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%I #4 Jul 08 2012 13:04:34
%S 0,0,5,2,4,4,2,3,3,1,1,5,2,4,4,2,3,3,1,2,5,2,4,4,2,3,3,1,2,5,2,4,4,2,
%T 3,3,1,2,5,2,4,4,2,3,3,1,2,5,2,2,4,2,3,3,1,2,5,2,2,3,2,3,3,1,2,5,2,2,
%U 3,2,3,3,1,2,5,2,2,3,2,2,3,1,2,5,2,2,3,2,2,4,1,2,5,2,2,3,2,2,4,2,1,5,2,4,4
%N a(n) = number of squares of summations of digits before reaching the end of the cycle.
%C a(n) is the number of times you form the square of the sum of the digits before reaching the last number of the cycle. According to the computations with Maple program, this last number belong to the set {1,81,100,169,256}, and a(n) < = 5.
%C Example : 4 -> 16 -> (1+6)^2 = 49 -> (4+9)^2 = 169 -> (1+6+9)^2 = 256, and 256 is the last number of the list because 256 -> (2+5+6)^2 = 169 belong to the list.
%e 0 is in the sequence twice because 0->0 and 1 -> 1 ;
%e 5 is in the sequence because : 5 -> 25 -> 49 -> 169 -> 256, last number because 256 -> 169.
%p A177148 := proc(n)
%p local traj ,a,m,c;
%p traj := n ;
%p c := [n] ;
%p while true do
%p traj := A118881(traj) ;
%p if member(traj,c) then
%p return nops(c)-1 ;
%p end if;
%p c := [op(c),traj] ;
%p end do:
%p end proc: # _R. J. Mathar_, Jul 08 2012
%Y Cf. A118881.
%K nonn,base
%O 0,3
%A _Michel Lagneau_, May 03 2010
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