A357450 a(n) is the smallest integer having exactly n odd square divisors (A298735).

1, 9, 81, 225, 6561, 2025, 531441, 11025, 50625, 164025, 3486784401, 99225, 282429536481, 13286025, 4100625, 893025, 1853020188851841, 2480625, 150094635296999121, 8037225, 332150625, 87169610025, 984770902183611232881, 12006225, 2562890625, 7060738412025, 121550625

Offset: 1

Author

Bernard Schott, Sep 29 2022

Keywords

• nonn

Comments

All terms are odd and squares (A016754).

Examples

			2025 has 6 divisors that are odd squares: {1, 9, 25, 81, 225, 2025}; also, 2025 is the smallest integer that has 6 odd squares divisors, hence a(6) = 2025.
		

Crossrefs

Cf. A000290, A016754, A038547, A130279, A147516, A298735.

Programs

PARI

f(n) = factorback(apply(e->e\2+1, factor(n/2^valuation(n, 2))[, 2])); \\ A298735
a(n) = my(k=1); while (f(k)!=n, k++); k; \\ Michel Marcus, Sep 29 2022

Formula

a(n) = A038547(n)^2. - Thomas Scheuerle, Sep 30 2022

Proof: Suppose a(n) = Product p_i^(2*e_i), where the p_i are odd primes. Then the n odd 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 odd square divisors of a(n) and odd divisors of sqrt(a(n)). We conclude that sqrt(a(n)) is the smallest integer that has exactly n odd divisors. - Bernard Schott, Oct 01 2022

a(p) = 3^(2*(p-1)) for primes p. - Bernard Schott, Oct 03 2022

Extensions

a(7)-a(10) from Michel Marcus, Sep 29 2022

More terms from Amiram Eldar, Sep 29 2022