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a(n) is the smallest number with n divisors d such that sigma(d) / tau(d) is an integer.
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%I #12 Sep 08 2022 08:46:25

%S 1,3,6,15,45,30,42,60,132,264,270,378,594,210,462,780,1050,420,924,

%T 660,2100,840,3060,1848,3300,1890,2970,2520,9702,2310,5544,3780,11592,

%U 8316,18216,5460,5940,7980,16830,7140,11550,4620,21252,10920,23760,22440,49500

%N a(n) is the smallest number with n divisors d such that sigma(d) / tau(d) is an integer.

%C a(n) is the smallest number m with n arithmetic divisors d (terms of A003601).

%C See A338170, A338171 and A338172 for number, sum and product of such divisors for n>=1.

%F 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.

%t 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 *)

%o (Magma) [Min([m: m in[1..10^5] | #[d: d in Divisors(m) | IsIntegral(&+Divisors(d) / #Divisors(d))] eq n]): n in [1..30]]

%o (PARI) isok(m, n) = sumdiv(m, d, !(sigma(d) % numdiv(d))) == n;

%o a(n) = my(m=1); while(!isok(m,n), m++); m; \\ _Michel Marcus_, Oct 21 2020

%Y Cf. A000005 (tau), A000203 (sigma), A003601 (arithmetic numbers).

%Y Cf. A334421 (smallest number with n divisors d such that sigma(d)/tau(d) is an integer for all divisors).

%K nonn

%O 1,2

%A _Jaroslav Krizek_, Oct 20 2020