A345704 Zumkeller numbers k (A083207) such that the next Zumkeller number is k + 12.

282, 840, 1596, 1794, 1920, 2496, 2928, 3108, 3522, 3540, 3594, 4008, 4188, 4602, 4620, 4998, 5268, 5862, 6060, 6708, 6888, 7086, 7788, 7968, 8382, 8400, 9048, 9840, 10362, 10542, 10920, 11100, 11568, 12126, 12162, 12180, 13422, 14106, 14322, 14394, 14880, 15348

Offset: 1

Author

Amiram Eldar, Jun 24 2021

Keywords

• nonn

Comments

Frank Buss and T. D. Noe conjectured (see A083207) and Robert Gerbicz proved that the largest possible gap between Zumkeller numbers is 12 (SeqFan post, 2010). A proof was also published by Mahanta et al. (2020).

Examples

			282 is a term since it is a Zumkeller number, and the next Zumkeller number is 282 + 12 = 294.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..648
• Robert Gerbicz, A083207 On an observation of Frank Buss, posts to the SeqFan list, July 2010.
• Pankaj Jyoti Mahanta, Manjil P. Saikia and Daniel Yaqubi, Some properties of Zumkeller numbers and k-layered numbers, Journal of Number Theory, Vol. 217 (2020), pp. 218-236.

Crossrefs

Cf. A083207, A179529, A328327.

Programs

Maple

iszum:= proc(n) local D,s,P,d;
   D:= numtheory:-divisors(n);
   s:= convert(D,`+`);
   if s::odd then return false fi;
   P:= mul(1+x^d,d=D);
   coeff(P,x,s/2) > 0
end proc:
last:= 6: R:= NULL: count:= 0:
for i from 7 while count < 60 do
  if iszum(i) then
     if i-last = 12 then R:= R, last; count:= count+1 fi;
     last:= i;
  fi
od:
R; # Robert Israel, Feb 13 2023

Mathematica

zumQ[n_] := Module[{d = Divisors[n], sum, x}, sum = Plus @@ d;  EvenQ[sum] && CoefficientList[Product[1 + x^i, {i, d}], x][[1 + sum/2]] > 0]; z = Select[Range[5000], zumQ]; z[[Position[Differences[z], 12] // Flatten]]

Python

from itertools import count, islice
from sympy import divisors
def A345704_gen(startvalue=1): # generator of terms >= startvalue
    m = -20
    for n in count(max(startvalue,1)):
        d = divisors(n)
        s = sum(d)
        if s&1^1 and n<<1<=s:
            d = d[:-1]
            s2, ld = (s>>1)-n, len(d)
            z = [[0 for  in range(s2+1)] for  in range(ld+1)]
            for i in range(1, ld+1):
                y = min(d[i-1], s2+1)
                z[i][:y] = z[i-1][:y]
                for j in range(y,s2+1):
                    z[i][j] = max(z[i-1][j],z[i-1][j-y]+y)
                if z[i][s2] == s2:
                    if m == n-12:
                        yield m
                    m = n
                    break
A345704_list = list(islice(A345704_gen(),10)) # Chai Wah Wu, Feb 13 2023