[seqfan] Re: Help with A002793 needed/Karol A. Penson

[Richard Mathar]

Lisons http://list.seqfan.eu/pipermail/seqfan/2010-July/005230.html :

kp>  a(2)=4    =3+1
kp>  a(3)=20  =11+8+1
kp>  a(4)=124=50+58+15+1
kp>  a(5)=920=274+444+177+24+1,

One can get an interpretation if one follows the advice
given in  A002793 to look at this as the bisection of A056952.

The recurrence of A056952(n) can be looked at as a path
with jumps/skips of length 1 or 2 totaling n, where jumps of length 1
get weight 1, which is the factor 1 in front of a(n-1),
and jumps of length 2 get weight floor(n/2), which is the factor in front
of a(n-2).

Obtain A056952(n) as follows: write down all compositions
of n such that all addends are either 1 or 2, and such that
the first addend is 1 (this last restriction comes form A056952(0)=0).
Then associate with each composition a product of weights
as computed above, so the n in the weight is the product of
all floor(s/2) where s is the "partial sum" along the composition at
the point one sees a 2. Then the terms in the triangle are the
sums of the weights for all compositions of fixed common length.

Example of n=8:
[1, 2, 2, 2, 1], 6 = floor(3/2)*floor(5/2)*floor(7/2)
[1, 2, 2, 1, 2], 8   = floor(3/2)*floor(5/2)*floor(8/2)
[1, 2, 1, 2, 2], 12  = floor(3/2)*floor(6/2)*floor(8/2)
[1, 1, 2, 2, 2], 24
   50 = 6+8+12+24
[1, 2, 2, 1, 1, 1], 2
[1, 2, 1, 2, 1, 1], 3
[1, 1, 2, 2, 1, 1], 6
[1, 2, 1, 1, 2, 1], 3
[1, 1, 2, 1, 2, 1], 6
[1, 1, 1, 2, 2, 1], 6
[1, 2, 1, 1, 1, 2], 4
[1, 1, 2, 1, 1, 2], 8
[1, 1, 1, 2, 1, 2], 8
[1, 1, 1, 1, 2, 2], 12
    58 = 2+3+6+3+6+6+4+8+8+12
[1, 2, 1, 1, 1, 1, 1], 1
[1, 1, 2, 1, 1, 1, 1], 2
[1, 1, 1, 2, 1, 1, 1], 2
[1, 1, 1, 1, 2, 1, 1], 3
[1, 1, 1, 1, 1, 2, 1], 3
[1, 1, 1, 1, 1, 1, 2], 4
    15 = 1+2+2+3+3+4
[1, 1, 1, 1, 1, 1, 1, 1], 1
    1 =1
124

Example of n=10:

[1, 2, 2, 2, 2, 1], 24
[1, 2, 2, 2, 1, 2], 30
[1, 2, 2, 1, 2, 2], 40
[1, 2, 1, 2, 2, 2], 60
[1, 1, 2, 2, 2, 2], 120
   274 = 24+30+40+60+120
[1, 2, 2, 2, 1, 1, 1], 6
[1, 2, 2, 1, 2, 1, 1], 8
[1, 2, 1, 2, 2, 1, 1], 12
[1, 1, 2, 2, 2, 1, 1], 24
[1, 2, 2, 1, 1, 2, 1], 8
[1, 2, 1, 2, 1, 2, 1], 12
[1, 1, 2, 2, 1, 2, 1], 24
[1, 2, 1, 1, 2, 2, 1], 12
[1, 1, 2, 1, 2, 2, 1], 24
[1, 1, 1, 2, 2, 2, 1], 24
[1, 2, 2, 1, 1, 1, 2], 10
[1, 2, 1, 2, 1, 1, 2], 15
[1, 1, 2, 2, 1, 1, 2], 30
[1, 2, 1, 1, 2, 1, 2], 15
[1, 1, 2, 1, 2, 1, 2], 30
[1, 1, 1, 2, 2, 1, 2], 30
[1, 2, 1, 1, 1, 2, 2], 20
[1, 1, 2, 1, 1, 2, 2], 40
[1, 1, 1, 2, 1, 2, 2], 40
[1, 1, 1, 1, 2, 2, 2], 60
   444 = 6+8+12+24+8+...+60
[1, 2, 2, 1, 1, 1, 1, 1], 2
[1, 2, 1, 2, 1, 1, 1, 1], 3
[1, 1, 2, 2, 1, 1, 1, 1], 6
[1, 2, 1, 1, 2, 1, 1, 1], 3
[1, 1, 2, 1, 2, 1, 1, 1], 6
[1, 1, 1, 2, 2, 1, 1, 1], 6
[1, 2, 1, 1, 1, 2, 1, 1], 4
[1, 1, 2, 1, 1, 2, 1, 1], 8
[1, 1, 1, 2, 1, 2, 1, 1], 8
[1, 1, 1, 1, 2, 2, 1, 1], 12
[1, 2, 1, 1, 1, 1, 2, 1], 4
[1, 1, 2, 1, 1, 1, 2, 1], 8
[1, 1, 1, 2, 1, 1, 2, 1], 8
[1, 1, 1, 1, 2, 1, 2, 1], 12
[1, 1, 1, 1, 1, 2, 2, 1], 12
[1, 2, 1, 1, 1, 1, 1, 2], 5
[1, 1, 2, 1, 1, 1, 1, 2], 10
[1, 1, 1, 2, 1, 1, 1, 2], 10
[1, 1, 1, 1, 2, 1, 1, 2], 15
[1, 1, 1, 1, 1, 2, 1, 2], 15
[1, 1, 1, 1, 1, 1, 2, 2], 20
   177
[1, 2, 1, 1, 1, 1, 1, 1, 1], 1
[1, 1, 2, 1, 1, 1, 1, 1, 1], 2
[1, 1, 1, 2, 1, 1, 1, 1, 1], 2
[1, 1, 1, 1, 2, 1, 1, 1, 1], 3
[1, 1, 1, 1, 1, 2, 1, 1, 1], 3
[1, 1, 1, 1, 1, 1, 2, 1, 1], 4
[1, 1, 1, 1, 1, 1, 1, 2, 1], 4
[1, 1, 1, 1, 1, 1, 1, 1, 2], 5
   24
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1], 1
   1
920
274+444+177+24+1=920

Example of n=12:
1764+3708+2016+416+35+1 = 7940
Example of n=14:
13068+33984+23533+6560+835+48+1 = 78040

Apparently the first but last term in each row is A005563.
Apparently the first column in the triangle is A000254
Apparently the second column in the triangle is A002538
The last term in each row is 1.

In Maple:
n := 10; #or whatever the even index in A056952 may be
a := 0 ;
for c from 1 to n do
        fpar := 0 ;
        com := combinat[composition](n,c) ;
        for co in com do
                if max(op(co))<3 and op(1,co) = 1 then
                        psu := 1 ;
                        f := 1 ;
                        for pos from 2 to nops(co) do
                                psu := psu+op(pos,co) ;
                                if op(pos,co) = 2 then
                                        f := f*floor(psu/2) ;
                                end if;
                        end do:
                        a := a+f ;
                        fpar := fpar+f ;
                        print(co,f) ;
                end if;
        end do:
        print(fpar) ;
end do:
print(a) ;