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%I #15 Jan 10 2024 00:26:32
%S 1,56,1463,24320,293930,2785552,21737254,144538624,839848450,
%T 4347450800,20355385710,87265194240,345992859975,1279301331000,
%U 4442249264625,14573017267200,45398364338250,134897996890800,383822534859750,1049290591104000,2764459117589400
%N Dimensions of multiples of minimal representation of complex Lie algebra E7.
%C From _Alexander R. Povolotsky_, Nov 19 2007: (Start)
%C After adjustment for the fact that a(n) is indexed from 0 while A121736 is indexed from 1, it appears that in many cases (with some exceptions) (a(n) - A121736(n+1))/133 (where A121736(3) = 133) yields integral values:
%C (1 - 1)/133 = 0
%C (56 - 56)/133 = 0
%C (1463 - 133) / 133 = 10
%C (24320 - 912) / 133 = 176
%C (293930 - 1463) / 133 = 2199
%C (2785552 - 1539) / 133 = 146527/7
%C (21737254 - 6480) / 133 = 21730774/133
%C (144538624 - 7371) / 133 = 144531253/133
%C (839848450 - 8645) / 133 = 6314585
%C (4347450800 - 24320) / 133 = 228811920/7
%C (20355385710 - 27664) / 133 = 153047805
%C (87265194240 - 40755) / 133 = 656128974
%C (345992859975 - 51072) / 133 = 2601449691
%C (1279301331000 - 86184) / 133 = 9618806352
%C (4442249264625 - 150822) / 133 = 233802584937/7
%C (14573017267200 - 152152)/133 = 109571557256
%C (45398364338250 - 238602)/133 = 341341083456
%C (134897996890800 - 253935)/133 = 1014270651405
%C (383822534565820 - 293930)/133 = 2885883718540
%C (1049290591104000 - 320112)/133 = 1049290590783888/133
%C ...
%C Note that 133 is also the dimension of the Lie algebra E_7. (End)
%D Onishchik and Vinberg, Seminar on Lie Groups and Algebraic Groups, Springer Verlag 1990, see Table 5.
%H G. C. Greubel, <a href="/A030649/b030649.txt">Table of n, a(n) for n = 0..1000</a>
%H J. M. Landsberg and L. Manivel, <a href="https://doi.org/10.1016/j.aim.2005.02.001">The sextonions and E7 1/2</a>, Adv. Math. 201 (2006), 143-179. [Th. 7.2(ii), case a=4]
%F a(n) = (1/10950439500)*(n+9)*binomial(n+17, 4)*binomial(n+4, 4)*binomial(n+13, 9)^2.
%p b:=binomial; t72b:= proc(a,k) ((a+k+1)/(a+1)) * b(k+2*a+1,k)*b(k+3*a/2+1,k)/(b(k+a/2,k)); end; [seq(t72b(8,k),k=0..28)];
%t Table[(1/10950439500)*(n + 9)*Binomial[n + 17, 4]*Binomial[n + 4, 4]* Binomial[n + 13, 9]^2, {n,0,50}] (* _G. C. Greubel_, Feb 19 2017 *)
%o (PARI) for(n=0,25, print1((1/10950439500)*(n+9)*binomial(n+17, 4)*binomial(n+4, 4)*binomial(n+13, 9)^2, ", ")) \\ _G. C. Greubel_, Feb 19 2017
%Y Cf. A121736.
%K nonn
%O 0,2
%A Paolo Dominici (pl.dm(AT)libero.it)
%E Entry revised by _N. J. A. Sloane_, Oct 20 2007
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