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.

10, 10, 20, 60, 240, 1200, 8400, 75600, 831600, 10810800, 183783600, 3491888400, 73329656400, 1686582097200, 43851134527200, 1227831766761600, 36834953002848000, 1289223355099680000, 51568934203987200000, 2372170973383411200000, 123352890615937382400000

Offset: 0

Author

Jamie Robert Creasey, May 09 2021

Keywords

• base
• nonn

Comments

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.

Examples

			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.
a(6) is 1200 multiplied by 7, which is the number of 0's present so far, thus giving 8400.
		

Crossrefs

Cf. A130224, A130232.

Programs

Mathematica

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 *)

Python

A344104_list, c = [10], 1
for _ in range(20):
    A344104_list.append(A344104_list[-1]*c)
    c += str(A344104_list[-1]).count('0') # Chai Wah Wu, Jun 06 2021

Formula

a(n+1) = a(n)*#_0[a(n)...a(0)], where #_0(n) is the number of 0's in n.