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A370145 Expansion of ( (1 + x)*(1 - 2*x)*(1 + 4*x) )^(1/3).

%I A370145 #17 Oct 04 2024 16:52:37
%S A370145 1,1,-3,3,-12,30,-102,318,-1083,3657,-12747,44715,-159222,571332,
%T A370145 -2068608,7538664,-27646374,101915850,-377496030,1404077790,
%U A370145 -5242135728,19637862132,-73793090676,278068062756,-1050503580534,3977985415746,-15096209345958,57403753019238,-218683959367908
%N A370145 Expansion of ( (1 + x)*(1 - 2*x)*(1 + 4*x) )^(1/3).
%C A370145 The cube root of F(x) = (1 + x)*(1 - 2*x)*(1 + 4*x) = (1 + 3*x - 6*x^2 - 8*x^3) is an integer series because F(x) == (1+x)^3 (mod 9).
%H A370145 Paul D. Hanna, <a href="/A370145/b370145.txt">Table of n, a(n) for n = 0..630</a>
%H A370145 N. Heninger, E. M. Rains and N. J. A. Sloane, <a href="http://arXiv.org/abs/math.NT/0509316">On the Integrality of n-th Roots of Generating Functions</a>, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.
%F A370145 G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
%F A370145 (1) A(x)^3 = (1 + x)*(1 - 2*x)*(1 + 4*x) = (1 + 3*x - 6*x^2 - 8*x^3).
%F A370145 (2) Product_{n>=1} A( 2^(n-1)*x^n )^3 = Sum_{n>=0} 2^(n*(n-1)/2)*(1 + 2^(2*n+1))/3 * x^(n*(n+1)/2), which is the g.f. of A370015.
%F A370145 (3) A(x) = 1/B(x/A(x)) where B(x) = 1/A(x/B(x)) = x/Series_Reversion(x/A(x)) equals the g.f. of A370146.
%F A370145 a(n) ~ (-1)^(n+1) * 2^(2*n-1) / (3^(1/3) * Gamma(2/3) * n^(4/3)). - _Vaclav Kotesovec_, Feb 23 2024
%e A370145 G.f.: A(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 - 159222*x^12 + ...
%e A370145 where A(x)^3 = (1 + 3*x - 6*x^2 - 8*x^3).
%e A370145 RELATED SERIES.
%e A370145 The following infinite product equals the g.f. of A370015:
%e A370145 A(x)^3 * A(2*x^2)^3 * A(4*x^3)^3 * A(8*x^4)^3 * ... * A(2^(n-1)*x^n)^3 * ... = 1 + 3*x + 22*x^3 + 344*x^6 + 10944*x^10 + 699392*x^15 + 89489408*x^21 + 22907191296*x^28 + ... + 2^(n*(n-1)/2)*(1 + 2^(2*n+1))/3 * x^(n*(n+1)/2) + ... by the Jacobi triple product identity.
%e A370145 If A(x) = 1/B(x/A(x)) then B(x) = 1/A(x/B(x)) begins
%e A370145 B(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 + 431943435*x^32 - 8192222064*x^36 + ... + A370146(n)*x^n + ...
%t A370145 CoefficientList[Series[Surd[(1+x)(1-2x)(1+4x),3],{x,0,30}],x] (* _Harvey P. Dale_, Oct 04 2024 *)
%o A370145 (PARI) {a(n) = polcoeff( (1 + 3*x - 6*x^2 - 8*x^3 +x*O(x^n))^(1/3), n)}
%o A370145 for(n=0,40,print1(a(n),", "))
%Y A370145 Cf. A370015, A370016, A370146, A370147, A370149.
%K A370145 sign
%O A370145 0,3
%A A370145 _Paul D. Hanna_, Feb 23 2024