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

%I A357450 #37 Oct 03 2022 08:45:09
%S A357450 1,9,81,225,6561,2025,531441,11025,50625,164025,3486784401,99225,
%T A357450 282429536481,13286025,4100625,893025,1853020188851841,2480625,
%U A357450 150094635296999121,8037225,332150625,87169610025,984770902183611232881,12006225,2562890625,7060738412025,121550625
%N A357450 a(n) is the smallest integer having exactly n odd square divisors (A298735).
%C A357450 All terms are odd and squares (A016754).
%F A357450 a(n) = A038547(n)^2. - _Thomas Scheuerle_, Sep 30 2022
%F A357450 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
%F A357450 a(p) = 3^(2*(p-1)) for primes p. - _Bernard Schott_, Oct 03 2022
%e A357450 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.
%o A357450 (PARI) f(n) = factorback(apply(e->e\2+1, factor(n/2^valuation(n, 2))[, 2])); \\ A298735
%o A357450 a(n) = my(k=1); while (f(k)!=n, k++); k; \\ _Michel Marcus_, Sep 29 2022
%Y A357450 Cf. A000290, A016754, A038547, A130279, A147516, A298735.
%K A357450 nonn
%O A357450 1,2
%A A357450 _Bernard Schott_, Sep 29 2022
%E A357450 a(7)-a(10) from _Michel Marcus_, Sep 29 2022
%E A357450 More terms from _Amiram Eldar_, Sep 29 2022