A325806 -id:A325806 - cp's OEIS frontend

A083206 a(n) is the number of ways of partitioning the divisors of n into two disjoint sets with equal sum.

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 3, 0, 0, 0, 1, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 17, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 3, 0, 0, 0, 14, 0, 0, 0, 1, 0, 13, 0, 0, 0, 0, 0, 11, 0, 0, 0, 0, 0, 2, 0, 1

Offset: 1

Reinhard Zumkeller, Apr 22 2003

nonn

a(n)=0 for deficient numbers n (A005100), but the converse is not true, as 18 is abundant (A005101) and a(18)=0, see A083211;
a(n)=1 for perfect numbers n (A000396), see A083209 for all numbers with a(n)=1;
records: A083213(k)=a(A083212(k)).
In order that a(n)>0, the sum of divisors of n must be even by definition: a(n) = half the number of partitions of A000203(n)/2 into divisors of n, see formula. [Reinhard Zumkeller, Jul 10 2010]

			a(24)=3: 1+2+3+4+8+12=6+24, 1+3+6+8+12=2+4+24, 4+6+8+12=1+2+3+24.

Antti Karttunen, Table of n, a(n) for n = 1..101632 (terms 1..800 from Reinhard Zumkeller, terms 801..10000 from T. D. Noe).
Reinhard Zumkeller, Illustration of initial terms

Cf. A083208 [= a(A083207(n))], A083211, A000005, A000203, A082729, A378446 (inverse Möbius transform), A378449.
Cf. A083207 (positions of terms > 0), A083210 (positions of 0's), A083209 (positions of 1's), A378652 (of 2's).
Cf. also A033630, A065205, A058377, A325806.

a[n_] := (s = DivisorSigma[1, n]; If[Mod[s, 2] == 1, 0, f[n, s/2, 2]]); f[n_, m_, k_] := f[n, m, k] = If[k <= m, f[n, m, k+1] + f[n, m-k, k+1]*Boole[Mod[n, k] == 0], Boole[m == 0]]; Array[a, 105] (* Jean-François Alcover, Jul 29 2015, after Reinhard Zumkeller *)

A083206(n) = { my(s=sigma(n),p=1); if(s%2 || s < 2*n, 0, fordiv(n, d, p *= ('x^d + 'x^-d)); (polcoeff(p, 0)/2)); }; \\ Antti Karttunen, Dec 02 2024, after Ilya Gutkovskiy

a(n) = if sigma(n) mod 2 = 1 then 0 else f(n,sigma(n)/2,2), where sigma=A000203 and f(n,m,k) = if k<=m then f(n,m,k+1)+f(n,m-k,k+1)*0^(n mod k) else 0^m, cf. A033630, also using f. [Reinhard Zumkeller, Jul 10 2010]

a(n) is half the coefficient of x^0 in Product_{d|n} (x^d + 1/x^d). - Ilya Gutkovskiy, Feb 04 2024

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.)

Original entry on oeis.org

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, 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, 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

Offset: 1

Antti Karttunen, May 24 2019

nonn

			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.
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.
For n = 6, its divisor set [1, 2, 3, 6] can be partitioned in eight ways as:
  [] and [1, 2, 3, 6] (sums 0 and 12, gcd(6-0, 6-12) = 6),
  [1, 2] and [3, 6]   (sums 3 and 9,  gcd(6-3, 6-9) = 3),
  [1, 3] and [2, 6]   (sums 4 and 8,  gcd(6-4, 6-8) = 2),
  [2] and [1, 3, 6]   (sums 2 and 10, gcd(6-2, 6-10) = 4),
  [3] and [1, 2, 6]   (sums 3 and 9,  gcd(6-3, 6-9) = 3),
  [6] and [1, 2, 3]   (sums 6 and 6,  gcd(6-6, 6-6) = 0),
  [1] and [2, 3, 6]   (sums 1 and 11, gcd(6-1, 6-11) = 5),
  [1, 6] and [2, 3]   (sums 7 and 5,  gcd(6-7, 6-5) = 1),
with only the last partitioning satisfying the required condition, thus a(6) = 1.
For n = 10, its divisor set [1, 2, 5, 10] can be partitioned in eight ways as:
  [] and [1, 2, 5, 10] (sums 0 and 18, gcd(10-0, 10-18) = 2),
  [1, 2] and [5, 10]   (sums 3 and 15, gcd(10-3, 10-15) = 1),
  [1, 5] and [2, 10]   (sums 6 and 12, gcd(10-6, 10-12) = 2),
  [2] and [1, 5, 10]   (sums 2 and 16, gcd(10-2, 10-16) = 2),
  [5] and [1, 2, 10]   (sums 5 and 13, gcd(10-5, 10-13) = 1),
  [10] and [1, 2, 5]   (sums 10 and 8, gcd(10-10, 10-8) = 2),
  [1] and [2, 5, 10]   (sums 1 and 17, gcd(10-1, 10-17) = 1),
  [1, 10] and [2, 5]   (sums 11 and 7, gcd(10-11, 10-7) = 1),
of which four satisfy the required condition, thus a(10) = 4.

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

Cf. A000203, A000396, A065091, A100577, A324213, A325806, A325809.

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

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))); };
sumbybits(v,b) = { my(s=0,i=1); while(b>0,s += (b%2)*v[i]; i++; b >>= 1); (s); };

For all n >= 1:
a(n) <= A100577(n).
a(A065091(n)) = 1, a(A000396(n)) = 1.
a(A228058(n)) = A325809(n).

cp's OEIS Frontend

A083206 a(n) is the number of ways of partitioning the divisors of n into two disjoint sets with equal sum.

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