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%I #26 Sep 08 2022 08:46:25
%S 1,3,49,15,923521,147,88245939632761,105,8281,2770563,
%T 4345096786921664259621718196367601,735,
%U 9024585590445680759701490904755712009585829774768244676951841,264737818898283,45252529,1155
%N a(n) is the smallest number m with n divisors d such that sigma(d)/tau(d) is an integer for all divisors d.
%C Sequences of numbers m with n such divisors for 1 < n < 6:
%C n = 2: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ... (A065091 - odd primes).
%C n = 3: 49, 169, 361, 961, 1369, 1849, 3721, 4489, 5329, 6241, 9409, ...
%C n = 4: 15, 21, 33, 35, 39, 51, 55, 57, 65, 69, 77, 85, 87, 91, 93, ...
%C n = 5: 923521, 13845841, 519885601, 1073283121, 1982119441, ...
%C Other terms: a(18) = 24843, a(24) = 8085, a(32) = 15015.
%C a(p) = (k*p# + 1)^(p-1) for some k > 0 and p# is the product of primes <= k. - _David A. Corneth_, May 01 2020 [Proof: a(p) must be of the form q^(p-1), where q is a prime. Thus r | (q^r - 1)/(q - 1) if r <= p is prime. Suppose that q - 1 is not divisible by r, then 0 == q^r - 1 == q - 1 (mod r), a contradiction! Therefore, q - 1 is divisible by any primes r <= p. In conclusion, q = k*p# + 1 for some k > 0. - _Jinyuan Wang_, May 02 2020]
%C a(17) = 4084081^16 and a(19) = 106696591^18 are too large to be included. - _Amiram Eldar_, May 02 2020
%e a(4) = 15 because number 15 with divisors 1, 3, 5 and 15 is the smallest number with 4 such divisors: sigma(1)/tau(1) = 1/1 = 1, sigma(3)/tau(3) = 4/2 = 2, sigma(5)/tau(5) = 6/2 = 3, sigma(15)/tau(15) = 24/4 = 6.
%o (Magma) [Min([m: m in[1..1000] | &+[&+Divisors(d) mod #Divisors(d): d in Divisors(m)] eq 0 and #Divisors(m) eq k]): k in [1..4]]
%Y Cf. A000005 (tau), A000203 (sigma), A003601, A306639, A324499, A324500, A334420.
%K nonn
%O 1,2
%A _Jaroslav Krizek_, May 01 2020
%E a(13)-a(14) from _Amiram Eldar_, May 02 2020
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