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A230403
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a(n) = the largest k such that (k+1)! divides n; the number of trailing zeros in the factorial base representation of n (A007623(n)).
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14
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0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 3, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 3, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 3, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1
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OFFSET
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1,6
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COMMENTS
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Many of the comments given in A055881 apply also here.
The asymptotic density of the occurrences of k is (k+1)/(k+2)!.
The asymptotic mean of this sequence is e - 2 = 0.718281... (A001113 - 2). (End)
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LINKS
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FORMULA
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EXAMPLE
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In factorial number base representation (A007623), the numbers from 1 to 9 are represented as:
n A007623(n) a(n) (gives the number of trailing zeros)
1 1 0
2 10 1
3 11 0
4 20 1
5 21 0
6 100 2
7 101 0
8 110 1
9 111 0
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MATHEMATICA
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With[{b = MixedRadix[Range[12, 2, -1]]}, Array[LengthWhile[Reverse@ IntegerDigits[#, b], # == 0 &] &, 105]] (* Michael De Vlieger, Jun 03 2020 *)
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PROG
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(Scheme)
(define (A230403 n) (if (zero? n) 0 (let loop ((n n) (i 2)) (cond ((not (zero? (modulo n i))) (- i 2)) (else (loop (/ n i) (1+ i)))))))
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CROSSREFS
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Analogous sequence for binary system: A007814.
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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