A182009 a(n) = ceiling(sqrt(2n*log(2))+(3-2*log(2))/6).

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, 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, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11

Offset: 1

Author

David Brink, Apr 06 2012

Keywords

• nonn

Comments

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.

Links

• Gheorghe Coserea, Table of n, a(n) for n = 1..10000
• D. Brink, A (probably) exact solution to the Birthday Problem, Ramanujan Journal, 2012, pp 223-238.

Crossrefs

Approximates A033810.

Programs

Maple

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

Mathematica

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

PARI

a(n) = { ceil((2*n*log(2))^(1/2) + (3-2*log(2))/6) };
apply(n->a(n), vector(84, i, i))  \\ Gheorghe Coserea, Aug 23 2015