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

1,2

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

Links

Table of n, a(n) for n=1..67.

Example

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.

Mathematica

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

Prog

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

Crossrefs

Cf. A000005, A000040 (subsequence), A000203.

Sequence in context: A152073 A331046 A329150 * A230606 A117289 A339817

Adjacent sequences: A384351 A384352 A384353 * A384355 A384356 A384357

Keyword

nonn

Author

Ivan N. Ianakiev, May 27 2025

Status

approved