A384354 Numbers k such that the arithmetic mean of the divisors of k evenly divides k+1.

1, 2, 3, 5, 7, 11, 13, 17, 19, 20, 23, 29, 31, 35, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 104, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 207, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293

Offset: 1

Author

Ivan N. Ianakiev, May 27 2025

Keywords

• nonn

Comments

A term k with a fractional arithmetic mean of divisors is allowed as long as that arithmetic mean evenly divides k+1.

There exist triples such as (19,20,21) and quadruples such as (1,2,3,4) of consecutive numbers where the arithmetic mean of the divisors of every earlier number evenly divides the immediately following number. Are there similar quintuples?

Contains every prime p since (1+p)/2 evenly divides 1+p. - Michael S. Branicky, May 29 2025

Examples

			2 is a term since (1+2)/2 = 3/2 and 3/2 evenly divides 3.
19 is a term since (1+19)/2 is 10 and 10 evenly divides 20.
20 is a term since (1+2+4+5+10+20)/6 = 7 and 7 evenly divides 21.
		

Crossrefs

Cf. A000005, A000040 (subsequence), A000203.

Programs

Mathematica

fQ[n_]:=Divisible[n+1,Mean[Divisors[n]]]; Select[Range[300],fQ]

PARI

isok(k) = my(f=factor(k)); denominator((k+1)/(sigma(f)/numdiv(f))) == 1; \\ Michel Marcus, May 31 2025

Python

from sympy import divisors
def ok(n): return n and (n+1)*len(d:=divisors(n))%sum(d) == 0
print([k for k in range(300) if ok(k)]) # Michael S. Branicky, May 29 2025