A361206 Lexicographically earliest infinite sequence of distinct imperfect numbers such that the sum of the abundance of all terms is never < 1.

12, 1, 2, 4, 18, 3, 8, 20, 10, 24, 5, 7, 16, 30, 9, 14, 32, 36, 11, 13, 40, 15, 42, 17, 48, 19, 21, 54, 22, 44, 56, 50, 60, 23, 25, 52, 64, 66, 26, 70, 72, 27, 29, 34, 78, 45, 80, 33, 68, 84, 31, 35, 88, 90, 37, 38, 96, 39, 41, 100, 46, 102, 76, 104, 108, 43, 58

Offset: 1

Author

John Tyler Rascoe, Mar 04 2023

Keywords

• nonn
• easy

Comments

The abundance of each term is A033880(a(n)) and s = Sum_{i=1..n} A033880(a(i)).

All imperfect numbers A132999 will appear in this sequence and the abundant numbers A005101 will appear in order.

Examples

			The sequence starts with a(1) = 12, since 12 is the first imperfect number with abundance greater than 0. Then the next term not yet in the sequence, such that s is not less than 1, is 1.
a(5) is the next abundant number 18, since any deficient number would bring s below 1.
n   :  1  2  3  4   5  6  7   8   9  10
a(n): 12  1  2  4  18  3  8  20  10  24
s   :  4  3  2  1   4  2  1   3   1  13
		

Crossrefs

Cf. A005101, A033879, A033880, A132999.

Programs

Python

from sympy.ntheory import abundance
from itertools import count, filterfalse
def A361206_list(nmax):
    A,s = [],0
    for n in range(1,nmax+1):
        A2 = set(A)
        for y in filterfalse(A2._contains_,count(1)):
            ab = abundance(y)
            if ab != 0 and ab + s >= 1:
                A.append(y)
                s += ab
                break
    return(A)