A297672 - OEIS

%I #26 Nov 14 2019 22:09:40

%S 1,0,0,0,1,1,1,0,3,1,1,1,6,6,6,2,20,10,10,10,50,50,50,25,175,105,105,

%T 105,490,490,490,294,1764,1176,1176,1176,5292,5292,5292,3528,19404,

%U 13860,13860

%N Array with four columns read by rows: T(n,k) = number of n step walks in the first octant on a square lattice with last step being right (k=1), left (k=2), up (k=3) or down (k=4).

%C Sum_{k=1..4} T(n,k) = A005558(n).

%C For n >= 1, the ratio of the numbers of right or up last steps to left or down last steps is floor((n+2)/2): floor(n/2). - _Roger Ford_, Oct 28 2019

%F n=1: T(1,1)=1, T(1,2)=0, T(1,3)=0, T(1,4)=0;

%F n>1: T(n,1) = C(n,floor(n/2))*C(n-1,floor((n-1)/2)) - C(n,floor((n-1)/2))*C(n-1,floor((n-2)/2));

%F T(n,2) = T(n,3) = C(n-1,floor(n/2)-1)*C(n,floor(n/2)-1)/floor(n/2);

%F n odd:  T(n,4) = T(n,2);

%F n even: T(n,4) = T(n,2)*((n/2-1)/(n/2+1));

%F For n > 1, T(n,2) = T(n,3) = A001263(n,floor(n/2)).

%e k=    1    2    3    4    total

%e N   right left up   down  walks

%e 1     1    0    0    0    =1

%e 2     1    1    1    0    =3

%e 3     3    1    1    1    =6

%e 4     6    6    6    2    =20

%e There are 6 walks of 4 steps in the octant with the last step right. T(4,1)=6 RRRR, RRLR, RLRR, RUDR, RURR, RRUR.

%Y Cf. A005558, A001263.

%K nonn,tabf,walk,more

%O 1,9

%A _Roger Ford_, Jan 02 2018