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 1-6 of 6 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

A060648 Number of cyclic subgroups of the group C_n X C_n (where C_n is the cyclic group of order n).

Original entry on oeis.org

1, 4, 5, 10, 7, 20, 9, 22, 17, 28, 13, 50, 15, 36, 35, 46, 19, 68, 21, 70, 45, 52, 25, 110, 37, 60, 53, 90, 31, 140, 33, 94, 65, 76, 63, 170, 39, 84, 75, 154, 43, 180, 45, 130, 119, 100, 49, 230, 65, 148, 95, 150, 55, 212, 91, 198, 105, 124, 61, 350, 63, 132, 153, 190
Offset: 1

Views

Author

Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 04 2001

Keywords

Comments

The group U(n) of units modulo n acts on the direct product (Z_n)^k by multiplication. The number g(n,k) of orbits of U(n) acting on Z/(n)^k is g(n,k) = (1/phi(n))*Sum(gcd(n,a-1)^k) where the sum is over a in U(n) and phi(n) is the Euler totient function. A060648 gives g(n,2). - W. Edwin Clark, Jul 20 2001
a(n) is also the number of orbits of length n for the map TxT (Cartesion product) where T is a map with one orbit of each length. - Thomas Ward, Apr 08 2009

Examples

			The cycle index of C_4 X C_4 is (x(1)^4 + x(2)^2 + 2*x(4))^2 = x(1)^8 + 2*x(1)^4*x(2)^2 + 4*x(1)^4*x(4) + x(2)^4 + 4*x(2)^2*x(4) + 4*x(4)^2 and C_4 X C_4 has 1 element of order 1, 3 elements of order 2 and 12 elements of order 4. So a(4) = 1/phi(1) + 3/phi(2) + 12/phi(4) = 10, where phi = Euler totient function, cf. A000010. - _Vladeta Jovovic_, Jul 05 2001
For a(4) the pairs (e,d) are (1,4),(2,4),(4,4),(4,2),(4,1) with gcds 1,2,4,2,1 resp. giving 10 in total. - _Thomas Ward_, Apr 08 2009
		

Crossrefs

Programs

  • Maple
    for n from 1 to 200 do:ans := 1:for i from 1 to nops(ifactors(n)[2]) do p := ifactors(n)[2][i][1]:e := ifactors(n)[2][i][2]:ans := ans*(p^(e+1)+p^e-2)/(p-1):od:printf(`%d,`,ans):od:
  • Mathematica
    Table[ Plus @@ Map[ Times @@ (EulerPhi /@ #)/EulerPhi[ LCM @@ # ] &, Flatten[ Outer[ {##} &, Divisors[ i ], Divisors[ i ] ], 1 ] ], {i, 1, 100} ]
    f[p_, e_] := (p^(e+1)+p^e-2)/(p-1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 20 2020 *)
  • PARI
    a(n) = sumdiv(n, d,  2^omega(d)*(n/d) ); \\ Joerg Arndt, Sep 16 2012
  • Sage
    def A060648(n) :
        def dedekind_psi(n) : return n*mul(1+1/p for p in prime_divisors(n))
        return reduce(lambda x,y: x+y, [dedekind_psi(d) for d in divisors(n)])
    [A060648(n) for n in (1..64)]  # Peter Luschny, Sep 10 2012
    

Formula

a(n) is multiplicative: if the canonical factorization of n is the product of p^e(p) over primes then a(n) = product a(p^e(p)). If n = p^e, p prime, a(n) = (p^(e+1)+p^e-2)/(p-1).
a(n) = Sum_{i|n, j|n} phi(i)*phi(j)/phi(lcm(i, j)). - Vladeta Jovovic, Jul 07 2001
a(n) = Sum_{i|n, j|n} phi(gcd(i, j)).
a(n) = Sum_{d|n} phi(n/d)*tau(d^2).
a(n) = sum(d|n, sigma(d)*moebius(n/d)^2 ). - Benoit Cloitre, Sep 08 2002
Inverse Euler transform of A156302. - Vladeta Jovovic, Feb 14 2009
Moebius transform of A060724. - Vladeta Jovovic, Apr 05 2009
Also a(n) = (1/n)*Sum_{d|n} sigma(d)^2*moebius(n/d). - Vladeta Jovovic, Mar 31 2009
Inverse Moebius transform of A001615. - Vladeta Jovovic, Apr 05 2009
From Thomas Ward, Apr 08 2009: (Start)
a(n) = Sum_{lcm(e,d)=n} gcd(e,d).
Dirichlet g.f.: zeta(s)^2*zeta(s-1)/zeta(2s). (End)
For the proofs of these formulas see the papers of L. Toth.
a(n) = Sum_{d|n} psi(d), where psi is Dedekind's psi function A001615. - Peter Luschny, Sep 10 2012
a(n) = Sum_{d|n} 2^omega(d)*(n/d). - Peter Luschny, Sep 15 2012
Sum_{k=1..n} a(k) ~ (5/4) * n^2. - Amiram Eldar, Oct 19 2022
a(n) = Sum_{k=1..n} tau(gcd(n,k)^2). - Ridouane Oudra, Apr 10 2023
a(n) = Sum_{d divides n} J_2(d)/phi(d) = Sum_{1 <= i, j <= n} 1/phi(n/gcd(i,j,n)), where the Jordan totient function J_2(n) = A007434(n). - Peter Bala, Jan 23 2024

Extensions

More terms and additional comments from Vladeta Jovovic, Jul 05 2001

A061502 a(n) = Sum_{k<=n} tau(k)^2, where tau = number of divisors function A000005.

Original entry on oeis.org

1, 5, 9, 18, 22, 38, 42, 58, 67, 83, 87, 123, 127, 143, 159, 184, 188, 224, 228, 264, 280, 296, 300, 364, 373, 389, 405, 441, 445, 509, 513, 549, 565, 581, 597, 678, 682, 698, 714, 778, 782, 846, 850, 886, 922, 938, 942, 1042, 1051, 1087
Offset: 1

Views

Author

N. J. A. Sloane, Jun 14 2001

Keywords

References

  • R. Ayoub, An Introduction to the Analytic Theory of Numbers, Amer. Math. Soc., 1963; Chapter II, Problem 56.

Crossrefs

Programs

  • Magma
    [&+[NumberOfDivisors(k^2)*Floor(n/k): k in [1..n]]: n in [1..60]]; // Vincenzo Librandi, Sep 10 2016
  • Mathematica
    Table[Sum[DivisorSigma[0, k^2]*Floor[n/k], {k, 1, n}], {n, 1, 50}] (* Vaclav Kotesovec, Aug 30 2018 *)
    Accumulate[Table[DivisorSigma[0, n]^2, {n, 1, 50}]] (* Vaclav Kotesovec, Sep 10 2018 *)
  • PARI
    for (n=1, 1024, write("b061502.txt", n, " ", sum(k=1, n, numdiv(k)^2)) ) \\ Harry J. Smith, Jul 23 2009
    
  • PARI
    vector(60, n, sum(k=1, n, numdiv(k)^2)) \\ Michel Marcus, Jul 23 2015
    
  • PARI
    first(n)=my(v=vector(n),s); forfactored(k=1,n, v[k[1]] = s += numdiv(k)^2); v; \\ Charles R Greathouse IV, Nov 28 2018
    

Formula

a(n) = Sum_{k=1..n} tau(k^2)*floor(n/k).
Asymptotic to A*n*log(n)^3 + B*n*log(n)^2 + C*n*log(n) + D*n + O(n^(1/2+eps)) where A = 1/Pi^2 and B = (12*gamma-3)/Pi^2 - 36*zeta'(2)/Pi^4. [corrected by Vaclav Kotesovec, Aug 30 2018]
C = 36*gamma^2/Pi^2 - (288*z1/Pi^4 + 24/Pi^2)*gamma + (864*z1^2/Pi^6 + 72*z1/Pi^4 - 72/Pi^4*z2 + 6/Pi^2) - 24*g1/Pi^2 and D = 24*gamma^3/Pi^2 - (432*z1 /Pi^4+ 36/Pi^2)*gamma^2 + (3456*z1^2/Pi^6 + 288*(z1-z2)/Pi^4 + 24/Pi^2 - 72*g1/Pi^2)*gamma + g1*(288*z1/Pi^4 + 24/Pi^2)-10368*z1^3/Pi^8 - 864*z1^2/Pi^6 + 1728*z2*z1/Pi^6 + 72*(z2-z1)/Pi^4- 48*z3/Pi^4 + (12*g2-6)/Pi^2, where gamma is the Euler-Mascheroni constant A001620, z1 = Zeta'(2) = A073002, z2 = Zeta''(2) = A201994, z3 = Zeta'''(2) = A201995 and g1, g2 are the Stieltjes constants, see A082633 and A086279. - Vaclav Kotesovec, Sep 10 2018
See Cully-Hugill & Trudgian, Theorem 2, for an explicit version of the asymptotic given above. - Charles R Greathouse IV, Nov 19 2019

Extensions

Definition corrected by N. J. A. Sloane, May 25 2008

A353551 a(n) = Sum_{k=1..n} tau(k^3), where tau is the number of divisors function A000005.

Original entry on oeis.org

0, 1, 5, 9, 16, 20, 36, 40, 50, 57, 73, 77, 105, 109, 125, 141, 154, 158, 186, 190, 218, 234, 250, 254, 294, 301, 317, 327, 355, 359, 423, 427, 443, 459, 475, 491, 540, 544, 560, 576, 616, 620, 684, 688, 716, 744, 760, 764, 816, 823, 851, 867, 895, 899, 939, 955, 995
Offset: 0

Views

Author

Karl-Heinz Hofmann, May 07 2022

Keywords

Examples

			  A048785(0) = 0
+ A048785(1) = 1
+ A048785(2) = 4
+ A048785(3) = 4
------------------
= A353551(3) = 9
		

Crossrefs

Partial sums of A048785.
Cf. A000005, A006218, A061503 (squares).

Programs

  • Maple
    a:= proc(n) option remember; `if`(n=0, 0, a(n-1)+numtheory[tau](n^3)) end:
    seq(a(n), n=0..100);  # Alois P. Heinz, May 08 2022
  • Mathematica
    Accumulate[Join[{0}, Table[DivisorSigma[0, k^3], {k, 1, 50}]]] (* Amiram Eldar, May 08 2022 *)
  • PARI
    a(n) = sum(k=1, n, numdiv(k^3)); \\ Michel Marcus, May 08 2022
    
  • Python
    from sympy import divisor_count
    def A048785(n): return divisor_count(n**3)
    def A353551(n): return sum(A048785(n) for n in range(1, n))
    print([A353551(n) for n in range(1, 58)])
    
  • Python
    from math import prod
    from sympy import factorint
    def A353551(n): return sum(prod(3*e+1 for e in factorint(k).values()) for k in range(1,n+1)) # Chai Wah Wu, May 10 2022

Formula

a(n) = Sum_{k=1..n} tau(k^3).
a(n) = a(n-1) + A048785(n) for n >= 1, a(0) = 0.

A182082 Number of pairs, (x,y), with x >= y, whose LCM does not exceed n.

Original entry on oeis.org

1, 3, 5, 8, 10, 15, 17, 21, 24, 29, 31, 39, 41, 46, 51, 56, 58, 66, 68, 76, 81, 86, 88, 99, 102, 107, 111, 119, 121, 135, 137, 143, 148, 153, 158, 171, 173, 178, 183, 194, 196, 210, 212, 220, 228, 233, 235, 249, 252, 260, 265, 273, 275, 286, 291, 302, 307, 312
Offset: 1

Views

Author

Walt Rorie-Baety, Apr 10 2012

Keywords

Comments

Note that this is the asymmetric count. If all pairs (x,y) are counted, A061503 is obtained. - T. D. Noe, Apr 10 2012

Examples

			a(1000000) = 37429395, according to Project Euler problem #379.
		

Crossrefs

Partial sums of A018892.
Cf. A000005, A007875, A013661, A061503 (symmetric case).

Programs

  • Haskell
    a n = length [(x,y)| x <- [1..n], y <- [x..n], lcm x y <= n]
    
  • Mathematica
    Table[Count[Flatten[Table[LCM[i, j], {i, n}, {j, i, n}]], ?(# <= n &)], {n, 60}] (* _T. D. Noe, Apr 10 2012 *)
    nn = 100; (Accumulate[Table[DivisorSigma[0, n^2], {n, nn}]] + Range[nn])/2 (* T. D. Noe, Apr 10 2012 *)
  • PARI
    a(n)=(sum(k=1,n,numdiv(k^2))+n)/2 \\ Charles R Greathouse IV, Apr 10 2012

Formula

a(n) = Sum_{k=1..n} (d(k^2)+1)/2, where d is the number of divisors function (A000005). - Charles R Greathouse IV, Apr 10 2012
a(n) = Sum_{k=1..n} A007875(k) * floor(n/k). - Daniel Suteu, Jan 08 2021
a(n) ~ n * log(n)^2 /(4*zeta(2)) (see A018892 for a more accurate asymptotic formula). - Amiram Eldar, Feb 01 2025

A306775 Partial sums of A060648: sum of the inverse Moebius transform of the Dedekind psi function from 1 to n.

Original entry on oeis.org

1, 5, 10, 20, 27, 47, 56, 78, 95, 123, 136, 186, 201, 237, 272, 318, 337, 405, 426, 496, 541, 593, 618, 728, 765, 825, 878, 968, 999, 1139, 1172, 1266, 1331, 1407, 1470, 1640, 1679, 1763, 1838, 1992, 2035, 2215, 2260, 2390, 2509, 2609, 2658, 2888, 2953, 3101
Offset: 1

Views

Author

Daniel Suteu, Mar 09 2019

Keywords

Comments

In general, for m >= 1, Sum_{k=1..n} Sum_{d|k} psi_m(d) = Sum_{k=1..n} k^m * A064608(floor(n/k)), where psi_m(d) is the generalized Dedekind psi function.
Additionally, for m >= 1, Sum_{k=1..n} Sum_{d|k} psi_m(d) ~ (n^(m+1) * zeta(m+1)^2) / ((m+1) * zeta(2*(m+1))).

Crossrefs

Programs

  • Maple
    with(numtheory): psi := n -> n*mul(1+1/p, p in factorset(n)):
    seq(add(psi(i)*floor(n/i), i=1..n), n=1..80); # Ridouane Oudra, Aug 27 2019
  • Mathematica
    Accumulate[Table[Sum[EulerPhi[n/d] * DivisorSigma[0, d^2], {d, Divisors[n]}], {n, 1, 100}]] (* Vaclav Kotesovec, Oct 09 2019 *)
  • PARI
    a(n) = sum(k=1, n, 2^omega(k) * (n\k) * (1+n\k))/2;

Formula

a(n) ~ (5/4) * n^2.
a(n) = Sum_{k=1..n} A060648(k).
a(n) = Sum_{k=1..n} Sum_{d|k} A001615(d).
a(n) = Sum_{k=1..n} k * A064608(floor(n/k)).
a(n) = (1/2)*Sum_{k=1..n} 2^omega(k) * floor(n/k) * floor(1+n/k).
a(n) = Sum_{k=1..n} A001615(k)*floor(n/k). - Ridouane Oudra, Aug 27 2019
Showing 1-6 of 6 results.