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%I #29 Sep 11 2022 10:10:10
%S 9,18,25,27,45,49,50,54,63,75,81,90,98,99,117,121,125,126,135,147,150,
%T 153,162,169,171,175,189,198,207,225,234,242,243,245,250,261,270,275,
%U 279,289,294,297,306,315,325,333,338,342,343,350,351,361,363,369,375,378,387,405
%N Numbers k such that the sum of the squares of the odd non-unitary divisors of k divides the sum of the squares of the non-unitary divisors of k.
%C Conjecture: For any nonnegative integer power p the sum of the p-th powers of the odd non-unitary divisors of a(n) divides the sum of the p-th powers of the non-unitary divisors of a(n).
%C The start of this sequence, including the 58 terms currently shown in the data section, is consistent with a definition "nonsquarefree numbers not divisible by 4", but some larger terms are divisible by 4: for example, a(1305) = 9216 = 2^10 * 3^2. - _Peter Munn_, Sep 21 2020
%C It seems that, for every n and odd prime p, p*a(n) is a term. - _Ivan N. Ianakiev_, Sep 11 2022
%H Amiram Eldar, <a href="/A319927/b319927.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/UnitaryDivisor.html">Unitary Divisor</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Unitary_divisor">Unitary divisor</a>
%e The non-unitary divisors of 18 are 3 and 6, of which only 3 is odd. 3^2=9 divides 3^3 + 6^2 = 45 and therefore 18 is in the sequence.
%t sigmaNU[n_,p_]:=Total[Select[Divisors[n],GCD[#,n/#]>1&]^p];
%t sigmaNUOdd[n_,p_]:=Total[Select[Divisors[n],OddQ[#]&&GCD[#,n/#]>1&]^p];
%t p=2;(*if you want to check the conjecture for the power of 99, replace 2 with 99*)
%t Select[Range[1000],IntegerQ[sigmaNU[#,p]/sigmaNUOdd[#,p]]&]//Quiet
%o (PARI) su2(n) = sumdiv(n, d, if(gcd(d, n/d)!=1, d^2));
%o suo2(n) = sumdiv(n, d, if ((d%2) && (gcd(d, n/d)!=1), d^2));
%o isok(n) = my(suo = suo2(n)); if (suo, (su2(n) % suo) == 0); \\ _Michel Marcus_, Oct 28 2018
%Y Cf. A048105 (number, i.e., sum of the 0th powers, of non-unitary divisors of n), A034444 (number of unitary divisors of n).
%K nonn
%O 1,1
%A _Ivan N. Ianakiev_, Oct 02 2018