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A099563
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Final nonzero number in the sequence n, f(n,2), f(f(n,2),3), f(f(f(n,2),3),4),..., where f(n,d)=Floor(n/d).
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3
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1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,4
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COMMENTS
| Records in {a(n)} occur at {1,4,18,96,600,4320,35280,322560,3265920,...}, which appears to be n*n!=A001563.
The most significant digit in the factorial expansion of n (A007623). Here we can assume that a(0)=0. Proof: The algorithm that computes the factorial expansion of n, generates the successive digits by repeatedly dividing the previous quotient with successively larger divisors (the remainders give the digits), starting from n itself and divisor 2. As a corollary we find that A001563 indeed gives the positions of the records. - Antti Karttunen (his-firstname.his-surname(AT)gmail.com), Jan 01 2007.
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EXAMPLE
| For n=15, f(15,2)=Floor(15/2)=7, f(7,3)=2, f(2,4)=0, so a(15)=2.
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CROSSREFS
| Cf. A001563, A007623, A099564.
Sequence in context: A114139 A029884 A118164 * A099564 A194606 A126389
Adjacent sequences: A099560 A099561 A099562 * A099564 A099565 A099566
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KEYWORD
| nonn
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AUTHOR
| John W. Layman (layman(AT)math.vt.edu), Oct 22 2004
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