

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) 1515 2000 CMS.
D. Brink, A (probably) exact solution to the Birthday Problem, Ramanujan Journal, June 2012, Volume 28, Issue 2, pp 223238.  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



