A359873 Minimum number of consecutive positive integers to guarantee n abundant or perfect numbers.

6, 12, 18, 24, 30, 36, 38, 42, 48, 52, 56, 60, 66, 72, 78

Offset

1,1

Comments

It is rather easy to show that a(n) <= 6n. Marc Deléglise (1998)'s bounds on the density of abundant numbers gives that a(n)/n approaches a value between 4.032 and 4.043.

Links

Table of n, a(n) for n=1..15.

Marc Deléglise, Bounds for the density of abundant integers, Experimental Mathematics, Vol. 7, No. 2 (1998), pp. 137-143; alternative link.

Example

For n=8, there must be 7 multiples of 6 within the 42 numbers, which are all abundant/perfect. There must also be 2 multiples of 20 that are abundant/perfect. Note that between the multiples of 6 and the multiples of 20, there can be at most one repetition, thus giving 8 abundant numbers. 41 fails with 115, 116, ..., 155.

Prog

(Python)
def f(n):
    ans = 0
    for i in range(1, n):
        if n % i ==0:
            ans += i
    return ans
l=[]
for i in range(1, 100000):
    if f(i) - i > -1:
        l.append(i)
def listing(k, f):
    for i in range(len(l)-k):
        if l[i+k]-l[i] > f:
            return (l[i], l[i+k])
def a(n):
    a = 0
    while not listing(n, a) == None:
        a += 1
    return a

Crossrefs

Cf. A023196 (abundant or perfect numbers).

Sequence in context: A315768 A315769 A315770 * A315771 A315772 A121827

Adjacent sequences: A359870 A359871 A359872 * A359874 A359875 A359876

Keyword

nonn,more

Author

Lincoln Liu, May 20 2023

Status

approved