A396381 Bi-unitary half-Zumkeller numbers: numbers whose set of bi-unitary proper divisors can be partitioned into two disjoint subsets of equal sum.

6, 12, 20, 24, 30, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 102, 104, 108, 114, 120, 138, 150, 160, 162, 168, 174, 180, 186, 192, 210, 216, 220, 222, 224, 240, 246, 252, 258, 264, 270, 280, 282, 288, 294, 312, 318, 320, 330, 336, 352, 354, 360, 364, 366, 378, 384

Offset

1,1

Comments

The first term that is not: (a) bi-unitary Zumkeller number (A335215) is 12, (b) unitary Zumkeller number (A290466) is 12, (c) unitary half-Zumkeller number (A290467) is 24, (d) Zumkeller number (A083207) is 72, and (e) half-Zumkeller number (A246198) is 72.

All bi-unitary perfect numbers; i.e., 6, 60 and 90, are terms.

There are no odd terms. Suppose there was one. Since the set of its bi-unitary proper divisors must have an odd number of elements and all of them must be odd, then their sum must be odd. Contradiction.

Links

Table of n, a(n) for n=1..61.

Example

The set of bi-unitary proper divisors of 72 is a union of the disjoint subsets {1,2,36} and {4,8,9,18}, whose sums both equal 39.

Mathematica

uDiv[n_]:=Select[Divisors[n], GCD[#, n/#]==1&];
buDiv[n_]:=Select[Divisors[n], Intersection[uDiv[#], uDiv[n/#]]=={1}&];
buHZNQ[n_/; n>1]:=Module[{d=Most[buDiv[n]], t, ds, x}, ds=Plus@@d; If[Mod[ds, 2]>0, False, t=CoefficientList[Product[1+x^i, {i, d}], x]; t[[1+ds/2]]>0]]; Select[Range[500], buHZNQ]

Crossrefs

Cf. A083207, A246198, A290466, A290467, A335215.

Sequence in context: A378541 A378599 A370205 * A304391 A145278 A094371

Adjacent sequences: A396378 A396379 A396380 * A396382 A396383 A396384

Keyword

nonn

Author

Ivan N. Ianakiev, May 24 2026

Status

approved