A081374 Size of "uniform" Hamming covers of distance 1, that is, Hamming covers in which all vectors of equal weight are treated the same, included or excluded from the cover together.

1, 2, 2, 5, 10, 22, 43, 86, 170, 341, 682, 1366, 2731, 5462, 10922, 21845, 43690, 87382, 174763, 349526, 699050, 1398101, 2796202, 5592406, 11184811, 22369622, 44739242, 89478485, 178956970, 357913942, 715827883, 1431655766, 2863311530, 5726623061

Offset: 1

Author

David Applegate, Aug 22 2003

Keywords

• nonn

Comments

Motivation: consideration of the "hats" problem (which boils down to normal hamming covering codes) in the case when the people are indistinguishable or unlabeled.

From Paul Curtz, May 26 2011: (Start)

If we add a(0)=1 in front and build the table of a(n) and iterated differences in further rows we get:

1, 1, 2, 2, 5, 10,

0, 1, 0, 3, 5, 12,

1, -1, 3, 2, 7, 9,

-2, 4, -1, 5, 2, 13,

6, -5, 6, -3, 11, 6

-11, 11, -9, 14, -5, 21.

The first column is the inverse binomial transform, which is 1,0 followed by (-1)^n*A083322(n-1), n>=2.

The main diagonal in the table above is A001045, the adjacent upper diagonals are A078008, A048573 and A062092. (End)

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (2,0,-1,2).

Crossrefs

Cf. A083322.

Programs

Magma

I:=[1,2,2,5]; [n le 4 select I[n] else 2*Self(n-1)-Self(n-3)+2*Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jul 08 2016

Maple

hatwork := proc(n,i,covered) local val, val2; options remember;
# computes the minimum cover of the i-bit through n-bit words.
# if covered is true the i-bit words are already covered (by the (i-1)-bit words)
if (i>n or (i = n and covered)) then 0; elif (i = n and not covered) then 1; else
# one choice is to include the i-bit words in the cover
val := hatwork(n, i+1, true) + binomial(n,i);
# the other choice is not to include the i-bit words in the cover
if (covered) then val2 := hatwork (n, i+1, false); if (val2 < val) then val := val2; fi; else
# if the i-bit words were not covered by (i-1), they must be covered by the (i+1)-bit words
if (i <= n) then val2 := hatwork (n, i+2, true) + binomial(n,i+1); if (val2 < val) then val := val2; fi; fi; fi; val; fi; end proc;
A081374 := proc (n) hatwork(n, 0, false); end proc;

Mathematica

LinearRecurrence[{2,0,-1,2},{1,2,2,5},40] (* Harvey P. Dale, Feb 11 2015 *)

Formula

If (n mod 6 = 5) then sum(binomial(n, 3*i+1), i=0..n/3); elif (n mod 6 = 2) then sum(binomial(n, 3*i), i=0..n/3)+1; else sum(binomial(n, 3*i), i=0..n/3); fi;

G.f.: x*(2*x^3-2*x^2+1)/( (1-2*x)*(1+x)*(1-x+x^2) ).

a(n)=2*a(n-1)-a(n-3)+2*a(n-4).

From Paul Curtz, May 26 2011: (Start)

a(n+1) - 2*a(n) has period length 6: repeat 0, -2, 1, 0, 2, -1 (see A080425).

a(n) - A083322(n-1) = A010892(n-1) has period length 6.

a(n) + a(n+3) = 3*2^n = A007283(n).

a(n+6)-a(n) = 21*2^n = A175805(n).

a(n) - A131708(n) = -A131531(n). (End)