A368215 - OEIS

%I #11 Jan 22 2024 06:09:56

%S 1,4,16,36,256,100,200,576,400,800,2600,900,3200,1800,16900,6400,

%T 12800,3600,20800,7200,11700,36000,67600,14400,23400,28800,32400,

%U 88200,397800,64800,270400,46800,152100,115200,234000,93600,1258400,230400,259200,352800,1081600

%N a(n) is the smallest number k >= 1 that has exactly n divisors in A020487.

%C a(n) exists for each n because 4^(n-1) has n antiharmonic divisors.

%e a(1) = 1 because 1 has only one divisor 1 = A020487(1).

%e The numbers 2 and 3 have only the divisor 1 in A020487 and 4 has the divisors 1 = A020487(1) and 4 = A020487(2), so a(2) = 4.

%t f[n_] := DivisorSum[n, 1 &, Divisible[DivisorSigma[2, #], DivisorSigma[1, #]] &]; seq[len_] := Module[{s = Table[0, {len}], c = 0, n = 1}, While[c < len, If[(i = f[n]) <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[25] (* _Amiram Eldar_, Dec 18 2023 *)

%o (Magma) f:=func<n|DivisorSigma(2,n) mod DivisorSigma(1,n) eq 0>; a:=[]; for n in [1..41] do k:=1; while #[d:d in Divisors(k)|f(d)] ne n do k:=k+1; end while; Append(~a,k); end for; a;

%o (PARI) a(n) = my(k=1); while(sumdiv(k, d, sigma(d, 2)%sigma(d)==0) != n, k++); k; \\ _Michel Marcus_, Dec 18 2023

%Y Cf. A000005, A000203, A001157, A020487, A337326.

%K nonn

%O 1,2

%A _Marius A. Burtea_, Dec 17 2023