%I #6 Jul 19 2015 10:14:58
%S 1,1,5,1,5,7,11,1,4,5,7,8,11,1,5,7,9,11,13,17,1,6,9,11,13,15,17,19,1,
%T 7,11,13,17,19,23,29,1,11,17,19,23,29,31,51,55
%N Exponents m(i) for exceptional groups with best guesses for E7 1/2 and E9 added (there is a problem with the dimension of E9 as no sum of odd numbers will equal the 484, I get 483): triangular sequence is: A1,G2,F4,E6,E7 E7 1/2,E8,E9.
%C Betti number row sums: Table[Apply[Plus, CoefficientList[Expand[Product[(1 + t^(2*a[i][[n]] + 1)), {n, 1, Length[a[i]]}]], t]], {i, 0, 7}] {2, 4, 16, 64, 128, 256, 256, 512} Group dimensions sums: b[n_] = 2*a[n] + 1 Table[Apply[Plus, b[n]], {n, 0, 7}] {3, 14, 52, 78, 133, 190, 248, 483}.
%C From these exponents it is possible to get PoincarĂ© polynomial estimates for the new E7 1/2 and E8 that best fit the pattern of the known exponents.
%D J. M. Landsberg, The sextonions and E_{7 1/2} (with L.Manivel) (Advances in Math 201(2006) p143  179) page 22; J. M. Landsberg, http://www.math.tamu.edu/~jml/LMsexpub.pdf: The sextonions and E_{7 1/2}
%D Armand Borel's Essays in History of Lie Groups and Algebraic Groups: gives G2 PoincarĂ© polynomial, History of Mathematics, V. 21; http://www.amazon.com/EssaysHistoryGroupsAlgebraicMathematics/dp/0821802887/ref=pd_rhf_p_3/10400296170633535
%F a(0) = {1}; a(1) = {1, 5}; a(2) = {1, 5, 7, 11}; a(3) = {1, 4, 5, 7, 8, 11}; a(4) = {1, 5, 7, 9, 11, 13, 17}; a(5) = {1, 6, 9, 11, 13, 15, 17, 19}; a(6) = {1, 7, 11, 13, 17, 19, 23, 29}; a(7) = {1, 11, 17, 19, 23, 29, 31, 51, 55};
%t a[0] = {1}; a[1] = {1, 5}; a[2] = {1, 5, 7, 11}; a[3] = {1, 4, 5, 7, 8, 11}; a[4] = {1, 5, 7, 9, 11, 13, 17}; a[5] = {1, 6, 9, 11, 13, 15, 17, 19}; a[6] = {1, 7, 11, 13, 17, 19, 23, 29}; a[7] = {1, 11, 17, 19, 23, 29, 31, 51, 55};
%Y Cf. A005556, A005763, A005776.
%K nonn,uned
%O 1,3
%A _Roger L. Bagula_, May 16 2007
