|
%I #6 Jul 14 2016 17:37:24
%S 1,1,3,12,52,246,1224,6300,33300,179643,985014,5473611,30756638,
%T 174460232,997620618,5744867987,33286011182,193909247415,
%U 1135089596103,6673249089811,39385246833252,233269982839202,1386037324968892,8259646628699352,49352953593386343,295622321452540572,1774808912507223393,10677836739621878302,64367558119301035734,388726705119505499253,2351595589979028349894
%N G.f. A(x) satisfies: A(x)^3 - 3*A(x)^4 = A(x^3).
%H Paul D. Hanna, <a href="/A274396/b274396.txt">Table of n, a(n) for n = 1..300</a>
%F G.f. A(x) satisfies: A(B(x)^3) = x^3 - 3*x^4, where A(B(x)) = x.
%e G.f.: A(x) = x + x^2 + 3*x^3 + 12*x^4 + 52*x^5 + 246*x^6 + 1224*x^7 + 6300*x^8 + 33300*x^9 + 179643*x^10 + 985014*x^11 + 5473611*x^12 + 30756638*x^13 + 174460232*x^14 + 997620618*x^15 + 5744867987*x^16 + 33286011182*x^17 +...
%e such that A(x)^3 - 3*A(x)^4 = A(x^3).
%e RELATED SERIES.
%e A(x)^3 = x^3 + 3*x^4 + 12*x^5 + 55*x^6 + 264*x^7 + 1329*x^8 + 6915*x^9 + 36846*x^10 + 200100*x^11 + 1103367*x^12 + 6160500*x^13 + 34758036*x^14 + 197858104*x^15 + 1134939672*x^16 + 6553596816*x^17 + 38064666342*x^18 +...
%e A(x)^4 = x^4 + 4*x^5 + 18*x^6 + 88*x^7 + 443*x^8 + 2304*x^9 + 12282*x^10 + 66700*x^11 + 367785*x^12 + 2053500*x^13 + 11586012*x^14 + 65952684*x^15 + 378313224*x^16 + 2184532272*x^17 + 12688222032*x^18 +...
%e where
%e A(x)^3 - 3*A(x)^4 = x^3 + x^6 + 3*x^9 + 12*x^12 + 52*x^15 + 246*x^18 + 1224*x^21 +...
%e Let B(x) be the series reversion of g.f. A(x), so that A(B(x)) = x, then
%e B(x) = x - x^2 - x^3 - 2*x^4 - 2*x^5 - 8*x^6 - 18*x^7 - 39*x^8 - 111*x^9 - 266*x^10 - 683*x^11 - 1859*x^12 - 4824*x^13 - 13091*x^14 - 35642*x^15 - 96917*x^16 - 267206*x^17 - 738605*x^18 - 2048769*x^19 - 5720874*x^20 - 16009584*x^21 -...
%e where B(x^3 - 3*x^4) = B(x)^3.
%e Further, we have
%e A(B(x)^3) = C(x) = x^3 - 3*x^4,
%e A(B(x)^9) = C(C(x)) = x^9 - 9*x^10 + 27*x^11 - 30*x^12 + 36*x^13 - 162*x^14 + 324*x^15 - 243*x^16, ...
%e so that A(B(x)^(3^n)) = C^n(x), the n-th iteration of C(x) = x^3 - 3*x^4.
%o (PARI) /* From A(B(x)^3) = x^3 - 3*x^4, where A(B(x)) = x: */
%o {a(n) = my(A=[1, 1], F, B); for(i=1, n, A=concat(A, 0); F=x*Ser(A); B=serreverse(F); A[#A] = Vec(subst(F, x, B^3))[#A]/3); A[n]}
%o for(n=1, 40, print1(a(n), ", "))
%Y Cf. A273095.
%K nonn
%O 1,3
%A _Paul D. Hanna_, Jul 14 2016
| 