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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).
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%I #18 Feb 12 2023 17:26:28

%S 1,3,5,7,11,13,15,17,19,23,29,31,33,37,41,43,47,49,51,53,59,61,67,69,

%T 71,73,79,83,87,89,91,97,101,103,107,109,113,123,127,131,133,137,139,

%U 141,149,151,157,159,163,167,169,173,177,179,181,191,193,197,199,211

%N 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).

%C All the terms are arithmetic numbers (A003601).

%C All the terms are odd numbers.

%C All the odd primes are terms.

%C 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.

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

%H Amiram Eldar, <a href="/A359260/b359260.txt">Table of n, a(n) for n = 1..10000</a>

%e 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.

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

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

%Y Subsequence of A003601.

%Y Subsequences: A065091, A343022 \ {81}.

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

%K nonn,easy

%O 1,2

%A _Amiram Eldar_, Dec 23 2022