This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A217794 #6 Mar 12 2025 18:23:22
%S A217794 1,1,5,50,705,12225,244076,5408850,130087875,3345893240,91094645450,
%T A217794 2606026116783,77908655011205,2423758771361300,78209895085897845,
%U A217794 2610730390260224350,89962745874096081282,3194452580744306873560,116712765368638543260825
%N A217794 G.f. satisfies: A(x/A(x)^3) = 1 + x*A(x)^2.
%C A217794 More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then
%C A217794 A(x) = 1 + x*G(x)^(m+k) where G(x) = A(x*G(x)^k) and G(x/A(x)^k) = A(x);
%C A217794 thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1.
%F A217794 G.f.: A(x) = 1 + x*G(x)^5 where G(x) = A(x*G(x)^3) and A(x) = G(x/A(x)^3).
%F A217794 a(n) = [x^(n-1)] 5*A(x)^(3*n+2)/(3*n+2) for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in 5*A(x)^(3*n+2)/(3*n+2) for n>=1 (see comment).
%e A217794 G.f.: A(x) = 1 + x + 5*x^2 + 50*x^3 + 705*x^4 + 12225*x^5 + 244076*x^6 +...
%e A217794 Related expansions:
%e A217794 A(x)^2 = 1 + 2*x + 11*x^2 + 110*x^3 + 1535*x^4 + 26360*x^5 + 522152*x^6 +...
%e A217794 A(x)^3 = 1 + 3*x + 18*x^2 + 181*x^3 + 2505*x^4 + 42630*x^5 + 837968*x^6 +...
%e A217794 A(x/A(x)^3) = 1 + x + 2*x^2 + 11*x^3 + 110*x^4 + 1535*x^5 + 26360*x^6 +...
%e A217794 Note that A(x) = 1 + x*G(x)^5 where G(x) = A(x*G(x)^3):
%e A217794 G(x) = 1 + x + 8*x^2 + 107*x^3 + 1840*x^4 + 36973*x^5 + 828944*x^6 +...
%e A217794 G(x)^3 = 1 + 3*x + 27*x^2 + 370*x^3 + 6378*x^4 + 127608*x^5 +...
%e A217794 G(x)^5 = 1 + 5*x + 50*x^2 + 705*x^3 + 12225*x^4 + 244076*x^5 +...
%e A217794 To illustrate the formula a(n) = [x^(n-1)] 5*A(x)^(3*n+2)/(3*n+2),
%e A217794 form a table of coefficients in A(x)^(3*n+2) as follows:
%e A217794 A^5: [(1), 5, 35, 360, 4930, 82576, 1599555, 34576120, ...];
%e A217794 A^8: [1,(8), 68, 736, 10050, 165336, 3139976, 66675504, ...];
%e A217794 A^11: [1, 11,(110), 1265, 17435, 283712, 5299448, 110675125, ...];
%e A217794 A^14: [1, 14, 161,(1974), 27706, 448882, 8275827, 170250124, ...];
%e A217794 A^17: [1, 17, 221, 2890,(41565), 674373, 12316568, 250050450, ...];
%e A217794 A^20: [1, 20, 290, 4040, 59795,(976304), 17727230, 355906600, ...]; ...
%e A217794 in which the main diagonal forms the initial terms of this sequence:
%e A217794 [5/5*(1), 5/8*(8), 5/11*(110), 5/14*(1974), 5/17*(41565), 5/20*(976304), ...].
%o A217794 (PARI) {a(n)=local(A=1+x, G); for(i=0, n, G=(serreverse(x/(A+x*O(x^n))^3)/x)^(1/3); A=1+x*G^5); polcoeff(A, n)}
%o A217794 (PARI) /* This sequence is generated when k=3, m=2: A(x/A(x)^k) = 1 + x*A(x)^m */
%o A217794 {a(n, k=3, m=2)=local(A=sum(i=0, n-1, a(i, k, m)*x^i)+x*O(x^n)); if(n==0, 1, polcoeff((m+k)/(m+k*n)*A^(m+k*n), n-1))}
%o A217794 for(n=0, 20, print1(a(n), ", "))
%Y A217794 Cf. A145349, A145350.
%K A217794 nonn
%O A217794 0,3
%A A217794 _Paul D. Hanna_, Oct 11 2012