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A325807 Number of ways to partition the divisors of n into complementary subsets x and y for which gcd(n-Sum(x), n-Sum(y)) = 1. (Here only distinct unordered pairs of such subsets are counted.)

%I A325807 #18 May 28 2019 19:34:31
%S A325807 1,2,1,4,1,1,1,8,3,4,1,16,1,4,2,16,1,16,1,16,4,4,1,40,3,3,4,1,1,40,1,
%T A325807 32,2,4,4,244,1,4,4,48,1,40,1,16,8,3,1,220,3,27,2,10,1,32,4,64,4,4,1,
%U A325807 672,1,4,14,64,4,40,1,13,2,64,1,1205,1,4,16,10,4,40,1,236,15,4,1,864,4,3,2,64,1,640,2,16,4,4,2,537,1,26,8,241,1,40,1,64,40
%N A325807 Number of ways to partition the divisors of n into complementary subsets x and y for which gcd(n-Sum(x), n-Sum(y)) = 1. (Here only distinct unordered pairs of such subsets are counted.)
%H A325807 Antti Karttunen, <a href="/A325807/b325807.txt">Table of n, a(n) for n = 1..1079</a>
%F A325807 For all n >= 1:
%F A325807 a(n) <= A100577(n).
%F A325807 a(A065091(n)) = 1, a(A000396(n)) = 1.
%F A325807 a(A228058(n)) = A325809(n).
%e A325807 For n = 1, its divisor set [1] can be partitioned only to an empty set [] and set [1], with sums 0 and 1 respectively, and gcd(1-0,1-1) = gcd(1,0) = 1, thus this partitioning is included, and a(1) = 1.
%e A325807 For n = 3, its divisor set [1, 3] can be partitioned as [] and [1,3] (sums 0 and 4, thus gcd(3-0,3-4) = 1), [1] and [3] (sums 1 and 3, thus gcd(3-1,3-3) = 2), thus a(3) = 1, and similarly a(p) = 1 for any other odd prime p as well.
%e A325807 For n = 6, its divisor set [1, 2, 3, 6] can be partitioned in eight ways as:
%e A325807   [] and [1, 2, 3, 6] (sums 0 and 12, gcd(6-0, 6-12) = 6),
%e A325807   [1, 2] and [3, 6]   (sums 3 and 9,  gcd(6-3, 6-9) = 3),
%e A325807   [1, 3] and [2, 6]   (sums 4 and 8,  gcd(6-4, 6-8) = 2),
%e A325807   [2] and [1, 3, 6]   (sums 2 and 10, gcd(6-2, 6-10) = 4),
%e A325807   [3] and [1, 2, 6]   (sums 3 and 9,  gcd(6-3, 6-9) = 3),
%e A325807   [6] and [1, 2, 3]   (sums 6 and 6,  gcd(6-6, 6-6) = 0),
%e A325807   [1] and [2, 3, 6]   (sums 1 and 11, gcd(6-1, 6-11) = 5),
%e A325807   [1, 6] and [2, 3]   (sums 7 and 5,  gcd(6-7, 6-5) = 1),
%e A325807 with only the last partitioning satisfying the required condition, thus a(6) = 1.
%e A325807 For n = 10, its divisor set [1, 2, 5, 10] can be partitioned in eight ways as:
%e A325807   [] and [1, 2, 5, 10] (sums 0 and 18, gcd(10-0, 10-18) = 2),
%e A325807   [1, 2] and [5, 10]   (sums 3 and 15, gcd(10-3, 10-15) = 1),
%e A325807   [1, 5] and [2, 10]   (sums 6 and 12, gcd(10-6, 10-12) = 2),
%e A325807   [2] and [1, 5, 10]   (sums 2 and 16, gcd(10-2, 10-16) = 2),
%e A325807   [5] and [1, 2, 10]   (sums 5 and 13, gcd(10-5, 10-13) = 1),
%e A325807   [10] and [1, 2, 5]   (sums 10 and 8, gcd(10-10, 10-8) = 2),
%e A325807   [1] and [2, 5, 10]   (sums 1 and 17, gcd(10-1, 10-17) = 1),
%e A325807   [1, 10] and [2, 5]   (sums 11 and 7, gcd(10-11, 10-7) = 1),
%e A325807 of which four satisfy the required condition, thus a(10) = 4.
%t A325807 Table[Function[d, Count[DeleteDuplicates[Sort /@ Map[{#, Complement[d, #]} &, Subsets@ d]], _?(CoprimeQ @@ (n - Total /@ #) &)]]@ Divisors@ n, {n, 105}] (* _Michael De Vlieger_, May 27 2019 *)
%o A325807 (PARI)
%o A325807 A325807(n) = { my(divs=divisors(n), s=sigma(n),r); sum(b=0,(2^(-1+length(divs)))-1,r=sumbybits(divs,2*b);(1==gcd(n-(s-r),n-r))); };
%o A325807 sumbybits(v,b) = { my(s=0,i=1); while(b>0,s += (b%2)*v[i]; i++; b >>= 1); (s); };
%Y A325807 Cf. A000203, A000396, A065091, A100577, A324213, A325806, A325809.
%K A325807 nonn
%O A325807 1,2
%A A325807 _Antti Karttunen_, May 24 2019