%I #38 Sep 08 2022 08:46:18
%S 0,1,1,2,9,2,3,17,29,3,4,25,56,67,4,5,33,83,131,129,5,6,41,110,195,
%T 254,221,6,7,49,137,259,379,437,349,7,8,57,164,323,504,653,692,519,8,
%U 9,65,191,387,629,869,1035,1031,737,9,10,73,218,451,754,1085,1378,1543,1466,1009,10
%N Triangle read by rows: T(n, k) = (n - k)*(k + 1)^3 + k, 0 <= k <= n.
%F Row sums sum_{k>=0} T(n,k) = n*(n+1)*(3*n^3+12*n^2+13*n+32)/60. - _R. J. Mathar_, Mar 19 2017
%e Triangle begins:
%e 0;
%e 1, 1;
%e 2, 9, 2;
%e 3, 17, 29, 3;
%e 4, 25, 56, 67, 4;
%e 5, 33, 83, 131, 129, 5;
%e 6, 41, 110, 195, 254, 221, 6;
%e 7, 49, 137, 259, 379, 437, 349, 7;
%e 8, 57, 164, 323, 504, 653, 692, 519, 8;
%e 9, 65, 191, 387, 629, 869, 1035, 1031, 737, 9;
%e 10, 73, 218, 451, 754, 1085, 1378, 1543, 1466, 1009, 10;
%e ...
%t t[n_, k_] := (n - k)*(k + 1)^3 + k; Table[ t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Robert G. Wilson v_, Feb 09 2017 *)
%o (Magma) /* As triangle */ [[(n-k)*(k+1)^3+k: k in [1..n]]: n in [0..10]];
%o (PARI) for(n=0,10,for(k=0,n,print1((n-k)*(k+1)^3+k,", "))) \\ _Derek Orr_, Feb 26 2017
%Y Cf. Triangle read by rows: T(n,k) = (n-k)*(k+1)^m+k: A003056 (m = 0), A059036 (m = 1), A274602 (m = 2), this sequence (m = 3).
%Y Cf. A001477 (column 0), A017077 (column 1), A281546 (column 2), A242604 (middle diagonal).
%K nonn,tabl,easy
%O 1,4
%A _Juri-Stepan Gerasimov_, Jan 23 2017