OFFSET
1,2
COMMENTS
a(n) exists for each n because 4^(n-1) has n antiharmonic divisors.
EXAMPLE
MATHEMATICA
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 *)
PROG
(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;
(PARI) a(n) = my(k=1); while(sumdiv(k, d, sigma(d, 2)%sigma(d)==0) != n, k++); k; \\ Michel Marcus, Dec 18 2023
CROSSREFS
KEYWORD
nonn
AUTHOR
Marius A. Burtea, Dec 17 2023
STATUS
approved