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A357450
a(n) is the smallest integer having exactly n odd square divisors (A298735).
5
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
1,2
COMMENTS
All terms are odd and squares (A016754).
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
EXAMPLE
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.
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, Sep 29 2022
EXTENSIONS
a(7)-a(10) from Michel Marcus, Sep 29 2022
More terms from Amiram Eldar, Sep 29 2022
STATUS
approved