

A122878


Periodic sequence of period 21 related to a simple scheduling problem.


1



12, 12, 12, 12, 13, 13, 13, 23, 23, 23, 23, 12, 12, 12, 13, 13, 13, 13, 23, 23, 23, 12, 12, 12, 12, 13, 13, 13, 23, 23, 23, 23, 12, 12, 12, 13, 13, 13, 13, 23, 23, 23, 12, 12, 12, 12, 13, 13, 13, 23, 23, 23, 23, 12, 12, 12, 13, 13, 13, 13, 23, 23, 23, 12, 12, 12, 12, 13, 13
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OFFSET

1,1


COMMENTS

Each term is 12, 13, or 23, which can be interpreted as the presence of two of three resources or people: {1,2}, {1,3}, or {2,3}. (If, instead, 0, 1 and 2 are used for these unordered pairs, sequence A122879 results.). This sequence is one symmetric solution to a scheduling problem to balance an ongoing workload simply while maximizing the lengths of scheduled breaks and meeting requirement (*) below. Note that each of 1, 2 and 3 is present in 14 terms of each period and in seven consecutive terms whenever present. Each of 1, 2 and 3 is absent from three consecutive terms or four consecutive terms alternately. Also, if and only if, a(3+7k), where k>=0, is chosen to correspond to the first day of a 7day workweek, then (*) this schedule requires no more than five workdays per person per workweek. Finally, each resource or person is scheduled to be absent exactly once per named day of the week per 21day period (and in the same pattern: alternately, i) the three middle days of the week and ii) the last two days of one week and the first two days of the next week).


LINKS

Table of n, a(n) for n=1..69.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,1,1,0,0,0,0,0,1,1).


FORMULA

a(n+21) = a(n).
a(n)=(1/105)*{63*(n mod 21)+8*[(n+1) mod 21]+8*[(n+2) mod 21]42*[(n+3) mod 21]+8*[(n+4) mod 21]+8*[(n+5) mod 21]+8*[(n+6) mod 21]+3*[(n+7) mod 21]+8*[(n+8) mod 21]+8*[(n+9) mod 21)+63*[(n+10) mod 21]+8*[(n+11) mod 21]+8*[(n+12) mod 21]+8*[(n+13) mod 21]42*[(n+14) mod 21]+8*[(n+15) mod 21]+8*[(n+16) mod 21]+3*[(n+17) mod 21]+8*[(n+18) mod 21]+8*[(n+19) mod 21]+8*[(n+20) mod 21]}, with n>=0.  Paolo P. Lava, Oct 23 2008
G.f.: x*(23*x^1410*x^11+22*x^7+x^4+12) / ((x1)*(x^2+x+1)*(x^12x^11 +x^9 x^8+x^6x^4+x^3x+1)).  Colin Barker, Dec 21 2012


CROSSREFS

Cf. A122879.
Sequence in context: A010851 A343194 A123896 * A302341 A064162 A028996
Adjacent sequences: A122875 A122876 A122877 * A122879 A122880 A122881


KEYWORD

nonn,easy


AUTHOR

Rick L. Shepherd, Sep 16 2006


STATUS

approved



