OFFSET
1,2
COMMENTS
This sequence is infinite because k can be of the form m*10^n for some m and n > 0.
FORMULA
EXAMPLE
a(4) = 1112, because a(3) = 111 and tau(111) = 4, so a(n) must be equal to 111//k ("//" denotes concatenation) and tau(a(4)) > 4, therefore the smallest k that satisfies this condition is 2.
MATHEMATICA
f[1] = {1, 1}; f[n_] := f[n] = Module[{k = 1, t = f[n - 1][[1]], d = IntegerDigits[f[n - 1][[2]]]}, While[(t1 = DivisorSigma[0, (n1 = FromDigits @ Join[d, IntegerDigits[k]])]) <= t, k++]; {t1, n1}]; a[n_] := f[n][[2]]; Array[a, 14] (* Amiram Eldar, Sep 26 2020 *)
PROG
(PARI) a(n) = if(n == 1, l = m = vector(100); return(l[1] = 1), for(k = 1, oo, j = eval(concat(Str(l[n-1]), k)); if((m[n] = numdiv(j)) > m[n-1], return(l[n] = j))))
for(k=1, 10, print1(a(k), ", "));
CROSSREFS
KEYWORD
base,nonn,less
AUTHOR
Eder Vanzei, Sep 25 2020
STATUS
approved