A110731 a(n) is the least term of the period of the sequence defined by t(1)=n; t(k) = t(k-1)/sod(t(k-1)) if t(k-1) is a multiple of sod(t(k-1)) or t(k-1)*sod(t(k-1)) otherwise, where sod(m)=A007953(m) is the sum of digits of m.

1, 1, 1, 1, 1, 1, 1, 1, 1, 10, 1144, 1, 28, 10, 10, 28, 136, 1, 19, 10, 1, 1144, 7522840, 1, 28, 208, 1, 28, 3016, 10, 217, 28, 1144, 32032, 28, 1, 37, 174813842944, 208, 10, 20653750, 1, 172, 352, 1, 46, 4888, 1, 532, 10, 32032, 28, 424, 1, 55, 5824, 174813842944

Offset

1,10

Comments

Even period lengths from 2 to 16 arise for the first time from n equal to 11, 263, 149, 3244, 7339, 6929, 31201, and 47397, respectively. Are periods of odd length greater than 1 possible? - Giovanni Resta, Nov 27 2019

Links

Table of n, a(n) for n=1..57.

Example

Some terms and corresponding sequences:
a(1)=1          1,1,1,(1),...
a(11)=1144      11,22,88,1408,18304,1144,11440,(1144,11440),...
a(14)=10        14,70,10,10,(10),...
a(263)=1374175  263,2893,63646,1591150,72325,(1374175,38476900,1423645300,50844475),...

Mathematica

a[k_] := Block[{aa, w, t, p}, aa[1] = k; aa[n_] := aa[n] = Block[{s = Plus @@ IntegerDigits@ aa[n-1]}, aa[n-1] If[Mod[aa[n-1], s] == 0, 1/s, s]]; w = Array[aa, 50]; While[({t, p} = FindTransientRepeat[w, 2])[[2]] == {}, w = Array[aa, Length[w] + 50]]; Min[p]]; Array[a, 60]; (* Giovanni Resta, Nov 27 2019 *)

Crossrefs

Sequence in context: A130598 A267595 A266621 * A223119 A233252 A249849

Adjacent sequences: A110728 A110729 A110730 * A110732 A110733 A110734

Keyword

nonn,base

Author

Amarnath Murthy, Aug 09 2005

Extensions

Edited and extended by Giovanni Resta, Nov 27 2019

Status

approved