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Let f(n) = average number of comparisons needed for sorting n elements using merge insertion. Sequence gives n!*f(n).
3

%I #8 Jan 02 2015 22:07:45

%S 0,2,16,112,832,6912,62784,623232

%N Let f(n) = average number of comparisons needed for sorting n elements using merge insertion. Sequence gives n!*f(n).

%D D. E. Knuth, TAOCP, Vol. 3, Section 5.3.1.

%Y A117627 is a lower bound for any comparison-based sorting algorithm.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Oct 06 2006