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A033810 Number of people needed so that probability of at least two sharing a birthday out of n possible days is at least 50%. 11
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

a(365) = 23 is the solution to the Birthday Problem.

LINKS

T. D. Noe and Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (366 terms from T. D. Noe)

S. E. Ahmed & R. J. McIntosh, An Asymptotic Approximation for the Birthday Problem, Crux Mathematicorum 26(3) 151-5 2000 CMS.

D. Brink, A (probably) exact solution to the Birthday Problem, Ramanujan Journal, June 2012, Volume 28, Issue 2, pp 223-238. - From N. J. A. Sloane, Oct 08 2012

Mathforum, The Birthday Problem

Eric Weisstein's World of Mathematics, Birthday Problem

FORMULA

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.

MATHEMATICA

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 *)

CROSSREFS

Essentially the same as A088141. See also A182008, A182009, A182010.

Cf. A014088 (n people on 365 days), A225852 (3 on n days), A225871 (4 people on n days).

Sequence in context: A194287 A194303 A124755 * A253272 A023965 A274094

Adjacent sequences:  A033807 A033808 A033809 * A033811 A033812 A033813

KEYWORD

nonn,easy,nice

AUTHOR

Eric W. Weisstein

STATUS

approved

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Last modified October 13 21:59 EDT 2019. Contains 327981 sequences. (Running on oeis4.)