A105125 - OEIS

%I #17 Oct 12 2018 23:33:39

%S 0,1,2,8,9,16,27,28,35,54,64,65,72,91,128,125,126,133,152,189,250,216,

%T 217,224,243,280,341,432,343,344,351,370,407,468,559,686,512,513,520,

%U 539,576,637,728,855,1024,729,730,737,756,793,854,945,1072,1241,1458,1000,1001,1008,1027

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

%F T(n,k) = n^3 + k^3, n >= 0, 0 <= k <= n.

%F 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 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 Triangle begins (modulo 2 plot is a checkerboard):

%e   {0}

%e   {1, 2}

%e   {8, 9, 16}

%e   {27, 28, 35, 54}

%e   {64, 65, 72, 91, 128}

%e   {125, 126, 133, 152, 189, 250}

%e   ...

%e The identity for T(2, 1): 9 = 3*(3^2 + 3*1^2)/4 = 3*12/4 = 9. - _Wolfdieter Lang_, May 15 2015

%p seq(seq(n^3+k^3,k=0..n),n=0..10); # _Robert Israel_, May 15 2015

%t 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 Cf. A069011. Different from A004999. A257238, A025581, A051162.

%K nonn,easy,tabl

%O 0,3

%A _Roger L. Bagula_, Apr 09 2005