%I #32 Jul 27 2024 01:26:30
%S 9,28,35,65,72,91,126,133,152,189,217,224,243,280,341,344,351,370,407,
%T 468,559,513,520,539,576,637,728,855,730,737,756,793,854,945,1072,
%U 1241,1001,1008,1027,1064,1125,1216,1343,1512,1729
%N Triangle read by rows, sums of 2 distinct nonzero cubes: T(n,k) = (n+1)^3+k^3, 1 <= k <= n.
%C When n=k: T(n,k) = (2n+1)(n^2+n+1). Therefore, T(n,k)/(2n+1) = A002061(n+1).
%C A002383 is the sequence of all primes of the form T(n,k)/(2n+1), n=k.
%C When starting at T(n,k) n=k, diagonal sums are n^2*(2n+1)^2. For example, starting at T(4,4) = 189: 189+243+351+513 = 4^2*9^2 = 1296.
%C Coefficients in T(n,k) are multiples of n+k+1; therefore, coefficients in all diagonals starting at T(n,1) are multiples of n+2.
%C Let T"(n,k) = T(n,k)/(n+k+1). Then reading T"(n,k) by rows:
%C i. Row sums are A162147(n). For example, T"(3,k) = [65/5, 72/6, 91/7] = [13,12,13]. 13+12+13 = 38; A162147(3) = 38.
%C ii. Extend the triangle in A215631 to a symmetric array by reflection about the main diagonal, and let that array be T"215631(n,k). Then the diagonal starting with T"215631(n,1) is row n in T"(n,k). For example, the diagonal starting at T"215631(4,1) = [21,19,19,21]; T"(4,k) = [126/6, 133/7, 152/8, 189/9] = [21,19,19,21].
%C iii. Coefficients in T"(n,k) are a permutation of A024612.
%F T(n,k) = (n+1)^3+k^3.
%F T(n,k) = (2k+1)(k^2+k+1) + Sum_{j=k+1..n} A003215(j), n>=k+1. For example, T(8,4) = 9*21 + 91 + 127 + 169 + 217 = 793.
%e Triangle starts:
%e n\k 1 2 3 4 5 6 7 8 9 10 ...
%e 1: 9
%e 2: 28 35
%e 3: 65 72 91
%e 4: 126 133 152 189
%e 5: 217 224 243 280 341
%e 6: 344 351 370 407 468 559
%e 7: 513 520 539 576 637 728 855
%e 8: 730 737 756 793 854 945 1072 1241
%e 9: 1001 1008 1027 1064 1125 1216 1343 1512 1729
%e 10: 1332 1339 1358 1395 1456 1547 1674 1843 2060 2331
%e ...
%e The successive terms are (2^3+1^3), (3^3+1^3), (3^3+2^3), (4^3+1^3), (4^3+2^3), (4^3+3^3), ...
%Y Cf. A024670.
%Y Related: A002061, A002383, A003215, A024612, A162147, A215631.
%K nonn,tabl
%O 1,1
%A _Bob Selcoe_, Mar 31 2015