A243841 Pair deficit of the most nearly equal partition of n into two parts using ceiling rounding of the expectations of n, floor(n/2) and n-floor(n/2), assuming equal likelihood of states defined by the number of 2-cycles.

0, 0, 1, 0, 0, 0, 0, 0, -1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, -1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 2, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1

Offset

0,41

Comments

A162970 and A000085 provide the numerator and the denominator for calculating the expected value.

Links

Table of n, a(n) for n=0..53.

Formula

a(n) = ceiling(A162970(n)/A000085(n)) - (ceiling(A162970(floor(n/2))/A000085(floor(n/2))) + ceiling(A162970(n-floor(n/2))/A000085(n-floor(n/2)))).

Example

Trivially, for n = 0,1 no pairs are possible so a(0) and a(1) are 0.
For n = 2, the expectation, E(n), equals 0.5. So a(2) = ceiling(E(2)) - (ceiling(E(1)) + ceiling(E(1))) = 1.
For n = 5 = 2 + 3, E(5) = 20/13, E(2) = 0.5 and E(3) = 0.75 and a(5) = ceiling(E(5)) - (ceiling(E(2)) + ceiling(E(3))) = 0.
Interestingly, for n = 8, E(8) = 532/191 and E(4) = 6/5, so a(n) = 3 - (2 + 2) = -1.

Crossrefs

Cf. A000085, A162970.

Sequence in context: A016366 A016427 A326170 * A131038 A016353 A016398

Adjacent sequences: A243838 A243839 A243840 * A243842 A243843 A243844

Keyword

sign,more

Author

Rajan Murthy, Jun 12 2014

Status

approved