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A131838 Multiplicative persistence of Woodall numbers.

%I A131838 #15 Mar 01 2025 11:17:06
%S A131838 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 A131838 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 A131838 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 A131838 Multiplicative persistence of Woodall numbers.
%C A131838 After the 111th term, all the numbers have some digits equal to zero, thus the persistence is equal to 1.
%F A131838 a(n) = A031346(A003261(n)). - _Michel Marcus_, Mar 01 2025
%e A131838 Woodall number 159 --> 1*5*9=45 --> 4*5=20 --> 2*0=0 thus persistence is 3.
%p A131838 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);
%t A131838 Table[wn=n*2^n-1;Length[NestWhileList[Times@@IntegerDigits[#]&, wn, #>=10&]], {n,  105}]-1  (* _James C. McMahon_, Mar 01 2025 *)
%Y A131838 Cf. A003261, A031346, A131841.
%K A131838 easy,nonn,base
%O A131838 1,4
%A A131838 _Paolo P. Lava_ and _Giorgio Balzarotti_, Jul 20 2007