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A139003
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Number of operations A000142 (i.e., x!) or A000196 (i.e., floor(sqrt(x))) needed to get n, starting with 3.
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2
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1, 2, 0, 20, 4, 1, 14, 17, 31, 6, 26, 41, 35, 20, 31, 31, 19, 28, 27, 38, 21, 33, 21, 21, 26, 3, 51, 38, 28, 26, 20, 35, 36, 36, 13, 23, 27, 62, 45, 50, 45, 40, 9, 15, 31, 8, 32, 52, 36, 13, 68, 69, 57, 33, 54, 36, 46, 34, 49, 63, 56, 68, 14, 63, 23, 33, 36, 47, 43, 16, 38, 66, 38
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OFFSET
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1,2
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COMMENTS
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Knuth conjectured that any number can be obtained in that way, starting from 4.
This seems also to be true using 3 as the starting value. Since 3 is the minimal possible choice, this variant could be considered to be more natural.
To ensure the sequence is well-defined, define a(n)=-1 if it is not possible to get n in the given way.
See A139004 for references and links.
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LINKS
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FORMULA
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a(3) = 0; a(n) = min { a(k)+1 ; n^2 <= k < (n+1)^2 or k! = n }
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EXAMPLE
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Representing the operation x -> floor(sqrt(x)) by "s" and x -> x! by "f",
we have:
a(1) = 1 since 1 = s3 is clearly the shortest way to obtain 1 from 3.
a(2) = 2 since 2 = sf3 is clearly the shortest way to obtain 2 from 3.
a(3) = 0 since no operation is required to get 3 which is there at the beginning.
a(5) = 4 since 5 = ssff3 is the shortest way to obtain 5 from 3.
a(6) = 1 since 6 = f3 is certainly the shortest way to get 6 from 3.
a(4) = 20 = 7+9+a(5) since 4 = ssssssfsssssssffssff3 = floor(35!^(1/2^6)), 35 = floor((5!)!^(1/2^7)).
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PROG
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(PARI) A139003( n, S=Set(3), LIM=10^5 )={ for( i=0, LIM, setsearch( S, n) & return(i); S=setunion( S, setunion( Set( vector( #S, j, sqrtint(eval(S[j])))), Set( vector( #S, j, if( LIM > j=eval(S[j]), j!))))))}
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Corrected formula, added terms from a(12) onward. - Jon E. Schoenfield, Nov 17 2008, Nov 19 2008
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STATUS
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approved
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