A359260 Numbers m such that the arithmetic mean of the first k divisors of m is an integer for all k in 1..d(m), where d(m) = A000005(m).

1, 3, 5, 7, 11, 13, 15, 17, 19, 23, 29, 31, 33, 37, 41, 43, 47, 49, 51, 53, 59, 61, 67, 69, 71, 73, 79, 83, 87, 89, 91, 97, 101, 103, 107, 109, 113, 123, 127, 131, 133, 137, 139, 141, 149, 151, 157, 159, 163, 167, 169, 173, 177, 179, 181, 191, 193, 197, 199, 211

Offset: 1

Author

Amiram Eldar, Dec 23 2022

Keywords

• nonn
• easy

Comments

All the terms are arithmetic numbers (A003601).

All the terms are odd numbers.

All the odd primes are terms.

There are infinitely many composite numbers in this sequence. For example, if p is a prime of the form 6*k-1 (A007528), then 3*p is a term. Also, if p is a prime of the form 6*k + 1 (A002476), then p^2 is a term.

prime(n)^k is a term for k = 0..A359262(n).

Examples

			15 is a term since its divisors are {1, 3, 5, 15}, 1/1 =1, (1 + 3)/2 = 2, (1 + 3 + 5)/3 = 3, and (1 + 3 + 5 + 15)/4 = 6 are all integers.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Subsequence of A003601.

Subsequences: A065091, A343022 \ {81}.

Cf. A000005, A002476, A007528, A065091, A359261, A359262.

Programs

Mathematica

q[n_] := AllTrue[Accumulate[(d = Divisors[n])]/Range[Length[d]], IntegerQ]; Select[Range[1, 200, 2], q]

PARI

is(n) = {my(s = k = 0); fordiv(n, d, k++; s += d; if(s%k, return(0))); 1;}