|
%I #21 Nov 27 2019 04:28:34
%S 1,1,1,1,1,1,1,1,1,10,1144,1,28,10,10,28,136,1,19,10,1,1144,7522840,1,
%T 28,208,1,28,3016,10,217,28,1144,32032,28,1,37,174813842944,208,10,
%U 20653750,1,172,352,1,46,4888,1,532,10,32032,28,424,1,55,5824,174813842944
%N 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.
%C 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
%e Some terms and corresponding sequences:
%e a(1)=1 1,1,1,(1),...
%e a(11)=1144 11,22,88,1408,18304,1144,11440,(1144,11440),...
%e a(14)=10 14,70,10,10,(10),...
%e a(263)=1374175 263,2893,63646,1591150,72325,(1374175,38476900,1423645300,50844475),...
%t 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 *)
%K nonn,base
%O 1,10
%A _Amarnath Murthy_, Aug 09 2005
%E Edited and extended by _Giovanni Resta_, Nov 27 2019
|
