, n^2] [From Enrique Perez Herrero (psychgeometry(AT)gmail.com), May 30 2010]
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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 (Maximilian.Hasler(AT)gmail.com), Dec 30 2007
(MuPad)numlib::tau (n^2)$ n=1..90 - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 13 2008
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CROSSREFS
| Cf. A035116, A061391, A000005, A018805, A002088, A015614, A018892, A063647.
Partial sums give A061503.
For similar LCM sequences, see A070919-A070921.
Sequence in context: A015909 A029620 A204100 * A071053 A094439 A122037
Adjacent sequences: A048688 A048689 A048690 * A048692 A048693 A048694
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KEYWORD
| nonn,mult
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AUTHOR
| David Johnson-Davies (David(AT)interface.co.uk)
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EXTENSIONS
| Additional comments from Vladeta Jovovic (vladeta(AT)eunet.rs), Apr 29 2001
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