A048691 a(n) = d(n^2), where d(k) = A000005(k) is the number of divisors of k.

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

Offset: 1

Author

David Johnson-Davies

Keywords

• nonn
• easy
• mult

Comments

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-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

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

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

References

• A. M. Gleason et al., The William Lowell Putnam Mathematical Competitions, Problems & Solutions:1938-1960 Soln. to Prob. 1 1960, p. 516, MAA, 1980.
• Ross Honsberger, More Mathematical Morsels, Morsel 43, pp. 232-3, DMA No. 10 MAA, 1991.
• Loren C. Larson, Problem-Solving Through Problems, Prob. 3.3.7, p. 102, Springer 1983.
• Alfred S. Posamentier and Charles 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.
• Wacław Sierpiński, Elementary Theory of Numbers, pp. 71-2, Elsevier, North Holland, 1988.
• James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 91.
• 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 3-4.
• Daniele A. Gewurz and Francesca Merola, Sequences realized as Parker vectors of oligomorphic permutation groups, J. Integer Seq., Vol. 6 (2003), Article 03.1.6.
• John Scholes, Problem A1 of 21st Putnam Competition 1960, 2002. [Wayback Machine link]
• Wacław Sierpiński, Elementary Theory of Numbers, Warszawa 1964.
• Index to sequences related to Olympiads and other Mathematical competitions.

Crossrefs

Cf. A000005, A034444, Mobius transf. of A035116, A048785, A061391, A018805, A002088, A015614, A018892, A063647, A182139.

Partial sums give A061503.

For similar LCM sequences, see A070919, A070920, A070921.

Cf. A005408, A124010.

For the earliest occurrence of 2n-1 see A016017.

Cf. A001620, A082633, A217584, A339055.

Programs

Haskell

a048691 = product . map (a005408 . fromIntegral) . a124010_row
-- Reinhard Zumkeller, Jul 12 2012

Mathematica

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

MuPAD

numlib::tau (n^2)$ n=1..90 // Zerinvary Lajos, May 13 2008

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

PARI

for(n=1, 100, print1(direuler(p=2, n, (1 - X^2)/(1 - X)^3)[n], ", ")) \\ Vaclav Kotesovec, Aug 21 2021

Sage

[sigma(n**2, 0) for n in range(1, 81)] # Stefano Spezia, Jul 14 2025

Formula

a(n) = A000005(A000290(n)).

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, Jul 23 2001

Also a(n) = Sum_{d|n} (tau(d)*moebius(n/d)^2), Dirichlet convolution of A000005 and A008966. - Benoit Cloitre, Sep 08 2002

a(n) = A055205(n) + A000005(n). - Reinhard Zumkeller, Dec 08 2009

Dirichlet g.f.: (zeta(s))^3/zeta(2s). - R. J. Mathar, Feb 11 2011

a(n) = Sum_{d|n} 2^omega(d). Inverse Mobius transform of A034444. - Enrique Pérez Herrero, Apr 14 2012

G.f.: Sum_{k>=1} 2^omega(k)*x^k/(1 - x^k). - Ilya Gutkovskiy, Mar 10 2018

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

Extensions

Additional comments from Vladeta Jovovic, Apr 29 2001