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A083207
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Zumkeller or integer-perfect numbers: numbers n whose divisors can be partitioned into two disjoint sets with equal sum.
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103
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6, 12, 20, 24, 28, 30, 40, 42, 48, 54, 56, 60, 66, 70, 78, 80, 84, 88, 90, 96, 102, 104, 108, 112, 114, 120, 126, 132, 138, 140, 150, 156, 160, 168, 174, 176, 180, 186, 192, 198, 204, 208, 210, 216, 220, 222, 224, 228, 234, 240, 246, 252, 258, 260, 264, 270, 272
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OFFSET
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1,1
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COMMENTS
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The 229026 Zumkeller numbers less than 10^6 have a maximum difference of 12. This leads to the conjecture that any 12 consecutive numbers include at least one Zumkeller number. There are 1989 odd Zumkeller numbers less than 10^6; they are exactly the odd abundant numbers that have even abundance, A174865. - T. D. Noe, Mar 31 2010
All 205283 odd abundant numbers less than 10^8 that have even abundance are Zumkeller numbers. - T. D. Noe, Nov 14 2010
Except for 1 and 2, all primorials (A002110) are Zumkeller numbers (follows from Fact 6 in the Rao/Peng paper). - Ivan N. Ianakiev, Mar 23 2016
Conjecture: Any 4 consecutive terms include at least one number k such that sigma(k)/2 is also a Zumkeller number (verified for the first 10^5 Zumkeller numbers). - Ivan N. Ianakiev, Apr 03 2017
LeVan studied these numbers using the equivalent definition of numbers n such that n = Sum_{d|n, d<n} alpha(d)*d, where alpha(d) is either 1 or -1, and named them "integer-perfect numbers". She also named the primitive Zumkeller numbers (A180332) "minimal integer-perfect numbers". - Amiram Eldar, Dec 20 2018
The numbers 3 * 2^k for k > 0 are all Zumkeller numbers: half of one such partition is {3*2^k, 3*2^(k-2), ...}, replacing 3 with 2 if it appears. With this and the lemma that the product of a Zumkeller number and a number coprime to it is again a Zumkeller number (see A179527), we have that all numbers divisible by 6 but not 9 (or numbers congruent to 6 or 12 modulo 18) are Zumkeller numbers, proving that the difference between consecutive Zumkeller numbers is at most 12. - Charlie Neder, Jan 15 2019
Improvements on the previous comment: 1) For every integer q > 0, every odd integer r > 0 and every integer s > 0 relatively prime to 6, the integer 2^q*3^r*s is a Zumkeller number, and therefore 2) there exist Zumkeller numbers divisible by 9 (such as 54, 90, 108, 126, etc.). - Ivan N. Ianakiev, Jan 16 2020
Conjecture: If d > 1, d|k and tau(d)*sigma(d) = k, then k is a Zumkeller number. - Ivan N. Ianakiev, Apr 24 2020
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REFERENCES
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Marijo O. LeVan, Integer-perfect numbers, Journal of Natural Sciences and Mathematics, Vol. 27, No. 2 (1987), pp. 33-50.
Marijo O. LeVan, On the order of nu(n), Journal of Natural Sciences and Mathematics, Vol. 28, No. 1 (1988), pp. 165-173.
J. Sandor and B. Crstici, Handbook of Number Theory, II, Springer Verlag, 2004, chapter 1.10, pp. 53-54.
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LINKS
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K. P. S. Bhaskara Rao and Yuejian Peng, On Zumkeller Numbers, Journal of Number Theory, Volume 133, Issue 4, April 2013, pp. 1135-1155.
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EXAMPLE
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Given n = 48, we can partition the divisors thus: 1 + 3 + 4 + 6 + 8 + 16 + 24 = 2 + 12 + 48, therefore 48 is a term (A083206(48) = 5).
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MAPLE
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with(numtheory): with(combstruct):
is_A083207 := proc(n) local S, R, Found, Comb, a, s; s := sigma(n);
if not(modp(s, 2) = 0 and n * 2 <= s) then RETURN(false) fi;
S := s / 2 - n; R := select(m -> m <= S, divisors(n)); Found := false;
Comb := iterstructs(Combination(R)):
while not finished(Comb) and not Found do
Found := add(a, a = nextstruct(Comb)) = S
od; Found end:
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MATHEMATICA
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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]]; Select[Range[1000], ZumkellerQ] (* T. D. Noe, Mar 31 2010 *)
znQ[n_]:=Length[Select[{#, Complement[Divisors[n], #]}&/@Most[Rest[ Subsets[ Divisors[ n]]]], Total[#[[1]]]==Total[#[[2]]]&]]>0; Select[Range[300], znQ] (* Harvey P. Dale, Dec 26 2022 *)
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PROG
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(Haskell)
a083207 n = a083207_list !! (n-1)
a083207_list = filter (z 0 0 . a027750_row) $ [1..] where
z u v [] = u == v
z u v (p:ps) = z (u + p) v ps || z u (v + p) ps
(PARI) part(n, v)=if(n<1, return(n==0)); forstep(i=#v, 2, -1, if(part(n-v[i], v[1..i-1]), return(1))); n==v[1]
is(n)=my(d=divisors(n), s=sum(i=1, #d, d[i])); s%2==0 && part(s/2-n, d[1..#d-1]) \\ Charles R Greathouse IV, Mar 09 2014
(Python3)
from sympy import divisors
from sympy.combinatorics.subsets import Subset
for n in range(1, 10**3):
d = divisors(n)
s = sum(d)
if not s % 2 and max(d) <= s/2:
for x in range(1, 2**len(d)):
if sum(Subset.unrank_binary(x, d).subset) == s/2:
print(n, end=', ')
break
(Python)
from sympy import divisors
import numpy as np
for n in range(2, 10**3):
d = divisors(n)
s = sum(d)
if not s % 2 and 2*n <= s:
d.remove(n)
s2, ld = int(s/2-n), len(d)
z = np.zeros((ld+1, s2+1), dtype=int)
for i in range(1, ld+1):
y = min(d[i-1], s2+1)
z[i, range(y)] = z[i-1, range(y)]
z[i, range(y, s2+1)] = np.maximum(z[i-1, range(y, s2+1)], z[i-1, range(0, s2+1-y)]+y)
if z[i, s2] == s2:
break
(Sage)
def is_Zumkeller(n):
s = sigma(n)
if not (2.divides(s) and n*2 <= s): return False
S = s // 2 - n
R = (m for m in divisors(n) if m <= S)
return any(sum(c) == S for c in Combinations(R))
A083207_list = lambda lim: [n for n in (1..lim) if is_Zumkeller(n)]
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CROSSREFS
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Cf. A083209, A083211, A000203, A005101, A000396, A005835, A048055, A171641, A027750, A175592, A221054, A293453.
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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