%I #24 Jul 11 2020 13:38:46
%S 1,3,4,7,11,10,14,23,26,20,25,42,51,50,35,41,70,88,94,85,56,63,109,
%T 140,156,155,133,84,92,161,210,240,250,237,196,120,129,228,301,350,
%U 375,374,343,276,165,175,312,416,490,535,550,532,476,375,220,231,415,558,664,735,771,770,728,639,495,286
%N Accumulation array, T, of the natural number array A000027, by antidiagonals.
%C Suppose that R={R(n,k) : n>=1, k>=1} is a rectangular array. The accumulation array of R is given by T(n,k) = Sum_{R(i,j): 1<=i<=n, 1<=j<=k}. (See A144112.)
%C The formula for the integer T(n,k) has denominator 12. The 2nd, 3rd, and 4th accumulation arrays of A000027 have formulas in which the denominators are 144, 2880, and 86400, respectively; see A185507, A185508, and A185509.
%H G. C. Greubel, <a href="/A185506/b185506.txt">Table of n, a(n) for the first 50 antidiagonals, flattened</a>
%F T(n,k) = k*n*(2*n^2 + 3*(k+1)*n + 2*k^2 - 3*k + 5)/12.
%e The natural number array A000027 starts with
%e 1, 2, 4, 7, ...
%e 3, 5, 8, 12, ...
%e 6, 9, 13, 18, ...
%e ...
%e T(n,k) is the sum of numbers in the rectangle with corners at (1,1) and (n,k) of A000027, so that a corner of T is as follows:
%e 1, 3, 7, 14, 25, 41
%e 4, 11, 23, 42, 70, 109
%e 10, 26, 51, 88, 140, 210
%e 20, 50, 94, 156, 240, 350
%e 35, 85, 155, 250, 375, 535
%t f[n_,k_]:=k*n*(2n^2+3(k+1)*n+2k^2-3k+5)/12;
%t TableForm[Table[f[n,k],{n,1,10},{k,1,15}]]
%t Table[f[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten
%Y Cf. A000027, A185507, A185508, A185509.
%Y Cf. A004006 (row 1), A000292 (col 1), A051925 (col 2), A185505 (1st diagonal).
%K nonn,tabl
%O 1,2
%A _Clark Kimberling_, Jan 29 2011