A359262 -id:A359262 - cp's OEIS frontend

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

Amiram Eldar, Dec 23 2022

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).

			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.

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

Subsequence of A003601.
Subsequences: A065091, A343022 \ {81}.
Cf. A000005, A002476, A007528, A065091, A359261, A359262.

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

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

A359261 a(n) is the least term of A359260 whose number of divisors is n.

Original entry on oeis.org

1, 3, 49, 15, 923521, 1519, 88245939632761, 3913, 1117249, 3131659711, 4345096786921664259621718196367601, 238483, 9024585590445680759701490904755712009585829774768244676951841, 2772760313554466311, 198528059518891985825881, 32748812641

Offset: 1

Amiram Eldar, Dec 23 2022

nonn

a(n) is the least number m whose number of divisors is A000005(m) = n such that the arithmetic mean of the first k divisors of m is an integer for all k in 1..n.
a(17) = 4084081^16 = 5.991...*10^105 is too large to include in the data section.
a(n) exists for all n >= 1. For n > 1, consider a prime p of the form m*lcm(1,2,...n-1) + 1, with m >= 1. Such a prime exists by Dirichlet's theorem on arithmetic progressions. Then, p^(n-1) has n divisors, and p^k == 1 (mod lcm(1..n-1)) for k = 0..(n-1). Therefore, Sum_{k=0..n-1} p^k == k (mod lcm(1,2,...n-1)), or equivalently, Sum_{k=0..n-1} p^k is divisible by k for k = 0..(n-1). Thus, p^(n-1) is in A359260.

			a(3) = 49 since 49 is the least number with 3 divisors in A359260. Its divisors are {1, 7, 49}, 1/1 = 1, (1+7)/2 = 4, and (1+7+49)/3 = 19 are all integers.

Cf. A000005, A003418, A359260, A359262.

Similar sequence: A334421.

cp's OEIS Frontend

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).

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