A293453 Zumkeller numbers k such that sigma(k)/2 is a Zumkeller number.

6, 24, 28, 42, 54, 56, 60, 78, 84, 88, 96, 102, 108, 114, 120, 126, 132, 140, 150, 168, 174, 176, 186, 198, 204, 216, 220, 222, 224, 228, 240, 246, 252, 258, 260, 264, 270, 276, 280, 294, 308, 312, 330, 336, 340, 342, 348, 350, 352, 354, 366, 368, 372, 378, 380, 384, 390, 396, 402

Offset: 1

Author

Ivan N. Ianakiev, Oct 09 2017

Keywords

• easy
• nonn

Comments

Subsequence of A083207 (Zumkeller numbers).

Conjecture: Any four consecutive Zumkeller numbers include at least one term of the present sequence (verified for the first 10^5 Zumkeller numbers).

The perfect numbers (A000396) are all trivially in this sequence.

Examples

			The fourth Zumkeller number is 24, since sigma(24) = A000203(24) = 60, 60/2 = 30, and the divisors of 24 can be partitioned as 1 + 2 + 3 + 4 + 8 + 12 = 6 + 24 = 30.
In turn, 30 is also a Zumkeller number, as sigma(30)/2 = 72/2 = 36 and 1 + 2 + 3 + 5 + 10 + 15 = 6 + 30 = 36.
Therefore 24 is in this sequence.
But since 36 is not a Zumkeller number at all, 30 is not in this sequence.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000203, A000396, A083207.

Programs

Mathematica

zumkellerQ[n_] := Module[{d = Divisors[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]]; zn = Select[Range[1000], zumkellerQ] (* a code by T. D. Noe at A083207 *); Select[zn, zumkellerQ[DivisorSigma[1, #]/2] &]