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Search: id:A135736
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| A135736 |
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Nearest integer to n*sum(1/k,k=1..n) = rounded expected coupon collection numbers. |
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+0
1
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| 0, 1, 3, 6, 8, 11, 15, 18, 22, 25, 29, 33, 37, 41, 46, 50, 54, 58, 63, 67, 72, 77, 81, 86, 91, 95, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 161, 166, 171, 176, 182, 187, 192, 198, 203, 209, 214, 219, 225, 230, 236, 242, 247, 253, 258, 264, 269, 275
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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Somewhat more realistic than A052488 but more optimistic than A060293, the expected number of boxes that must be bought to get the full collection of N objects, if each box contains any one of them at random. See also comments in A060293, A052488.
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FORMULA
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a(n) = round( n*A001008(n)/A002805(n)) = either A052488(n) or A060293(n).
a(n) ~ A060293[n] ~ A052488[n] ~ A050502[n] ~ A050503[n] ~ A050504[n] (asymptotically)
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EXAMPLE
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a(0) = 0 since nothing needs to be bought if nothing is to be collected.
a(1) = 1 since only 1 box needs to be bought if only 1 object is to be collected.
a(2) = 3 since the chance of getting the other object at the second purchase is only 1/2, so it takes 2 boxes on the average to get it.
a(3) = 6 since the chance of getting a new object at the second purchase is 2/3 so it takes 3/2 boxes in the mean, then the chance becomes 1/3 to get the 3rd, i.e. 3 other boxes on the average to get the full collection, and the rounded value of 1+3/2+3=5.5 is 6.
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PROGRAM
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(PARI) A135736(n)=round(n*sum(i=1, n, 1/i))
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CROSSREFS
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Cf. A060293, A052488, A050502-A050504.
Adjacent sequences: A135733 A135734 A135735 this_sequence A135737 A135738 A135739
Sequence in context: A050503 A133869 A085197 this_sequence A036434 A140400 A077148
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KEYWORD
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easy,nonn
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AUTHOR
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M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Nov 29 2007
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