cp's OEIS Frontend

A297672 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).

%I A297672 #26 Nov 14 2019 22:09:40
%S A297672 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 A297672 105,490,490,490,294,1764,1176,1176,1176,5292,5292,5292,3528,19404,
%U A297672 13860,13860
%N A297672 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 A297672 Sum_{k=1..4} T(n,k) = A005558(n).
%C A297672 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 A297672 n=1: T(1,1)=1, T(1,2)=0, T(1,3)=0, T(1,4)=0;
%F A297672 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 A297672 T(n,2) = T(n,3) = C(n-1,floor(n/2)-1)*C(n,floor(n/2)-1)/floor(n/2);
%F A297672 n odd:  T(n,4) = T(n,2);
%F A297672 n even: T(n,4) = T(n,2)*((n/2-1)/(n/2+1));
%F A297672 For n > 1, T(n,2) = T(n,3) = A001263(n,floor(n/2)).
%e A297672 k=    1    2    3    4    total
%e A297672 N   right left up   down  walks
%e A297672 1     1    0    0    0    =1
%e A297672 2     1    1    1    0    =3
%e A297672 3     3    1    1    1    =6
%e A297672 4     6    6    6    2    =20
%e A297672 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 A297672 Cf. A005558, A001263.
%K A297672 nonn,tabf,walk,more
%O A297672 1,9
%A A297672 _Roger Ford_, Jan 02 2018