cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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

Original entry on oeis.org

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

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Author

Keywords

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.

Crossrefs

Partial sums give A061503.
For similar LCM sequences, see A070919, A070920, A070921.
For the earliest occurrence of 2n-1 see A016017.

Programs

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

A070919 a(n) = Card{ (x,y,z) | lcm(x,y,z)=n }.

Original entry on oeis.org

1, 7, 7, 19, 7, 49, 7, 37, 19, 49, 7, 133, 7, 49, 49, 61, 7, 133, 7, 133, 49, 49, 7, 259, 19, 49, 37, 133, 7, 343, 7, 91, 49, 49, 49, 361, 7, 49, 49, 259, 7, 343, 7, 133, 133, 49, 7, 427, 19, 133, 49, 133, 7, 259, 49, 259, 49, 49, 7, 931, 7, 49, 133, 127, 49, 343, 7, 133
Offset: 1

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Author

Benoit Cloitre, May 20 2002

Keywords

Comments

A048691(n) gives Card{ (x,y) | lcm(x,y)=n }.

Crossrefs

Programs

  • Mathematica
    Join[{1},Table[Product[(k + 1)^3 - k^3, {k, FactorInteger[n][[All, 2]]}], {n,2, 68}]] (* Geoffrey Critzer, Jan 10 2015 *)
  • PARI
    for(n=1,100,print1(sumdiv(n,d,numdiv(d)^3*moebius(n/d)),","))
    
  • PARI
    a(n) = vecprod(apply(x->(x+1)^3-x^3, factor(n)[, 2])); \\ Amiram Eldar, Sep 03 2023

Formula

a(n) = Sum_{d|n} A000005(d)^3*A008683(n/d).
Sum_{k>0} a(k)/k^s = (1/zeta(s))*Sum_{k>0} tau(k)^3/k^s.
Multiplicative with a(p^e) = 1+3*e+3*e^2 for prime p and e >= 0. - Werner Schulte, Nov 30 2018

A070920 a(n) = Card{ (x,y,z,u) | lcm(x,y,z,u)=n }.

Original entry on oeis.org

1, 15, 15, 65, 15, 225, 15, 175, 65, 225, 15, 975, 15, 225, 225, 369, 15, 975, 15, 975, 225, 225, 15, 2625, 65, 225, 175, 975, 15, 3375, 15, 671, 225, 225, 225, 4225, 15, 225, 225, 2625, 15, 3375, 15, 975, 975, 225, 15, 5535, 65, 975, 225, 975, 15, 2625, 225
Offset: 1

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Author

Benoit Cloitre, May 20 2002

Keywords

Comments

A048691(n) gives Card{ (x,y) | lcm(x,y)=n }.

Crossrefs

Cf. A000005, A008683, A048691, A070919, A070921, A247516 (Mobius transform).

Programs

  • Mathematica
    Join[{1},Table[Product[(k + 1)^4 - k^4, {k, FactorInteger[n][[All, 2]]}], {n,2, 68}]] (* Geoffrey Critzer, Jan 10 2015 *)
  • PARI
    for(n=1,100,print1(sumdiv(n,d,numdiv(d)^4*moebius(n/d)),","))
    
  • PARI
    a(n) = vecprod(apply(x->(x+1)^4-x^4, factor(n)[, 2])); \\ Amiram Eldar, Sep 03 2023

Formula

a(n) = Sum_{d|n} A000005(d)^4*A008683(n/d).
Sum_{k>0} a(k)/k^s = (1/zeta(s))*Sum_{k>0} tau(k)^4/k^s.
Multiplicative with a(p^e) = (e+1)^4 - e^4. - Amiram Eldar, Sep 03 2023

A247517 Card{(x,y,z,t,u): 1<=x,y,z,t,u<=n, gcd(x,y,z,t,u)=1, lcm(x,y,z,t,u)=n}.

Original entry on oeis.org

1, 30, 30, 180, 30, 900, 30, 570, 180, 900, 30, 5400, 30, 900, 900, 1320, 30, 5400, 30, 5400, 900, 900, 30, 17100, 180, 900, 570, 5400, 30, 27000, 30, 2550, 900, 900, 900, 32400, 30, 900, 900, 17100, 30, 27000, 30, 5400, 5400, 900, 30, 39600, 180, 5400, 900
Offset: 1

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Author

Ovidiu Bagdasar, Sep 18 2014

Keywords

Comments

For given n and k positive integers, let L(n,k) represent the number of ordered k-tuples of positive integers, whose GCD is 1 and LCM is n. In this notation, the sequence corresponds to a(n) = L(n,5).

Crossrefs

L(n,2) produces A034444, A245019, A070921.
Cf. A247516.

Programs

  • Mathematica
    f[p_, e_] := 10*(2*e^3 + e); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 03 2023 *)
  • PARI
    a(n) = {f = factor(n); 10^omega(n)*prod(k=1, #f~, 2*f[k, 2]^3+f[k, 2]); }

Formula

For n = p_1^{n_1} p_2^{n_2}...p_r^{n_r} one has
a(n) = Product_{i=1..r} ((n_i+1)^5 - 2*n_i^5 + (n_i-1)^5).
a(n) = 10^omega(n)*Product_{i=1..r} (2n_i^3 + n_i).
Showing 1-4 of 4 results.