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a(n) = ceiling(sqrt(2n*log(2))+(3-2*log(2))/6).
5

%I #19 Aug 24 2015 04:47:33

%S 2,2,3,3,3,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,7,8,8,

%T 8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,

%U 10,10,10,10,10,11,11,11,11,11,11,11,11,11,11,11

%N a(n) = ceiling(sqrt(2n*log(2))+(3-2*log(2))/6).

%C This sequence approximates the sequence of solutions to the Birthday Problem, A033810. The two sequences agree for almost all n, i.e., on a set of integers n with density 1.

%H Gheorghe Coserea, <a href="/A182009/b182009.txt">Table of n, a(n) for n = 1..10000</a>

%H D. Brink, <a href="http://dx.doi.org/10.1007/s11139-011-9343-9">A (probably) exact solution to the Birthday Problem</a>, Ramanujan Journal, 2012, pp 223-238.

%p seq(ceil((2*n*log(2))^(1/2) + (3-2*log(2))/6), n=1..1000); # _Robert Israel_, Aug 23 2015

%t Table[Ceiling[Sqrt[2 n Log[2] + (3 - 2 Log[2])/6]], {n, 82}] (* _Michael De Vlieger_, Aug 24 2015 *)

%o (PARI)

%o a(n) = { ceil((2*n*log(2))^(1/2) + (3-2*log(2))/6) };

%o apply(n->a(n), vector(84, i, i)) \\ _Gheorghe Coserea_, Aug 23 2015

%Y Approximates A033810.

%K nonn

%O 1,1

%A _David Brink_, Apr 06 2012