A334421 a(n) is the smallest number m with n divisors d such that sigma(d)/tau(d) is an integer for all divisors d.

1, 3, 49, 15, 923521, 147, 88245939632761, 105, 8281, 2770563, 4345096786921664259621718196367601, 735, 9024585590445680759701490904755712009585829774768244676951841, 264737818898283, 45252529, 1155

Offset: 1

Author

Jaroslav Krizek, May 01 2020

Keywords

• nonn

Comments

Sequences of numbers m with n such divisors for 1 < n < 6:

n = 2: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ... (A065091 - odd primes).

n = 3: 49, 169, 361, 961, 1369, 1849, 3721, 4489, 5329, 6241, 9409, ...

n = 4: 15, 21, 33, 35, 39, 51, 55, 57, 65, 69, 77, 85, 87, 91, 93, ...

n = 5: 923521, 13845841, 519885601, 1073283121, 1982119441, ...

Other terms: a(18) = 24843, a(24) = 8085, a(32) = 15015.

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]

a(17) = 4084081^16 and a(19) = 106696591^18 are too large to be included. - Amiram Eldar, May 02 2020

Examples

			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.
		

Crossrefs

Cf. A000005 (tau), A000203 (sigma), A003601, A306639, A324499, A324500, A334420.

Programs

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]]

Extensions

a(13)-a(14) from Amiram Eldar, May 02 2020