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 1-8 of 8 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

A035116 a(n) = tau(n)^2, where tau(n) = A000005(n).

Original entry on oeis.org

1, 4, 4, 9, 4, 16, 4, 16, 9, 16, 4, 36, 4, 16, 16, 25, 4, 36, 4, 36, 16, 16, 4, 64, 9, 16, 16, 36, 4, 64, 4, 36, 16, 16, 16, 81, 4, 16, 16, 64, 4, 64, 4, 36, 36, 16, 4, 100, 9, 36, 16, 36, 4, 64, 16, 64, 16, 16, 4, 144, 4, 16
Offset: 1

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Author

Keywords

References

  • G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, Cambridge, University Press, 1940, p. 59.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, Theorem 304.

Crossrefs

Programs

Formula

Dirichlet g.f.: zeta(s)^4/zeta(2s).
tau(n)^2 = Sum_{d|n} tau(d^2), Dirichlet convolution of A048691 and A000012 (i.e.: inverse Mobius transform of A048691).
Multiplicative with a(p^e) = (e+1)^2. - Vladeta Jovovic, Dec 03 2001
G.f.: Sum_{n>=1} A000005(n^2)*x^n/(1-x^n). - Mircea Merca, Feb 25 2014
a(n) = A066446(n) + A184389(n). - Reinhard Zumkeller, Sep 08 2015
Let b(n), n > 0, be the Dirichlet inverse of a(n). Then b(n) is multiplicative with b(p^e) = (-1)^e*(Sum_{i=0..e} binomial(3,i)) for prime p and e >= 0, where binomial(n,k)=0 if n < k; abs(b(n)) is multiplicative and has the Dirichlet g.f.: (zeta(s))^4/(zeta(2*s))^3. - Werner Schulte, Feb 07 2021

Extensions

Additional comments from Vladeta Jovovic, Apr 29 2001

A321348 a(n) = Sum_{d|n} tau(d^n), where tau() is the number of divisors (A000005).

Original entry on oeis.org

1, 4, 5, 15, 7, 64, 9, 52, 30, 144, 13, 546, 15, 256, 289, 165, 19, 1140, 21, 1386, 529, 576, 25, 3848, 78, 784, 166, 2610, 31, 32768, 33, 486, 1225, 1296, 1369, 12321, 39, 1600, 1681, 10248, 43, 85184, 45, 6210, 6486, 2304, 49, 24250, 150, 7956
Offset: 1

Views

Author

Ilya Gutkovskiy, Nov 06 2018

Keywords

Comments

a(n) is prime iff n is in A001359, which makes the sequence a supersequence of A006512. - Ivan N. Ianakiev, Nov 07 2018

Crossrefs

Programs

  • Magma
    [&+[NumberOfDivisors(d^n): d in Divisors(n)]: n in [1..50]]; // Vincenzo Librandi, Nov 08 2018
    
  • Maple
    with(numtheory): seq(coeff(series(add(tau(k^n)*x^k/(1-x^k),k=1..n),x,n+1), x, n), n = 1 .. 50); # Muniru A Asiru, Nov 25 2018
  • Mathematica
    Table[Sum[DivisorSigma[0, d^n], {d, Divisors[n]}], {n, 50}]
    a[n_] := Times @@ ((#[[2]] + 1) (n #[[2]] + 2)/2 & /@ FactorInteger[n]); a[1] = 1; Table[a[n], {n, 50}]
  • PARI
    a(n) = sumdiv(n, d, numdiv(d^n)); \\ Michel Marcus, Nov 06 2018
    
  • Python
    from math import prod
    from sympy import factorint
    def A321348(n): return prod((e+1)*(n*e+2)>>1 for e in factorint(n).values()) # Chai Wah Wu, Dec 13 2022

Formula

a(n) = [x^n] Sum_{k>=1} tau(k^n)*x^k/(1 - x^k).
If n = Product (p_j^k_j) then a(n) = Product ((k_j + 1)*(n*k_j + 2)/2).
a(prime(n)) = prime(n) + 2 = A052147(n). - Michel Marcus, Nov 25 2018

A356574 a(n) = Sum_{d|n} tau(d^4), where tau(n) = number of divisors of n, cf. A000005.

Original entry on oeis.org

1, 6, 6, 15, 6, 36, 6, 28, 15, 36, 6, 90, 6, 36, 36, 45, 6, 90, 6, 90, 36, 36, 6, 168, 15, 36, 28, 90, 6, 216, 6, 66, 36, 36, 36, 225, 6, 36, 36, 168, 6, 216, 6, 90, 90, 36, 6, 270, 15, 90, 36, 90, 6, 168, 36, 168, 36, 36, 6, 540, 6, 36, 90, 91, 36, 216, 6, 90, 36, 216, 6, 420, 6, 36, 90, 90
Offset: 1

Views

Author

Seiichi Manyama, Dec 13 2022

Keywords

Crossrefs

Programs

  • Mathematica
    Array[DivisorSum[#, DivisorSigma[0, #^4] &] &, 120] (* Michael De Vlieger, Dec 13 2022 *)
    f[p_, e_] := 2*e^2 + 3*e + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Dec 14 2022 *)
  • PARI
    a(n) = sumdiv(n, d, numdiv(d^4));
    
  • PARI
    a(n) = sumdiv(n, d, numdiv(n*d^2));
    
  • PARI
    a(n) = sumdiv(n, d, numdiv(n^2));
    
  • PARI
    a(n) = numdiv(n)*numdiv(n^2);
    
  • PARI
    my(N=80, x='x+O('x^N)); Vec(sum(k=1, N, numdiv(k^4)*x^k/(1-x^k)))
    
  • Python
    from math import prod
    from sympy import factorint
    def A356574(n): return prod((e+1)*((e<<1)+1) for e in factorint(n).values()) # Chai Wah Wu, Dec 13 2022

Formula

a(n) = Sum_{d|n} tau(n * d^2) = Sum_{d|n} tau(n^2).
a(n) = tau(n) * tau(n^2).
G.f.: Sum_{k>=1} tau(k^4) * x^k/(1 - x^k).
Multiplicative with a(p^e) = 2*e^2 + 3*e + 1. - Amiram Eldar, Dec 14 2022

A359037 a(n) = Sum_{d|n} tau(d^6), where tau(n) = number of divisors of n, cf. A000005.

Original entry on oeis.org

1, 8, 8, 21, 8, 64, 8, 40, 21, 64, 8, 168, 8, 64, 64, 65, 8, 168, 8, 168, 64, 64, 8, 320, 21, 64, 40, 168, 8, 512, 8, 96, 64, 64, 64, 441, 8, 64, 64, 320, 8, 512, 8, 168, 168, 64, 8, 520, 21, 168, 64, 168, 8, 320, 64, 320, 64, 64, 8, 1344, 8, 64, 168, 133, 64, 512, 8, 168, 64, 512, 8, 840, 8
Offset: 1

Views

Author

Seiichi Manyama, Dec 13 2022

Keywords

Crossrefs

Programs

  • Mathematica
    Array[DivisorSum[#, DivisorSigma[0, #^6] &] &, 120] (* Michael De Vlieger, Dec 13 2022 *)
    f[p_, e_] := 3*e^2 + 4*e + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Dec 14 2022 *)
  • PARI
    a(n) = sumdiv(n, d, numdiv(d^6));
    
  • PARI
    a(n) = sumdiv(n, d, numdiv(n*d^4));
    
  • PARI
    a(n) = sumdiv(n, d, numdiv(n^2*d^2));
    
  • PARI
    a(n) = sumdiv(n, d, numdiv(n^3));
    
  • PARI
    a(n) = numdiv(n)*numdiv(n^3);
    
  • PARI
    my(N=80, x='x+O('x^N)); Vec(sum(k=1, N, numdiv(k^6)*x^k/(1-x^k)))

Formula

a(n) = Sum_{d|n} tau(n * d^4) = Sum_{d|n} tau(n^2 * d^2) = Sum_{d|n} tau(n^3).
a(n) = tau(n) * tau(n^3).
G.f.: Sum_{k>=1} tau(k^6) * x^k/(1 - x^k).
Multiplicative with a(p^e) = 3*e^2 + 4*e + 1. - Amiram Eldar, Dec 14 2022

A359038 a(n) = Sum_{d|n} tau(d^7), where tau(n) = number of divisors of n, cf. A000005.

Original entry on oeis.org

1, 9, 9, 24, 9, 81, 9, 46, 24, 81, 9, 216, 9, 81, 81, 75, 9, 216, 9, 216, 81, 81, 9, 414, 24, 81, 46, 216, 9, 729, 9, 111, 81, 81, 81, 576, 9, 81, 81, 414, 9, 729, 9, 216, 216, 81, 9, 675, 24, 216, 81, 216, 9, 414, 81, 414, 81, 81, 9, 1944, 9, 81, 216, 154, 81, 729, 9, 216, 81, 729, 9
Offset: 1

Views

Author

Seiichi Manyama, Dec 13 2022

Keywords

Crossrefs

Programs

  • Mathematica
    Array[DivisorSum[#, DivisorSigma[0, #^7] &] &, 120] (* Michael De Vlieger, Dec 13 2022 *)
    f[p_, e_] := 7*e^2/2 + 9*e/2 + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Dec 14 2022 *)
  • PARI
    a(n) = sumdiv(n, d, numdiv(d^7));
    
  • PARI
    a(n) = sumdiv(n, d, numdiv(n*d^5));
    
  • PARI
    a(n) = sumdiv(n, d, numdiv(n^2*d^3));
    
  • PARI
    a(n) = sumdiv(n, d, numdiv(n^3*d));
    
  • PARI
    a(n) = sumdiv(n, d, numdiv(n^4/d));
    
  • PARI
    my(N=80, x='x+O('x^N)); Vec(sum(k=1, N, numdiv(k^7)*x^k/(1-x^k)))
    
  • Python
    from math import prod
    from sympy import factorint
    def A359038(n): return prod((e+1)*(7*e+2)>>1 for e in factorint(n).values()) # Chai Wah Wu, Dec 13 2022

Formula

a(n) = Sum_{d|n} tau(n * d^5) = Sum_{d|n} tau(n^2 * d^3) = Sum_{d|n} tau(n^3 * d) = Sum_{d|n} tau(n^4 / d).
G.f.: Sum_{k>=1} tau(k^7) * x^k/(1 - x^k).
Multiplicative with a(p^e) = 7*e^2/2 + 9*e/2 + 1. - Amiram Eldar, Dec 14 2022

A358380 a(n) = Sum_{d|n} tau(d^5), where tau(n) = number of divisors of n, cf. A000005.

Original entry on oeis.org

1, 7, 7, 18, 7, 49, 7, 34, 18, 49, 7, 126, 7, 49, 49, 55, 7, 126, 7, 126, 49, 49, 7, 238, 18, 49, 34, 126, 7, 343, 7, 81, 49, 49, 49, 324, 7, 49, 49, 238, 7, 343, 7, 126, 126, 49, 7, 385, 18, 126, 49, 126, 7, 238, 49, 238, 49, 49, 7, 882, 7, 49, 126, 112, 49, 343, 7, 126, 49, 343, 7, 612, 7, 49, 126, 126
Offset: 1

Views

Author

Seiichi Manyama, Dec 13 2022

Keywords

Crossrefs

Programs

  • Mathematica
    Array[DivisorSum[#, DivisorSigma[0, #^5] &] &, 120] (* Michael De Vlieger, Dec 13 2022 *)
    f[p_, e_] := 5*e^2/2 + 7*e/2 + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Dec 14 2022 *)
  • PARI
    a(n) = sumdiv(n, d, numdiv(d^5));
    
  • PARI
    a(n) = sumdiv(n, d, numdiv(n*d^3));
    
  • PARI
    a(n) = sumdiv(n, d, numdiv(n^2*d));
    
  • PARI
    a(n) = sumdiv(n, d, numdiv(n^3/d));
    
  • PARI
    my(N=80, x='x+O('x^N)); Vec(sum(k=1, N, numdiv(k^5)*x^k/(1-x^k)))

Formula

a(n) = Sum_{d|n} tau(n * d^3) = Sum_{d|n} tau(n^2 * d) = Sum_{d|n} tau(n^3 / d).
G.f.: Sum_{k>=1} tau(k^5) * x^k/(1 - x^k).
Multiplicative with a(p^e) = 5*e^2/2 + 7*e/2 + 1. - Amiram Eldar, Dec 14 2022

A069546 a(n) = Sum_{d|n} sigma(n*d).

Original entry on oeis.org

1, 10, 17, 53, 37, 170, 65, 236, 174, 370, 145, 901, 197, 650, 629, 987, 325, 1740, 401, 1961, 1105, 1450, 577, 4012, 968, 1970, 1618, 3445, 901, 6290, 1025, 4026, 2465, 3250, 2405, 9222, 1445, 4010, 3349, 8732, 1765, 11050, 1937, 7685, 6438, 5770
Offset: 1

Views

Author

Vladeta Jovovic, Apr 17 2002

Keywords

Crossrefs

Programs

  • Magma
    [&+[DivisorSigma(1,n*d):d in Divisors(n)]:n in [1..50]]; // Marius A. Burtea, Sep 15 2019
    
  • Mathematica
    Table[ Apply[ Plus, DivisorSigma[1, n*Divisors[n]]], {n, 1, 50}]
    f[p_, e_] := (p^(e + 1)*(p^(e + 1) - 1) - (p - 1)*(e + 1))/(p - 1)^2; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Oct 28 2022 *)
  • PARI
    a(n) = sumdiv(n, d, sigma(n*d)); \\ Michel Marcus, Sep 15 2019

Formula

Multiplicative with a(p^e) = (p^(e+1)*(p^(e+1)-1)-(p-1)*(e+1))/(p-1)^2.
Sum_{k=1..n} a(k) ~ c * n^3, where c = ((zeta(2)*zeta(3)^2)/3) * Product_{p prime} (1 + 1/p^2 - 1/p^4 - 1/p^5) = 1.09461730308... . - Amiram Eldar, Oct 28 2022

Extensions

Edited and extended by Robert G. Wilson v, Apr 22 2002
Showing 1-8 of 8 results.