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%I #42 Jan 12 2020 13:54:04
%S 1,0,0,1,0,1,0,3,1,1,0,3,0,1,1,5,0,3,0,3,1,1,0,12,1,1,3,3,0,8,0,7,1,1,
%T 1,16,0,1,1,12,0,8,0,3,3,1,0,20,1,3,1,3,0,12,1,12,1,1,0,54,0,1,3,10,1,
%U 8,0,3,1,8,0,54,0,1,3,3,1,8,0,20
%N Number of squares s^2 that divide the product of the divisors of n, such that rad(s) = rad(n), where rad(n) = A007947.
%C This sequence describes the number of squares in the set of all possible products of divisors of n. In general one may ask how many other figurate numbers, polygonal numbers, polyhedral numbers are in the set of all possible products of divisors of n? A harder question: how many squares (or other figurate numbers) are in the set of all possible sums of divisors of n?
%H Antti Karttunen, <a href="/A216677/b216677.txt">Table of n, a(n) for n = 1..65537</a> (first 200 terms from R. J. Mathar)
%F An example: let n = p1^a1 * p2^a2 * p3^a3, p1, p2, p3 primes, a1, a2, a3 positive integers >= 1. Then f(n) = floor(a1 * (a1 + 1) * (a2 + a3 + a2 * a3 + 1) / 4) * floor(a2 * (a2 + 1) * (a1 + a3 + a1 * a3 + 1) / 4) * floor(a3 * (a3 + 1) * (a1 + a2 + a1 * a2 + 1) / 4).
%F For n = p1^a1 * ... * pk^ak, p1, ..., pk primes, a1, ..., ak positive integers >= 1 we have f(n) = Product_{s=1..k} (floor(as * (as + 1) * (1 + sum of all possible products of ai, i not equal to s) / 4)).
%e n = 4, divisors are 1, 2, 4. Product of divisors is 2^3. Floor(3/2) = 1 so 1 square is there in the form 2^2.
%e n = 28, divisors are 1, 2, 4, 7, 14, 28. Product of divisors is 2^6 * 7^3. Floor(6/2) * floor(3/2) = 3 so 3 squares are there : 2^6 * 7^2, 2^4 * 7^2, 2^2 * 7^2. Squares of the form 2^j or 7^k which do not have the full prime factor set {2,7} are not counted.
%t Table[With[{r = Times @@ FactorInteger[n][[All, 1]]}, DivisorSum[Times @@ Divisors@ n, 1 &, And[IntegerQ@ Sqrt@ #, Times @@ FactorInteger[#][[All, 1]] == r] &]], {n, 80}] (* _Michael De Vlieger_, Oct 30 2017 *)
%o (PARI) a(n)=my(p=1);fordiv(n,d,p*=d);p=factor(p)~;prod(i=1,#p,p[2,i]\2) \\ _M. F. Hasler_, Sep 18 2012
%Y Cf. A000005, A007955, A216864.
%K nonn
%O 1,8
%A _Ctibor O. Zizka_, Sep 14 2012
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