[seqfan] Re: Tatami

[Richard J. Mathar]

In answer to http://list.seqfan.eu/pipermail/seqfan/2016-April/016328.html :

I get rational values for S1(n), n>=0:
0, 2, 0, 0, 0, 4, 0, 4, 45/8, 6, 77/8, 12, 117/8, 14, 165/8, 24, 221/8..
and rational values for S5(n), n>=0:
0, 0, 0, 0, 0, 6, 0, 0, 0, 8, 0, 8, 85/8, 10, 133/8, 20, 189/8, 22, 253/8
and rational values for S3(n), n>=0:
0, 0, 0, 4, 4, 0, 4, 4, 4, 4, 9, 10, 11, 8, 13, 18, 165/8, 39/2, 213/8
and rational values for S1(n)+S5(n)+S3(n):
0, 2, 0, 4, 4, 10, 4, 8, 77/8, 18, 149/8, 30, 145/4, 32, 201/4, 62, 575/8
and rational values for S1(n)+S5(n)+S3(n) at the odd values n>=1:
2, 4, 10, 8, 18, 30, 32, 62, 151/2, 114, 319/2, 891/4
So looking at the last line there is somewhere a problem  with
these formulas, because counts of tilings need to have integer values.

The Maple implementation is this (please check!):

S1 := proc(n)
    local a,kstrt,k ;
    a := 0 ;
    if modp(n,4) = 1 then
        kstrt := 0 ;
    else
        kstrt := 1 ;
    end if;
    for k from kstrt to (n-1)/6 by 2 do
        a := a+ ((n+3)/4-1/2*k)*binomial((n-1)/4-1/2*k,k) ;
        a := simplify(a) ;
    end do:
    a*2 ;
end proc:
seq(S1(n),n=0..20) ;

S5 := proc(n)
    local a,kstrt,k ;
    a := 0 ;
    if modp(n,4) = 1 then
        kstrt := 0 ;
    else
        kstrt := 1 ;
    end if;
    for k from kstrt to (n-5)/6 by 2 do
        a := a+ ((n+7)/4-1/2*k)*binomial((n-5)/4-1/2*k,k) ;
        a := simplify(a) ;
    end do:
    a*2 ;
end proc:

seq(S5(n),n=0..20) ;
S3 := proc(n)
    local a,kstrt,k ;
    a := 0 ;
    if modp(n,4) = 1 then
        kstrt := 1 ;
    else
        kstrt := 0 ;
    end if;
    for k from kstrt to (n-3)/6 do
        a := a+ binomial((n-3)/4-1/2*k,k) ;
        a := simplify(a) ;
    end do:
    a*4 ;
end proc:
seq(S3(n),n=0..20) ;

seq(S1(n)+S5(n)+S3(n),n=0..20) ;
seq(S1(2*n+1)+S5(2*n+1)+S3(2*n+1),n=0..20) ;