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 1-2 of 2 results.

A365346 The sum of divisors of the smallest square divisible by n.

Original entry on oeis.org

1, 7, 13, 7, 31, 91, 57, 31, 13, 217, 133, 91, 183, 399, 403, 31, 307, 91, 381, 217, 741, 931, 553, 403, 31, 1281, 121, 399, 871, 2821, 993, 127, 1729, 2149, 1767, 91, 1407, 2667, 2379, 961, 1723, 5187, 1893, 931, 403, 3871, 2257, 403, 57, 217, 3991, 1281, 2863
Offset: 1

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Author

Amiram Eldar, Sep 02 2023

Keywords

Comments

The number of these divisors is A365345(n).
The sum of divisors of the square root of the smallest square divisible by n is A365347(n).

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := (p^(e + 1 + Mod[e, 2]) - 1)/(p - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, (f[i,1]^(f[i,2] + 1 + f[i,2]%2) - 1)/(f[i,1] - 1));}
    
  • PARI
    a(n) = sigma(n*core(n)); \\ Michel Marcus, Sep 02 2023

Formula

a(n) = A000203(A053143(n)).
Multiplicative with a(p^e) = (p^(e + 1 + (e mod 2)) - 1)/(p - 1).
Dirichlet g.f.: zeta(s) * zeta(2*s-2) * Product_{p prime} (1 + 1/p^(s-2) + 1/p^(s-1) - 1/p^(2*s-2)).
Sum_{k=1..n} a(k) ~ c * n^3, where c = (Pi^2/45) * zeta(3) * Product_{p prime} (1 + 1/p^2 - 1/p^3) = 0.344306233314... .

A365481 The sum of unitary divisors of the smallest number whose square is divisible by n.

Original entry on oeis.org

1, 3, 4, 3, 6, 12, 8, 5, 4, 18, 12, 12, 14, 24, 24, 5, 18, 12, 20, 18, 32, 36, 24, 20, 6, 42, 10, 24, 30, 72, 32, 9, 48, 54, 48, 12, 38, 60, 56, 30, 42, 96, 44, 36, 24, 72, 48, 20, 8, 18, 72, 42, 54, 30, 72, 40, 80, 90, 60, 72, 62, 96, 32, 9, 84, 144, 68, 54, 96
Offset: 1

Views

Author

Amiram Eldar, Sep 05 2023

Keywords

Comments

The number of unitary divisors of the smallest number whose square is divisible by n is the same as the number of unitary divisors of n, A034444(n).

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^Ceiling[e/2] + 1; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i=1, #f~, f[i,1]^ceil(f[i,2]/2) + 1);}
    
  • Python
    from math import prod
    from sympy import factorint
    def A365481(n): return prod(p**((e>>1)+(e&1))+1 for p,e in factorint(n).items()) # Chai Wah Wu, Sep 05 2023

Formula

a(n) = A034448(A019554(n)).
Multiplicative with a(p^e) = p^(ceiling(e/2)) + 1.
Dirichlet g.f.: zeta(s) * zeta(2*s-1) * Product_{p prime} (1 + 1/p^(s-1) - 1/p^(2*s-1) - 1/p^(3*s-1)).
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * zeta(2) * zeta(3) * Product_{p prime} (1 - 1/p^2 - 1/p^3 + 1/p^4 - 1/p^5 + 1/p^6) = 0.515959523197... .
Showing 1-2 of 2 results.