A325806 Number of ways to partition the divisors of n into two complementary sets whose sums are relatively prime. (Here only distinct unordered pairs of such subsets are counted.)

1, 1, 1, 3, 1, 2, 1, 4, 3, 3, 1, 13, 1, 2, 2, 15, 1, 15, 1, 9, 4, 3, 1, 33, 3, 2, 4, 12, 1, 40, 1, 18, 2, 3, 4, 201, 1, 2, 4, 33, 1, 40, 1, 9, 7, 3, 1, 245, 3, 20, 2, 15, 1, 25, 4, 34, 4, 3, 1, 577, 1, 2, 15, 63, 4, 40, 1, 9, 2, 44, 1, 951, 1, 2, 15, 10, 4, 34, 1, 164, 15, 3, 1, 864, 4, 2, 2, 34, 1, 592, 2, 9, 4, 3, 2, 577, 1, 21, 7, 210, 1, 40, 1, 29, 40

Offset: 1

Author

Antti Karttunen, May 24 2019

Keywords

• nonn

Comments

a(n) is the number of such subsets of divisors of n that include {1} and have sum that is coprime to the sum of their complement.

Records 1, 3, 4, 13, 15, 33, 40, 201, 245, 577, 951, 8672, 14595, 33904, 168904, 253694, 2057413, 2395584, 2396158, 2571028, 159504796, 572644864, ... occur at positions 1, 4, 8, 12, 16, 24, 30, 36, 48, 60, 72, 120, 144, 180, 240, 336, 360, 420, 480, 630, 720, 840, ...

Examples

			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(0,1) = 1, thus this partitioning is included, and a(1) = 1.
For n = 3, its divisor set [1, 3] can be partitioned as [] and [1,3] (sums 0 and 4), [1] and [3] (sums 1 and 3), and only in latter case the sums are coprime as gcd(1,3) = 1, thus a(3) = 1, and similarly a(p) = 1 for any other prime p as well.
For n = 6, its divisor set [1, 2, 3, 6] can be partitioned as [] and [1, 2, 3, 6] (sums 0 and 12), [1, 2] and [3, 6] (sums 3 and 9), [1, 3] and [2, 6] (sums 4 and 8), [2] and [1, 3, 6] (sums 2 and 10), [3] and [1, 2, 6] (sums 3 and 9), [6] and [1, 2, 3] (sums 6 and 6), and also as [1] and [2, 3, 6] (sums 1 and 11), and [1, 6] and [2, 3] (sums 7 and 5) and only in latter two cases their sums are coprime, thus a(6) = 2.
For n = 12, its divisor set [1, 2, 3, 4, 6, 12] can be partitioned altogether in 2^(6-1) = 32 ways, but of which only the following thirteen partitions have coprime sums:
  [1] and [2, 3, 4, 6, 12],
  [1, 2] and [3, 4, 6, 12],
  [1, 4] and [2, 3, 6, 12],
  [1, 2, 6] and [3, 4, 12],
  [1, 4, 6] and [2, 3, 12],
  [1, 2, 4, 6] and [3, 12],
  [1, 12] and [2, 3, 4, 6],
  [1, 2, 12] and [3, 4, 6],
  [1, 4, 12] and [2, 3, 6],
  [1, 2, 4, 12] and [3, 6],
  [1, 6, 12] and [2, 3, 4],
  [1, 4, 6, 12] and [2, 3],
  [1, 2, 4, 6, 12] and [3],
thus a(12) = 13.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..1079

Crossrefs

Cf. A000005, A000203, A100577, A325807.

Cf. A083206, A083207, A083209, A083210.

Programs

Mathematica

Array[Function[d, Count[DeleteDuplicates[Sort /@ Map[{#, Complement[d, #]} &, Subsets@ d]], ?(CoprimeQ @@ (Total /@ #) &)]]@ Divisors@ # &, 105] (* _Michael De Vlieger, May 27 2019 *)

PARI

A325806(n) = { my(divs=divisors(n), s=sigma(n)); sum(b=0,(2^(-1+length(divs)))-1,(1==gcd(s,sumbybits(divs,2*b)))); };
sumbybits(v,b) = { my(s=0,i=1); while(b>0,s += (b%2)*v[i]; i++; b >>= 1); (s); };

Formula

a(n) <= A100577(n).