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

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%I A359261 #6 Dec 23 2022 11:22:15
%S A359261 1,3,49,15,923521,1519,88245939632761,3913,1117249,3131659711,
%T A359261 4345096786921664259621718196367601,238483,
%U A359261 9024585590445680759701490904755712009585829774768244676951841,2772760313554466311,198528059518891985825881,32748812641
%N A359261 a(n) is the least term of A359260 whose number of divisors is n.
%C A359261 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.
%C A359261 a(17) = 4084081^16 = 5.991...*10^105 is too large to include in the data section.
%C A359261 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.
%e A359261 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.
%Y A359261 Cf. A000005, A003418, A359260, A359262.
%Y A359261 Similar sequence: A334421.
%K A359261 nonn
%O A359261 1,2
%A A359261 _Amiram Eldar_, Dec 23 2022