0,2

Comment from Alexander R. Povolotsky, Nov 19 2007 (Start):

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:

(1 - 1)/133 = 0

(56 - 56)/133 = 0

(1463 - 133) / 133 = 10

(24320 - 912) / 133 = 176

(293930 - 1463) / 133 = 2199

(2785552 - 1539) / 133 = 146527/7

(21737254 - 6480) / 133 = 21730774/133

(144538624 - 7371) / 133 = 144531253/133

(839848450 - 8645) / 133 = 6314585

(4347450800 - 24320) / 133 = 228811920/7

(20355385710 - 27664) / 133 = 153047805

(87265194240 - 40755) / 133 = 656128974

(345992859975 - 51072) / 133 = 2601449691

(1279301331000 - 86184) / 133 = 9618806352

(4442249264625 - 150822) / 133 = 233802584937/7

(14573017267200 - 152152)/133 = 109571557256

(45398364338250 - 238602)/133 = 341341083456

(134897996890800 - 253935)/133 = 1014270651405

(383822534565820 - 293930)/133 = 2885883718540

(1049290591104000 - 320112)/133 = 1049290590783888/133

...

Note that 133 is also the dimension of the Lie algebra E_7. (End)

J. M. Landsberg and L. Manivel, The sextonions and E7 1/2, Adv. Math. 201 (2006), 143-179. [Th. 7.2(ii), case a=4]

Onishchik and Vinberg, Seminar on Lie Groups and Algebraic Groups, Springer Verlag 1990, see Table 5.

a(n) = (1/10950439500)*(n+9)*binomial(n+17, 4)*binomial(n+4, 4)*binomial(n+13, 9)^2.

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)];

Cf. A121736.

Sequence in context: A025597 A034202 A160290 this_sequence A022081 A017772 A035723

Adjacent sequences: A030646 A030647 A030648 this_sequence A030650 A030651 A030652

nonn

Paolo Dominici (pl.dm(AT)libero.it)

Entry revised by N. J. A. Sloane (njas(AT)research.att.com), Oct 20 2007