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A318496 Scaled g.f. T(v) = Sum_{n>=0} a(n)*(v/16)^n satisfies 15*(189*v-80)*T + d/dv(4*v*(27*v-5)*(27*v-32)*T') = 0, and a(0)=1; sequence gives a(n).

%I A318496 #20 Dec 14 2024 07:08:30
%S A318496 1,30,1440,85260,5606100,391231080,28360117800,2110794125400,
%T A318496 160187289344100,12339496371120600,961855480344860640,
%U A318496 75700880007230883600,6005580964527420946800,479651805879329497831200,38529018420812424368031600,3110295017383730347887664560
%N A318496 Scaled g.f. T(v) = Sum_{n>=0} a(n)*(v/16)^n satisfies 15*(189*v-80)*T + d/dv(4*v*(27*v-5)*(27*v-32)*T') = 0, and a(0)=1; sequence gives a(n).
%C A318496 Period function T(v) measures precession of the J-vector along an algebraic sphere curve with local cyclic C_3 symmetry. For precise definitions, pictures, a proof certificate, and more information, see A318495.
%H A318496 É. Goursat, <a href="http://www.numdam.org/item/ASENS_1887_3_4__159_0">Étude des surfaces qui admettent tous les plans de symétrie d'un polyèdre régulier</a>, Annales scientifiques de l'École Normale Supérieure, Série 3 : Volume 4 (1887), 166-170.
%F A318496 10*n^2*a(n) - 3*(333*n^2-333*n+100)*a(n-1) + 324*(6*n-7)*(6*n-5)*a(n-2) = 0.
%F A318496 For n > 0, a(n) mod 30 = 0 (conjecture, tested up to n=10^6).
%F A318496 G.f.: hypergeom([1/12, 5/12],[1],-1728*(432*x-5)^2*x^3*(27*x-2)^5/(1-120*x+3240*x^2+174960*x^3)^3)/(1-120*x+3240*x^2+174960*x^3)^(1/4). - _Mark van Hoeij_, Dec 13 2024
%t A318496 RecurrenceTable[{10 n^2 a[n] - 3 (333 n^2 - 333 n + 100) a[n-1] + 324 (6*n - 7) (6 n - 5) a[n-2] == 0, a[0] == 1, a[1] == 30}, a, {n, 0, 15}]
%o A318496 (GAP) a:=[1,30];; for n in [3..20] do a[n]:=(1/(10*(n-1)^2))*(3*(333*(n^2-3*n+2)+100)*a[n-1]-(324*(6*n-13)*(6*n-11)*a[n-2])); od; a; # _Muniru A Asiru_, Sep 24 2018
%Y A318496 Cf. A318495. Periods: A186375, A318245, A318417.
%K A318496 nonn
%O A318496 0,2
%A A318496 _Bradley Klee_, Aug 27 2018