OFFSET
1,2
COMMENTS
FORMULA
a(3) = 6 because number 6 is the smallest number with 3 such divisors (1, 3 and 6): sigma(1) / tau(1) = 1 / 1 = 1; sigma(3) / tau(3) = 4 / 2 = 2; sigma(6) / tau(6) = 12 / 4 = 3.
MATHEMATICA
f[n_] := DivisorSum[n, 1 &, Divisible[DivisorSigma[1, #], DivisorSigma[0, #]] &]; m = 50; s = Table[0, {m}]; c = 0; n = 1; While[c < m, If[(i = f[n]) <= m && s[[i]] == 0, c++; s[[i]] = n]; n++]; s (* Amiram Eldar, Oct 21 2020 *)
PROG
(Magma) [Min([m: m in[1..10^5] | #[d: d in Divisors(m) | IsIntegral(&+Divisors(d) / #Divisors(d))] eq n]): n in [1..30]]
(PARI) isok(m, n) = sumdiv(m, d, !(sigma(d) % numdiv(d))) == n;
a(n) = my(m=1); while(!isok(m, n), m++); m; \\ Michel Marcus, Oct 21 2020
CROSSREFS
Cf. A334421 (smallest number with n divisors d such that sigma(d)/tau(d) is an integer for all divisors).
KEYWORD
nonn
AUTHOR
Jaroslav Krizek, Oct 20 2020
STATUS
approved