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A185779 Third accumulation array of Pascal's triangle (as a rectangle), by antidiagonals.

%I A185779 #8 Jul 12 2017 03:18:59
%S A185779 1,4,4,10,17,10,20,45,45,20,35,95,126,95,35,56,175,281,281,175,56,84,
%T A185779 294,546,662,546,294,84,120,462,966,1358,1358,966,462,120,165,690,
%U A185779 1596,2534,2941,2534,1596,690,165,220,990,2502,4410,5790,5790,4410,2502,990,220,286,1375,3762,7272,10620,12021,10620,7272,3762,1375,286,364,1859,5467,11484,18432,23229,23229,18432,11484,5467,1859,364,455
%N A185779 Third accumulation array of Pascal's triangle (as a rectangle), by antidiagonals.
%C A185779 Using "Axxxxxx < Ayyyyyy" to mean that Ayyyyyy is the accumulation array of Axxxxxx, as defined at A144112:
%C A185779 A185779 < A144225 < A007318 < A014430 < A077023 < A185779, where each of these is formatted as a rectangle (e.g., A007318 is Pascal's triangle).  See A185778.
%C A185779 row 1: A000292
%C A185779 row 2: A095667
%H A185779 G. C. Greubel, <a href="/A185779/b185779.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%F A185779 T(n,k) = C(n+k+4,n+2) - (k+3)*(k+4)/2 - (k+2)*n*(k*n+n+3*k+7)/4, for k>=1, n>=1.
%e A185779 Northwest corner:
%e A185779 1....4...10...20...35
%e A185779 4....17..45...95...175
%e A185779 10...45..126..281..546
%e A185779 20...95..281..662..1358
%t A185779 f[n_, k_] := Binomial[n + k + 4, n + 2] - (k + 3)*(k + 4)/2 - (k + 2)* n*(k*n + n + 3*k + 7)/4; TableForm[Table[f[n, k], {n, 1, 5}, {k, 1, 5}]]
%t A185779 Table[f[n - k + 1, k], {n, 10}, {k, n, 1, -1}] // Flatten (* _G. C. Greubel_, Jul 12 2017 *)
%Y A185779 Cf. A007318, A014430, A077023, A144112, A144225, A185778.
%K A185779 nonn,tabl
%O A185779 1,2
%A A185779 _Clark Kimberling_, Feb 03 2011