Search: id:A141388 Results 1-1 of 1 results found. %I A141388 %S A141388 0,6,6,23,60,23,56,208,208,56,109,496,713,496,109,184,968,1696,1696,968, %T A141388 184,279,1664,3311,4032,3311,1664,279,384,2616,5704,7872,7872,5704,2616, %U A141388 384,473,3840,9005,13568,15369,13568,9005,3840,473,488,5320,13312,21440 %N A141388 Cubic form (k=3) of the generalized neo-combinations: t(n,m,k)=(n - m)^k*(m + 1)^k - 2^(n - 1). %C A141388 Row sums are: %C A141388 {0, 12, 106, 528, 1923, 5696, 14540, 33152, 69141, 134096}; %C A141388 Note that the domain here is: %C A141388 {m, 0, n - 1}], {n, 1, 10} %C A141388 and not: %C A141388 {m, 0, n }], {n, 0, 10} %C A141388 The 2^(n-1) term was added to make the result generally symmetrical %C A141388 on the Dynkin / A_n weight domain.. %C A141388 Triangular coefficient sequences of the general form: %C A141388 t(n,m,k)=Floor[((n - m)^k*(m + 1)^k - 2^(n - 1))/n^k); %C A141388 have a Pascal triangle / binomial shape. %D A141388 R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 139. %F A141388 k=3; t(n,m,k)=(n - m)^k*(m + 1)^k - 2^(n - 1). %e A141388 {0}, %e A141388 {6, 6}, %e A141388 {23, 60, 23}, %e A141388 {56, 208, 208, 56}, %e A141388 {109, 496, 713, 496, 109}, %e A141388 {184, 968, 1696, 1696, 968, 184}, %e A141388 {279, 1664, 3311, 4032, 3311, 1664, 279}, %e A141388 {384, 2616, 5704, 7872, 7872, 5704, 2616, 384}, %e A141388 {473, 3840, 9005, 13568, 15369, 13568, 9005, 3840, 473}, %e A141388 {488, 5320, 13312, 21440, 26488, 26488, 21440, 13312, 5320, 488} %t A141388 Clear[T, n, m, a]; T[n_, m_] = (n - m)^3*(m + 1)^3 - 2^(n - 1); a = Table[Table[T[n, m], {m, 0, n - 1}], {n, 1, 10}]; Flatten[a] %Y A141388 Cf. A003991. %Y A141388 Sequence in context: A004983 A034695 A053168 this_sequence A087236 A077193 A056482 %Y A141388 Adjacent sequences: A141385 A141386 A141387 this_sequence A141389 A141390 A141391 %K A141388 nonn,uned,tabl %O A141388 1,2 %A A141388 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Aug 03 2008 Search completed in 0.001 seconds