A379505 a(n) is the number of ways of partitioning the divisors of n into two disjoint sets with equal sum, when an extra 1-divisor is added to the divisor set, and the two 1-divisors are considered indistinguishable.

1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 21, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 1

Author

Antti Karttunen, Jan 07 2025

Keywords

• nonn

Comments

Conjecture 1: For all n >= 1, a(A156942(n)) > 0. Also, if a(A156942(n)) > 1 is true for all n, it would imply that there are no quasiperfect numbers, numbers x with sigma(x) = 2x+1, as such numbers must all reside in A156942 and have a(x) = 1. (See references in A336700).

Conjecture 2: a(n) = 1 if and only if n = 2^k, with k >= 0. This claim is equal to the statement that there are neither quasiperfect numbers nor almost perfect (least deficient) numbers, numbers x with sigma(x) = 2x-1, others than those given by A000079.

Examples

			a(18) = 2 as its divisor set with an extra 1 is [1, 1, 2, 3, 6, 9, 18], and this can be partitioned to two sets with equal sums either as 1+1+3+6+9 = 2+18 or as 2+3+6+9 = 1+1+18.
a(36) = 5 as its divisor set with an extra 1 is [1, 1, 2, 3, 4, 6, 9, 12, 18, 36], and this can be partitioned in any of the following five ways, when two 1's are considered indistinguishable:
  1+1+2+6+36  = 3+4+9+12+18,
  1+2+3+4+36  = 1+6+9+12+18,
  1+3+6+36    = 1+2+4+9+12+18,
  1+9+36      = 1+2+3+4+6+12+18,
  4+6+36      = 1+1+2+3+9+12+18,
where each sum on the left and right hand side gives (sigma(36)+1)/2 = 46.
There are 42 partitions of (sigma(72)+1)/2 = 98 into the divisors of 72, [1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72], with an extra 1 allowed:
  [2, 24, 72],
  [1, 1, 24, 72],
  [8, 18, 72],
  [2, 6, 18, 72],
  [1, 1, 6, 18, 72],
  [1, 3, 4, 18, 72],
  [1, 1, 2, 4, 18, 72],
  [1, 4, 9, 12, 72],
  [2, 3, 9, 12, 72],
  [1, 1, 3, 9, 12, 72],
  [6, 8, 12, 72],
  [2, 4, 8, 12, 72],
  [1, 1, 4, 8, 12, 72],
  [1, 2, 3, 8, 12, 72],
  [1, 3, 4, 6, 12, 72],
  [1, 1, 2, 4, 6, 12, 72],
  [3, 6, 8, 9, 72],
  [1, 2, 6, 8, 9, 72],
  [2, 3, 4, 8, 9, 72],
  [1, 1, 3, 4, 8, 9, 72],
  [1, 1, 2, 3, 4, 6, 9, 72],
  [8, 12, 18, 24, 36],
  [2, 6, 12, 18, 24, 36],
  [1, 1, 6, 12, 18, 24, 36],
  [1, 3, 4, 12, 18, 24, 36],
  [1, 1, 2, 4, 12, 18, 24, 36],
  [3, 8, 9, 18, 24, 36],
  [1, 2, 8, 9, 18, 24, 36],
  [1, 4, 6, 9, 18, 24, 36],
  [2, 3, 6, 9, 18, 24, 36],
  [1, 1, 3, 6, 9, 18, 24, 36],
  [1, 1, 2, 3, 4, 9, 18, 24, 36],
  [2, 4, 6, 8, 18, 24, 36],
  [1, 1, 4, 6, 8, 18, 24, 36],
  [1, 2, 3, 6, 8, 18, 24, 36],
  [3, 6, 8, 9, 12, 24, 36],
  [1, 2, 6, 8, 9, 12, 24, 36],
  [2, 3, 4, 8, 9, 12, 24, 36],
  [1, 1, 3, 4, 8, 9, 12, 24, 36],
  [1, 1, 2, 3, 4, 6, 9, 12, 24, 36],
  [2, 3, 4, 6, 8, 9, 12, 18, 36],
  [1, 1, 3, 4, 6, 8, 9, 12, 18, 36],
therefore a(72) = 42/2 = 21.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..1799 [a larger b-file requested]

Crossrefs

Cf. A083206, A103977, A156942, A336700, A379502, A379503 (positions of nonzero terms), A379504 (version where two 1's are considered distinct).

Cf. A000079 (conjectured to give the positions of 1's).

Programs

PARI

partitions_into_distinct_parts_with_extra1allowed(n, parts, from=1) = if(n<=1, 1, if(from>#parts, 0, my(s=0); for(i=from, #parts, if(parts[i]<=n, s += partitions_into_distinct_parts_with_extra1allowed(n-parts[i], parts, i+1))); (s)));
A379505(n) = if(1==n, n, if(!issquare(n) && !issquare(2*n), 0, my(divs=divisors(n), s=sigma(n)); partitions_into_distinct_parts_with_extra1allowed((s+1)/2, vecsort(divs,,4))/2));

Formula

a(n) <= A379504(n).

A103977(n) = 1 <=> a(n) > 0.