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%I #7 Apr 16 2014 00:23:19
%S 0,0,1,2,3,3,2,1,1,1,2,2,1,2,2,1,5,2,2,1,1,8,3,1,1,1,2,2,2,2,1,2,2,2,
%T 1,1,1,1,1,1,1,2,2,1,2,2,1,1,2,2,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,
%U 1,1,1,1,1,1,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
%N Multiplicative persistence of Woodall numbers.
%C After the 111th term, all the numbers have some digits equal to zero, thus the persistence is equal to 1.
%e Woodall number 159 --> 1*5*9=45 --> 4*5=20 --> 2*0=0 thus persistence is 3.
%p P:=proc(n) local i,k,w,ok,cont; for i from 1 by 1 to n do w:=1; k:=i*2^i-1; ok:=1; if k<10 then print(0); else cont:=1; while ok=1 do while k>0 do w:=w*(k-(trunc(k/10)*10)); k:=trunc(k/10); od; if w<10 then ok:=0; print(cont); else cont:=cont+1; k:=w; w:=1; fi; od; fi; od; end: P(120);
%Y Cf. A003261, A131841.
%K easy,nonn,base
%O 1,4
%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Jul 20 2007
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