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A061338
Increase in maximal number of comparisons for sorting n elements by list merging.
4
0, 1, 2, 2, 4, 2, 3, 3, 8, 2, 3, 3, 5, 3, 4, 4, 16, 2, 3, 3, 5, 3, 4, 4, 9, 3, 4, 4, 6, 4, 5, 5, 32, 2, 3, 3, 5, 3, 4, 4, 9, 3, 4, 4, 6, 4, 5, 5, 17, 3, 4, 4, 6, 4, 5, 5, 10, 4, 5, 5, 7, 5, 6, 6, 64, 2, 3, 3, 5, 3, 4, 4, 9, 3, 4, 4, 6, 4, 5, 5, 17, 3, 4, 4, 6, 4, 5, 5, 10, 4, 5, 5, 7, 5, 6, 6, 33, 3, 4, 4
OFFSET
0,3
COMMENTS
Or, first differences of A003071. - Zak Seidov, Dec 28 2011
FORMULA
For n > 0: a(n) = A003071(n) - A003071(n - 1) = A006519(n) + A000120(n) - 1. If n is a power of 2 then a(n) = n, otherwise a(n) = a(A053645(n)) + 1 where A053645(n) = n - 2^floor(log_2(n)) is the amount by which n exceeds a power of 2.
G.f.: x/(1-x)^2 + (1/(1-x))*Sum_{k>=1} (-1 + (1-x)*2^(k-1))*x^2^k/(1-x^2^k). - Ralf Stephan, Apr 17 2003
MATHEMATICA
nn=100; s={1}; m = Ceiling[Log[2, nn]]; Do[s=Join[s, {2^n}, s+1], {n, m}]; Prepend[Take[s, nn], 0] (* Zak Seidov, Dec 28 2011 *)
PROG
(Haskell)
a061338 0 = 0
a061338 n = a006519 n + a000120 n - 1 -- Reinhard Zumkeller, Dec 29 2011
CROSSREFS
Cf. A003071.
Sequence in context: A221861 A057939 A163371 * A135714 A103274 A046820
KEYWORD
nonn
AUTHOR
Henry Bottomley, Apr 27 2001
STATUS
approved