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.

A130279 Smallest number having exactly n square divisors.

Original entry on oeis.org

1, 4, 16, 36, 256, 144, 4096, 576, 1296, 2304, 1048576, 3600, 16777216, 36864, 20736, 14400, 4294967296, 32400, 68719476736, 57600, 331776, 9437184, 17592186044416, 129600, 1679616, 150994944, 810000, 921600, 72057594037927936
Offset: 1

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Author

Reinhard Zumkeller, May 20 2007

Keywords

Comments

A046951(a(n)) = n and A046951(m) <> n for m < a(n);
all terms are smooth squares: if prime(k) is a factor of a(n) then also prime(i) are factors, i
a(p) = 2^(2*(p-1)) for primes p;
if prime(j) is the greatest prime factor of a(n) then a(2*n) = a(n)*prime(j+1)^2;
A001221(a(n)) = A122375(n); A001222(a(n)) = 2*A122376(n).
a(n+1) is the smallest nonsquarefree number m such that Diophantine equation S(x,y) = (x+y) + (x-y) + (x*y) + (x/y) = m has exactly n solutions, for n >= 0 (A353282); example: a(4) = 36 and 36 is the smallest number m such that equation S(x,y) = m has exactly 3 solutions: (9,1), (8,2), (5,5). - Bernard Schott, Apr 13 2022
a(n) is the square of the smallest integer having exactly n divisors (see formula with proof). - Bernard Schott, Oct 01 2022

Crossrefs

Cf. A357450 (similar, but with odd squares divisors).

Programs

  • PARI
    a(n) = my(k=1); while(sumdiv(k, d, issquare(d)) != n, k++); k; \\ Michel Marcus, Jul 15 2019

Formula

From Bernard Schott, Oct 01 2022: (Start)
a(n) = A005179(n)^2.
Proof: Suppose a(n) = Product p_i^(2*e_i), where the p_i are primes. Then the n square divisors are all of the form d = Product p_i^(2*k_i) with 0 <= k_i <= e_i. As a(n) = Product (p_i^e_i)^2 = (Product (p_i^e_i))^2, we get that sqrt(a(n)) = Product (p_i^e_i). This is the prime decomposition of sqrt(a(n)). As there is a bijection between prime factors p_i^(2*k_i) and (p_i^k_i), there is also bijection between square divisors of a(n) and divisors of sqrt(a(n)). We conclude that sqrt(a(n)) is the smallest integer that has exactly n divisors. (End)