1,2

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<k;

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

Amiram Eldar, Table of n, a(n) for n = 1..100

Eric Weisstein's World of Mathematics, Smooth Number

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)

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

Cf. A001221, A005179, A046951, A046952, A122375, A122376, A353282.

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

Sequence in context: A223403 A263385 A368215 * A030158 A054246 A173545

Adjacent sequences: A130276 A130277 A130278 * A130280 A130281 A130282

Reinhard Zumkeller, May 20 2007

approved