A085493 Numbers k having partitions into distinct divisors of k + 1.
1, 3, 5, 7, 11, 15, 17, 19, 23, 27, 29, 31, 35, 39, 41, 47, 53, 55, 59, 63, 65, 69, 71, 77, 79, 83, 87, 89, 95, 99, 103, 107, 111, 119, 125, 127, 131, 139, 143, 149, 155, 159, 161, 167, 175, 179, 191, 195, 197, 199, 203, 207, 209, 215, 219, 223, 227, 233, 239
Offset: 1
Keywords
Examples
The divisors of 42 are 1, 2, 3, 6, 7, 14, 21, 42. Since 6 + 14 + 21 = 41, 41 is in the sequence. The divisors of 43 are 1, 43. Since no selection of these divisors can possibly add up to 42, this means that 42 is not in the sequence.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..10000
- Paul K. Stockmeyer, Of camels, inheritance, and unit fractions, Math Horizons, 21 (2013), 8-11.
Programs
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Maple
q:= proc(m) option remember; local b, l; b, l:= proc(n, i) option remember; n=0 or i>=1 and (l[i]<=n and b(n-l[i], i-1) or b(n, i-1)) end, sort([numtheory[divisors](m+1)[]]); b(m, nops(l)-1) end: select(q, [$1..300])[]; # Alois P. Heinz, Feb 04 2023
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Mathematica
divNextableQ[n_] := TrueQ[Length[Select[Subsets[Divisors[n + 1]], Plus@@# == n &]] > 0]; Select[Range[100], divNextableQ] (* Alonso del Arte, Jan 07 2023 *)
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Scala
def divisors(n: Int): IndexedSeq[Int] = (1 to n).filter(n % _ == 0) def divPartSums(n: Int): List[Int] = divisors(n).toSet.subsets.toList.map(_.sum) (1 to 128).filter(n => divPartSums(n + 1).contains(n)) // Alonso del Arte, Jan 26 2023
Formula
{k > 0 : 0 < [x^k] Product_{d divides (k+1)} (1+x^d)}. - Alois P. Heinz, Feb 04 2023
Comments