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Expansion of x / Series_Reversion( x/(1 + 3*x - 6*x^2 - 8*x^3)^(1/3) ).
3

%I #15 Feb 24 2024 11:04:13

%S 1,-1,3,0,0,0,-9,0,27,0,0,0,-324,0,1215,0,0,0,-18711,0,75816,0,0,0,

%T -1301265,0,5484996,0,0,0,-100048689,0,431943435,0,0,0,-8192222064,0,

%U 35942240565,0,0,0,-700434986472,0,3108770417700,0,0,0,-61805774132388,0,276711654879477

%N Expansion of x / Series_Reversion( x/(1 + 3*x - 6*x^2 - 8*x^3)^(1/3) ).

%C The cube root of F(x) = (1 + 3*x - 6*x^2 - 8*x^3) = (1 + x)*(1 - 2*x)*(1 + 4*x) is an integer series because F(x) == (1+x)^3 (mod 9).

%H Paul D. Hanna, <a href="/A370146/b370146.txt">Table of n, a(n) for n = 0..630</a>

%F G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.

%F (1) A(x) = x / Series_Reversion( x/(1 + 3*x - 6*x^2 - 8*x^3)^(1/3) ).

%F (2) A(x) = 1 / B(x/A(x)) where B(x) = 1 / A(x/B(x)) = (1 + 3*x - 6*x^2 - 8*x^3)^(1/3) equals the g.f. of A370145.

%F (3) A(x) = 1/C(27*x^6)^(1/3) + 3*x^2*C(27*x^6)^(1/3) - x, where C(x) = 1 + x*C(x)^2 = (1-sqrt(1-4*x))/(2*x) is the Catalan function (A000108).

%e G.f.: A(x) = = 1 - x + 3*x^2 - 9*x^6 + 27*x^8 - 324*x^12 + 1215*x^14 - 18711*x^18 + 75816*x^20 - 1301265*x^24 + 5484996*x^26 - 100048689*x^30 + ...

%e RELATED SERIES.

%e If A(x) = 1/B(x/A(x)) then B(x) = (1 + 3*x - 6*x^2 - 8*x^3)^(1/3) begins

%e B(x) = 1 + x - 3*x^2 + 3*x^3 - 12*x^4 + 30*x^5 - 102*x^6 + 318*x^7 - 1083*x^8 + 3657*x^9 - 12747*x^10 + 44715*x^11 + ... + A370145(n)*x^n + ...

%e A(x) = 1/D(x^6) + 3*x^2*D(x^6) - x, where

%e 1/D(x) = 1 - 9*x - 324*x^2 - 18711*x^3 - 1301265*x^4 - 100048689*x^5 - 8192222064*x^6 - 700434986472*x^7 + ...

%e and D(x) = ( (1-sqrt(1-108*x))/(54*x) )^(1/3) begins

%e D(x) = 1 + 9*x + 405*x^2 + 25272*x^3 + 1828332*x^4 + 143981145*x^5 + 11980746855*x^6 + 1036256805900*x^7 + ... + 3^n*A008931(n)*x^n + ...

%o (PARI) {a(n) = polcoeff( x/serreverse( x/(1 + 3*x - 6*x^2 - 8*x^3 +x*O(x^n))^(1/3) ), n)}

%o for(n=0,50,print1(a(n),", "))

%Y Cf. A370016, A370145, A008931, A000108.

%K sign

%O 0,3

%A _Paul D. Hanna_, Feb 23 2024