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A344104 a(0) = 10; for n > 0, a(n) is a(n-1) multiplied by the number of 0's so far in the sequence.

%I A344104 #14 Jun 07 2021 00:21:30
%S A344104 10,10,20,60,240,1200,8400,75600,831600,10810800,183783600,3491888400,
%T A344104 73329656400,1686582097200,43851134527200,1227831766761600,
%U A344104 36834953002848000,1289223355099680000,51568934203987200000,2372170973383411200000,123352890615937382400000
%N A344104 a(0) = 10; for n > 0, a(n) is a(n-1) multiplied by the number of 0's so far in the sequence.
%C A344104 This sequence is the multiplicative counterpart to A130232, but the initial term 0 is replaced with 10 (0 obviously results in A000004). The data demonstrates that the number of 0's grows rapidly as a(n) increases, because each term is a multiple of 10, thus adding at least one 0 to successive terms.
%F A344104 a(n+1) = a(n)*#_0[a(n)...a(0)], where #_0(n) is the number of 0's in n.
%e A344104 To calculate a(5), multiply a(4)=240 by the number of 0's present in itself and previous terms, of which there are 5, thus yielding 1200.
%e A344104 a(6) is 1200 multiplied by 7, which is the number of 0's present so far, thus giving 8400.
%t A344104 a[0]=10;a[n_]:=a[n]=a[n-1]Count[Flatten[IntegerDigits/@Array[a,n,0]],0];Array[a,20,0] (* _Giorgos Kalogeropoulos_, May 09 2021 *)
%o A344104 (Python)
%o A344104 A344104_list, c = [10], 1
%o A344104 for _ in range(20):
%o A344104     A344104_list.append(A344104_list[-1]*c)
%o A344104     c += str(A344104_list[-1]).count('0') # _Chai Wah Wu_, Jun 06 2021
%Y A344104 Cf. A130224, A130232.
%K A344104 base,nonn
%O A344104 0,1
%A A344104 _Jamie Robert Creasey_, May 09 2021