OFFSET
1,1
COMMENTS
Conjecture: Any 4 consecutive terms include at least one Zumkeller number (verified for the first 10^5 terms). - Ivan N. Ianakiev, Oct 17 2019
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..10000
FORMULA
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]]; DivisorSigma[1, Select[ Range@ 275, ZumkellerQ]]/2 (* Robert G. Wilson v, Aug 03 2010 *)
PROG
(Python)
from sympy import divisors
import numpy as np
A175582 = []
for n in range(1, 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:
A175582.append(int(s/2))
break
# Chai Wah Wu, Aug 21 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladislav-Stepan Malakhovsky and Juri-Stepan Gerasimov, Jul 15 2010
EXTENSIONS
Inserted a(45) and corrected typo in a(49) and crossrefs by Chai Wah Wu, Aug 21 2014
STATUS
approved