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A105125 Triangle read by rows: T(n,k) = n^3 + k^3, n >= 0, 0 <= k <= n.

%I A105125 #17 Oct 12 2018 23:33:39
%S A105125 0,1,2,8,9,16,27,28,35,54,64,65,72,91,128,125,126,133,152,189,250,216,
%T A105125 217,224,243,280,341,432,343,344,351,370,407,468,559,686,512,513,520,
%U A105125 539,576,637,728,855,1024,729,730,737,756,793,854,945,1072,1241,1458,1000,1001,1008,1027
%N A105125 Triangle read by rows: T(n,k) = n^3 + k^3, n >= 0, 0 <= k <= n.
%F A105125 T(n,k) = n^3 + k^3, n >= 0, 0 <= k <= n.
%F A105125 T(n, k) = A051162(n, k)*(A051162(n, k)^2 + 3* A025581(n, k)^2)/4. See the comment on A051162 for this identity. - _Wolfdieter Lang_, May 15 2015
%F A105125 G.f. for triangle: -(9*x^5*y^3 - 8*x^4*y^3 - x^4*y^2 + 7*x^3*y^3 - 36*x^3*y^2 - 2*x^2*y^3 + 5*x^3*y + 27*x^2*y^2 + 12*x^2*y - 8*x*y^2 - x^2 + 3*x*y - 4*x - 2*y - 1)*x/((x-1)^4*(x*y-1)^4). - _Robert Israel_, May 15 2015
%e A105125 Triangle begins (modulo 2 plot is a checkerboard):
%e A105125   {0}
%e A105125   {1, 2}
%e A105125   {8, 9, 16}
%e A105125   {27, 28, 35, 54}
%e A105125   {64, 65, 72, 91, 128}
%e A105125   {125, 126, 133, 152, 189, 250}
%e A105125   ...
%e A105125 The identity for T(2, 1): 9 = 3*(3^2 + 3*1^2)/4 = 3*12/4 = 9. - _Wolfdieter Lang_, May 15 2015
%p A105125 seq(seq(n^3+k^3,k=0..n),n=0..10); # _Robert Israel_, May 15 2015
%t A105125 f[n_, m_, p_] := n^p + m^p p = 3 a = Table[Table[f[n, m, p], {n, 0, m}], {m, 0, 20}] aa = Flatten[a]
%Y A105125 Cf. A069011. Different from A004999. A257238, A025581, A051162.
%K A105125 nonn,easy,tabl
%O A105125 0,3
%A A105125 _Roger L. Bagula_, Apr 09 2005