A337844 a(1) = 1; a(n) = a(n-1) concatenated with the smallest number k>0 such that tau(a(n)) > tau(a(n-1)).

1, 11, 111, 1112, 11124, 1112410, 111241020, 11124102080, 11124102080130, 11124102080130102, 11124102080130102180, 11124102080130102180480, 11124102080130102180480160, 11124102080130102180480160116

Offset

1,2

Comments

This sequence is infinite because k can be of the form m*10^n for some m and n > 0.

Links

Table of n, a(n) for n=1..14.

Formula

a(1) = 1; a(n) = a(n-1) concatenated with the smallest number k>0, such that A000005(a(n)) > A000005(a(n-1)).

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

Cf. A000005.

Sequence in context: A346789 A134732 A166747 * A015456 A343355 A240367

Adjacent sequences: A337841 A337842 A337843 * A337845 A337846 A337847

Keyword

base,nonn,less

Author

Eder Vanzei, Sep 25 2020

Status

approved