%I #23 Dec 21 2023 10:21:50
%S 1,1,2,2,1,3,3,1,1,4,4,1,1,1,5,5,1,1,1,1,6,6,1,1,1,1,1,7,7,1,1,1,1,1,
%T 1,8,8,1,1,1,1,1,1,1,9,9,1,1,1,1,1,1,1,1,10,10,1,1,1,1,1,1,1,1,1,11,
%U 11,1,1,1,1,1,1,1,1,1,1,12,12,1,1,1,1,1,1,1,1,1,1,1,13,13,1,1,1,1,1,1,1,1,1
%N Weight array W={w(i,j)} of the natural number array A000027.
%C The lattice lines in the first quadrant (including the x and y axes) cut the plane into unit squares. Suppose a weight w(i,j) is assigned to the square that has as upper right corner the point (i,j). Let s(m,n) be the sum of the weights w(i,j) for 1<=i<=m, 1<=j<=n. We call the array W={w(i,j)} the weight array of the array S={s(m,n)} and S the accumulation array of W. For the case at hand, S is the array of natural numbers having the following antidiagonals: (1), then (2,3), then (4,5,6), then (7,8,9,10) and so on.
%C Contribution from _Clark Kimberling_, Sep 14 2008: (Start)
%C In general, the weight array W of an arbitrary rectangular array S={s(i,j): i>=1, j>=1} is defined in two steps:
%C (1) extend s by defining s(i,j)=0 if i=0 or j=0;
%C (2) then w(m,n)=s(m,n)+s(m-1,n-1)-s(m,n-1)-s(m-1,n) for m>=1, n>=1. (End)
%H Stefano Spezia, <a href="/A144112/b144112.txt">Table of n, a(n) for n = 1..11325</a>, (first 150 antidiagonals, flattened).
%F Row 1: 1 followed by A000027.
%F Row n: n followed by A000012, for n>1.
%F G.f.: x*y*(1 - (1 + x)*y + (1 - x + x^2)*y^2)/((1 - x)^2*(1 - y)^2). - _Stefano Spezia_, Oct 01 2023
%e From _Clark Kimberling_, Jan 31 2011: (Start)
%e Northwest corner:
%e 1 1 2 3 4 5
%e 2 1 1 1 1 1
%e 3 1 1 1 1 1
%e 4 1 1 1 1 1
%e 5 1 1 1 1 1.
%e so that the accumulation array has corner:
%e 1...2...4...7...11...16
%e 3...5...8...12..17...23
%e 6...9...13..18..24...31
%e 10..14..19..25..32...40
%e 15..20..26..33..41...50.
%e s(2,4)=1+1+2+3+2+1+1+1=12. (End)
%t Array[Append[PadRight[{#},#,1],#+1]&,15,0] (* _Paolo Xausa_, Dec 21 2023 *)
%Y Cf. A000012, A000027.
%K nonn,tabl
%O 1,3
%A _Clark Kimberling_, Sep 11 2008