This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A054120 #27 Jan 26 2022 10:04:02
%S A054120 1,1,1,1,3,1,1,6,6,1,1,9,18,9,1,1,12,39,39,12,1,1,15,69,114,69,15,1,1,
%T A054120 18,108,261,261,108,18,1,1,21,156,507,750,507,156,21,1,1,24,213,879,
%U A054120 1779,1779,879,213,24,1,1,27,279,1404,3672,5058
%N A054120 Triangular array T(n,k): start with T(n,0)=T(n,n)=1 for n >= 0; recursively, draw vertical lines through T(n-1,k-1) if present and T(n-1,k) if present; then T(n,k) is the sum of T(i,j) that lie on or between the lines and not below T(n,k).
%C A054120 Conjecture: T(n,k) = T(n-1,k-1) + 2*T(n-2,k-1) + T(n-1,k) (except for T(0,0) = 1 and T(2,1) = 3 and assuming that T(n,k) = 0 for elements outside the triangular array). - _Gerald McGarvey_, Sep 20 2007
%C A054120 Conjecture: T(n,k) = A081577(n,k) - A081577(n-2,k-1). (A081577 is Pascal-(1,2,1) array). - _Gerald McGarvey_, Sep 20 2007
%C A054120 From _Russell Jay Hendel_, Jun 02 2015: (Start)
%C A054120 We prove the 1st McGarvey conjecture, (A) T(n,k)=T(n-1,k-1)+2*T(n-2,k-1)+T(n-1,k), with McGarvey's added modifications that n >= 3 and that T(n,k)=0 for n<0, k<0, and k>n. We make explicit that the rows and columns of the triangle start at 0 and furthermore, that both diagonals are at 45 degrees to the perpendicular (in both directions). Note that we use letters (A), (B), (C) to indicate equations.
%C A054120 We indicate the set of numbers corresponding to the half-line starting at T(n,k) with(B) L(n,k)={T(n,k), T(n-2,k-1), T(n-4,k-2),T(n-6,k-3),...}. There is no loss of generality in extending this line upwards indefinitely since most of the numbers lying on a half-ray are 0 anyway. If S_1, S_2, S_3 are sets, then the notation SUM(S_1,S_2, S_3) will indicate the sum of all elements in S_1, S_2 and S_3.
%C A054120 Using two applications of (B) we have the identity (C) L(n,k)={T(n,k)} UNION L(n-2,k-1). The definition of T(n,k) is given by (D) T(n,k)=SUM(L(n-1,k-1),L(n-2,k-1),L(n-1,k)). Combining (D) with (C), we immediately have (E), T(n,k)=T(n-1,k-1)+T(n-2,k-1)+T(n-1,k)+SUM(L(n-3,k-2), L(n-4,k-2), L(n-3,k-1)). By another application of (D), we have (F), T(n-2,k-1)=SUM(L(n-3,k-2),L(n-4,k-2),L(n-3,k-1)). Combining (E) and (F), we obtain (A) as required.
%C A054120 We illustrate the proof with n=5 and k=3. We first calculate T(3,2). By (D), we have (G) T(3,2)=6=SUM(L(2,1),L(1,1),L(2,2)) with L(2,1)={3,1,0,0,0,...}, L(1,1)={1,0,0,....} and L(2,2)={1,0,0,...}. By another application of (D), we have T(5,3)=39=(SUM(L(4,2),L(3,2),L(4,3)) where L(4,2)={18,3,1,...}={18} UNION L(2,1), L(3,2)={6,1,0,...}={6} UNION L(1,1), and L(4,3) = {1,0,0,...}={1} UNION L(2,2). Combining this last equation with (G), we have T(5,3)=39=18+6+1+SUM(L(2,1),L(1,1),L(2,2))=18+6+1+6=18+2*6+1 as required.
%C A054120 Note: There are two parts to the definition of the triangle:(i) T(n,0)=1, and ii) equation (D). Since (D) does not apply to T(0,0), (A) does not hold there. Since (D) applied to T(2,1) refers back to T(0,0), (A) does not hold there either. This explains the McGarvey requirement that n>=3. (End)
%C A054120 Consider the array with g.f. (1-u*v)/(1-u-v-2*u*v). The triangle appears to be that symmetric array read by antidiagonals. - _R. J. Mathar_, Jan 26 2022
%e A054120 Rows:
%e A054120 1;
%e A054120 1,1;
%e A054120 1,3,1;
%e A054120 1,6,6,1;
%e A054120 1,9,18,9,1;
%e A054120 1,12,39,39,12,1;
%Y A054120 Row sums: A052945. A054122 (diagonal), A052392 (subdiagonal).
%K A054120 nonn,tabl,eigen
%O A054120 0,5
%A A054120 _Clark Kimberling_