OFFSET
0,2
COMMENTS
36*5^k has k-1 exponential abundant divisors for k >= 1. Therefore, a(n) exists for all n.
MATHEMATICA
f[p_, e_] := DivisorSum[e, p^# &]; esigma[1] = 1; esigma[n_] := Times @@ f @@@ FactorInteger[n];
s[n_] := Count[Divisors[n], _?(esigma[#] > 2*# &)];
seq[len_] := Module[{v = Table[0, {len}], c = 0, n = 1, i}, While[c < len, i = s[n] + 1; If[i <= len && v[[i]] == 0, c++; v[[i]] = n]; n++]; v]; seq[14]
PROG
(PARI) esigma(n) = {my(f = factor(n)); prod(k=1, #f~, sumdiv(f[k, 2], d, f[k, 1]^d)); }
s(n) = sumdiv(n, d, esigma(d) > 2*d);
list(len) = {my(v = vector(len), c = 0, n = 1, i); while(c < len, i = s(n) + 1; if(i <= len && v[i] == 0, c++; v[i] = n); n++); v; }
CROSSREFS
KEYWORD
nonn
AUTHOR
Amiram Eldar, Sep 29 2025
STATUS
approved
