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A033810
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Number of people needed so that probability of at least two sharing a birthday out of n possible days is at least 50%.
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11
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2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12
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OFFSET
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1,1
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COMMENTS
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a(365) = 23 is the solution to the Birthday Problem.
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LINKS
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FORMULA
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a(n) = ceiling(sqrt(2*n*log(2)) + (3 - 2*log(2))/6 + (9 - 4*log(2)^2) / (72*sqrt(2*n*log(2))) - 2*log(2)^2/(135*n)) for all n up to 10^18. It is conjectured that this formula holds for all n.
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MATHEMATICA
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lst = {}; s = 1; Do[Do[If[Product[(n - i + 1)/n, {i, j}] <= 1/2, If[j > s, s = j]; AppendTo[lst, j]; Break[]], {j, s, s + 1}], {n, 86}]; lst (* Arkadiusz Wesolowski, Apr 29 2012 *)
A033810[n_] := Catch@Do[If[1/2 >= n!/(n - m)!/n^m, Throw[m]], {m, 2, Infinity}]; Array[A033810, 86] (* JungHwan Min, Mar 27 2017 *)
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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