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A344104
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a(0) = 10; for n > 0, a(n) is a(n-1) multiplied by the number of 0's so far in the sequence.
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0
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10, 10, 20, 60, 240, 1200, 8400, 75600, 831600, 10810800, 183783600, 3491888400, 73329656400, 1686582097200, 43851134527200, 1227831766761600, 36834953002848000, 1289223355099680000, 51568934203987200000, 2372170973383411200000, 123352890615937382400000
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OFFSET
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0,1
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COMMENTS
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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.
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LINKS
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FORMULA
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a(n+1) = a(n)*#_0[a(n)...a(0)], where #_0(n) is the number of 0's in n.
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(Python)
for _ in range(20):
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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STATUS
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approved
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