cp's OEIS Frontend

A330617 Triangle read by rows: T(n,k) is the number of paths from node 0 to k in a directed graph with n+1 vertices labeled 0, 1, ..., n and edges leading from i to i+1 for all i, and from i to i+2 for even i and from i to i-2 for odd i.

%I A330617 #43 Jan 07 2024 16:31:29
%S A330617 1,1,1,1,1,2,1,2,2,2,1,2,2,2,4,1,3,2,4,4,4,1,3,2,4,4,4,8,1,4,2,6,4,8,
%T A330617 8,8,1,4,2,6,4,8,8,8,16,1,5,2,8,4,12,8,16,16,16,1,5,2,8,4,12,8,16,16,
%U A330617 16,32,1,6,2,10,4,16,8,24,16,32,32,32,1,6,2,10,4,16,8,24,16,32,32,32,64
%N A330617 Triangle read by rows: T(n,k) is the number of paths from node 0 to k in a directed graph with n+1 vertices labeled 0, 1, ..., n and edges leading from i to i+1 for all i, and from i to i+2 for even i and from i to i-2 for odd i.
%H A330617 Andrew Howroyd, <a href="/A330617/b330617.txt">Table of n, a(n) for n = 0..1325</a> (rows 0..50)
%H A330617 E. Krom and M. M. Roughan, <a href="http://girlsangle.org/page/bulletin-archive/GABv13n03E.pdf">Path Counting and Eulerian Numbers</a>, Girls' Angle Bulletin, Vol. 13, No. 3 (2020), 8-10.
%F A330617 For k odd: T(n, k) = 2^((k-1)/2)*(ceiling(n/2) - (k-1)/2).
%F A330617 For k even: T(n, k) = 2^(k/2).
%F A330617 T(2*n-1, 2*k-1) = A130128(n, k).
%e A330617 First few rows of the triangle are:
%e A330617   1;
%e A330617   1,  1;
%e A330617   1,  1,  2;
%e A330617   1,  2,  2,  2;
%e A330617   1,  2,  2,  2,  4;
%e A330617   1,  3,  2,  4,  4,  4;
%e A330617   1,  3,  2,  4,  4,  4,  8;
%e A330617   1,  4,  2,  6,  4,  8,  8,  8;
%e A330617   1,  4,  2,  6,  4,  8,  8,  8, 16;
%e A330617   1,  5,  2,  8,  4, 12,  8, 16, 16, 16;
%e A330617   1,  5,  2,  8,  4, 12,  8, 16, 16, 16, 32;
%e A330617   ...
%e A330617 For n=6 and k=3, T(6,3)=4 is the number of paths from node 0 to node 3 along the directed network: {0,1,2,3}, {0,2,3}, {0,2,4,5,3}, {0,1,2,4,5,3}.
%t A330617 Table[If[EvenQ@ k, 2^(k/2), 2^((k - 1)/2)*(Ceiling[n/2] - (k - 1)/2)], {n, 0, 12}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Mar 23 2020 *)
%o A330617 (PARI) T(n,k)={if(k%2, 2^(k\2)*((n+1)\2 - k\2), 2^(k/2))} \\ _Andrew Howroyd_, Mar 17 2020
%Y A330617 Cf. A130128.
%K A330617 nonn,easy,tabl,walk
%O A330617 0,6
%A A330617 _Grace Work_, Mar 01 2020