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A130279
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Smallest number having exactly n square divisors.
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15
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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
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OFFSET
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1,2
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COMMENTS
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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;
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
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LINKS
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FORMULA
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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)
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PROG
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(PARI) a(n) = my(k=1); while(sumdiv(k, d, issquare(d)) != n, k++); k; \\ Michel Marcus, Jul 15 2019
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CROSSREFS
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Cf. A357450 (similar, but with odd squares divisors).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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