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%I #25 Oct 30 2018 10:31:02
%S 2,7,4,5,7,4,5,4,3,2,8,4,5,8,4,5,4,3,3,7,5,5,8,4,5,4,3,3,8,4,5,8,4,5,
%T 4,3,3,8,5,5,8,5,5,4,3,3,9,4,5,7,5,8,4,4,3,8,5,8,9,4,8,6,4,3,8,4,5,8,
%U 4,5,6,4,3,8,5,5,8,4
%N Number of iterations of k = digitsum(n*k) until a number appears twice, starting with k = 1.
%C Take n, then starting with k=1, apply iteration k = digitsum(n*k), stop when the new k appears before in the row, length of the n-th row is a(n):
%C n=1: {1,1}, a(1)=2
%C n=2: {1,2,4,8,7,5,1}, a(2)=7
%C n=3: {1,3,9,9}, a(3)=4
%C n=4: {1,4,7,10,4}, a(4)=5.
%t Table[k = 1; s = {k}; While[k = Total[IntegerDigits[n*k]]; FreeQ[s, k], AppendTo[s, k]]; Length[s] + 1, {n, 100}]
%K nonn,base
%O 1,1
%A _Zak Seidov_, Oct 20 2011
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