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%I #7 Sep 17 2019 15:41:43
%S 1,1,24,1,36,612,1,48,1104,15912,1,60,1740,32130,417690,1,72,2520,
%T 56700,912492,11027016,1,84,3444,91350,1750014,25562628,292215924,1,
%U 96,4512,137808,3059856,52303968,710025264,7764594552,1,108,5724,197802,4992354
%N Triangle read by rows in which row n gives coefficients of polynomial R_n(y) that satisfies R_n(1/3) = 9^n, where R_n(y) forms the initial (n+1) terms of g.f. A097191(y)^(n+1).
%H G. C. Greubel, <a href="/A097190/b097190.txt">Rows n = 0..50 of triangle, flattened</a>
%F G.f.: A(x, y) = 3*y/((1-27*x*y) + (3*y-1)*(1-27*x*y)^(8/9)).
%F G.f.: A(x, y) = A097192(x*y)/(1 - x*A097193(x*y)).
%e Row polynomials evaluated at y=1/3 equals powers of 9:
%e 9^1 = 1 + 24/3;
%e 9^2 = 1 + 36/3 + 612/3^2;
%e 9^3 = 1 + 48/3 + 1104/3^2 + 15912/3^3;
%e 9^4 = 1 + 60/3 + 1740/3^2 + 32130/3^3 + 417690/3^4;
%e where A097191(y)^(n+1) has the same initial terms as the n-th row:
%e A097191(y) = 1 + 12y + 60y^2 + 90y^3 - 558y^4 - 2916y^5 + 2160y^6 +...
%e A097191(y)^2 = 1 + 24y +...
%e A097191(y)^3 = 1 + 36y + 612y^2 +...
%e A097191(y)^4 = 1 + 48y + 1104y^2 + 15912y^3 +...
%e A097191(y)^5 = 1 + 60y + 1740y^2 + 32130y^3 + 417690y^4 +...
%e Rows begin with n=0:
%e 1;
%e 1, 24;
%e 1, 36, 612;
%e 1, 48, 1104, 15912;
%e 1, 60, 1740, 32130, 417690;
%e 1, 72, 2520, 56700, 912492, 11027016;
%e 1, 84, 3444, 91350, 1750014, 25562628, 292215924;
%e 1, 96, 4512, 137808, 3059856, 52303968, 710025264, 7764594552; ...
%t Table[SeriesCoefficient[3*y/((1-27*x*y) + (3*y-1)*(1-27*x*y)^(8/9)), {x, 0,n}, {y,0,k}], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Sep 17 2019 *)
%o (PARI) {T(n,k)=if(n==0,1,if(k==0,1,if(k==n, 3^n*(9^n-sum(j=0,n-1, T(n,j)/3^j)), polcoeff((Ser(vector(n,i,T(n-1,i-1)),x) +x*O(x^k))^((n+1)/n),k,x))))}
%Y Cf. A097186, A097191, A097192, A097193, A097194, A097195.
%K nonn,tabl
%O 0,3
%A _Paul D. Hanna_, Aug 03 2004