A351771 - OEIS

%I #14 Mar 16 2022 16:47:02

%S 1,1,-1,3,-7,28,-79,350,-1075,5020,-16180,78023,-259417,1278340,

%T -4343642,21740636,-75065787,380161308,-1328887420,6792111260,

%U -23975385148,123448657904,-439228736887,2275311657814,-8148868193557,42427160829508,-152792221834364

%N Given g.f. A(x), the even bisections of both A(x) and A(x)^2 are equal, and the odd bisections of both A(x)^2 and A(x)^3 are equal (after the initial terms).

%C a(2*n+1) = A352383(n) for n >= 0.

%F G.f. A(x) satisfies:

%F (1a) [x^(2*n)] A(x) = [x^(2*n)] A(x)^2 for n >= 1.

%F (1b) [x^(2*n+1)] A(x)^2 = [x^(2*n+1)] A(x)^3 for n >= 1.

%F (1c) [x^(2*n+1)] A(x)^3 = [x^(2*n+1)] A(x)^4 for n >= 1.

%F (2) (A(x) - A(-x))/2 = x/(A(x)*A(-x)).

%F (3a) (A(x)^2 + A(-x)^2)/2 = (A(x) + A(-x))/2.

%F (3b) (A(x)^2 - A(-x)^2)/2 = 2*x + (A(x) - A(-x))^3/2.

%F (4) A(x)^2 = 2*x + (A(x) + A(-x))/2 + (A(x) - A(-x))^3/2.

%F (5a) A(x)^2 = 2*x + (A(x)^2 + A(-x)^2)/2 + (A(x) - A(-x))^3/2.

%F (5b) A(x)^3 = 3*x + (A(x)^3 + A(-x)^3)/2 + (A(x) - A(-x))^3/2.

%F (5c) A(x)^4 = 4*x + (A(x)^4 + A(-x)^4)/2 + (A(x) - A(-x))^3/2.

%F (6) A(x)^4 - A(x)^3 = x + x*(A(x) - A(-x)).

%F (7) A(-x) = (A(x)^2 + sqrt(A(x)^4 - 8*x*A(x)))/(2*A(x)).

%F (8) (A(x) - A(-x))^3/2 = 4*x*F(x^2), where F(x) = Series_Reversion( x*(1+x)^3/(1+2*x)^6 ).

%F (9) A(x)^2 - A(x) = Series_Reversion( x - x*(C(x) + C(-x))/2 ), where C(x) = x + C(x)^2 is the Catalan power series (A000108).

%F (10) A(x) = 1 + Series_Reversion( x*(1+x)*(3 + 2*x + sqrt(1-4*x-4*x^2))/4 ).

%F (11) 0 = 2*x^2 + A(x)*(1 - A(x))*(1 + 2*A(x))*x + A(x)^4*(1 - A(x))^2.

%e G.f. A(x) = 1 + x - 1*x^2 + 3*x^3 - 7*x^4 + 28*x^5 - 79*x^6 + 350*x^7 - 1075*x^8 + 5020*x^9 - 16180*x^10 + 78023*x^11 + ...

%e Compare A(x) with the coefficients in the following series expansions:

%e A(x)^2 = 1 + 2*x - 1*x^2 + 4*x^3 - 7*x^4 + 36*x^5 - 79*x^6 + 444*x^7 - 1075*x^8 + 6324*x^9 - 16180*x^10 + 97872*x^11 + ...

%e A(x)^3 = 1 + 3*x + 0*x^2 + 4*x^3 - 3*x^4 + 36*x^5 - 40*x^6 + 444*x^7 - 579*x^8 + 6324*x^9 - 9000*x^10 + 97872*x^11 + ...

%e A(x)^4 = 1 + 4*x + 2*x^2 + 4*x^3 + 3*x^4 + 36*x^5 + 16*x^6 + 444*x^7 + 121*x^8 + 6324*x^9 + 1040*x^10 + 97872*x^11 + ...

%e which illustrate the properties that the coefficients of x^k for even k in A(x) and A(x)^2 are equal, and that the coefficients of x^k for odd k > 1 in A(x)^2, A(x)^3, and A(x)^4 are equal.

%e Related series.

%e (1) Notice that A(x)^2 - A(x) forms an odd function:

%e A(x)^2 - A(x) = x + x^3 + 8*x^5 + 94*x^7 + 1304*x^9 + 19849*x^11 + 320600*x^13 + 5396108*x^15 + ...

%e such that the series reversion begins

%e Series_Reversion( A(x)^2 - A(x) ) = x - x^3 - 5*x^5 - 42*x^7 - 429*x^9 - 4862*x^11 - 58786*x^13 - ...

%e which equals x - x*(C(x) + C(-x))/2, where C(x) = x + C(x)^2:

%e C(x) = x + x^2 + 2*x^3 + 5*x^4 + 14*x^5 + 42*x^6 + 132*x^7 + 429*x^8 + 1430*x^9 + 4862*x^10 + ...

%e and is the Catalan power series C(x) = (1 - sqrt(1-4*x))/2.

%e (2) Also, the coefficients in the following series form a bisection of A(x):

%e (A(x)^4 - A(x)^3 - x)/2 = x^2 + 3*x^4 + 28*x^6 + 350*x^8 + 5020*x^10 + 78023*x^12 + 1278340*x^14 + ... + A352383(n)*x^(2*n+2) + ...

%e (3) Further, a series bisection of A(x)^2, A(x)^3, and A(x)^4 is

%e (A(x) - A(-x))^3/2 = 4*x^3 + 36*x^5 + 444*x^7 + 6324*x^9 + 97872*x^11 + 1598940*x^13 + 27136744*x^15 + ... + 4*A352384(n)*x^(2*n+3) + ...

%e which is equal to 4*x*F(x^2), where F( x*(1+x)^3/(1+2*x)^6 ) = x, and

%e F(x) = x + 9*x^2 + 111*x^3 + 1581*x^4 + 24468*x^5 + 399735*x^6 + 6784186*x^7 + ... + A352384(n)*x^(n+1) + ...

%e with

%e (F(x)/x)^(1/3) = 1 + 3*x + 28*x^2 + 350*x^3 + 5020*x^4 + 78023*x^5 + 1278340*x^6 + ... + A352383(n)*x^n + ...

%e (4) The above observations lead to the composition of functions

%e Series_Reversion(A(x) - 1) = [x - x*(C(x) + C(-x))/2] o (x + x^2)

%e which is equivalent to

%e Series_Reversion(A(x) - 1) = x*(1+x)*(3 + 2*x + sqrt(1-4*x-4*x^2))/4.

%t CoefficientList[1 + InverseSeries[Series[x*(1 + x)*(3 + 2*x + Sqrt[1 - 4*x - 4*x^2])/4, {x, 0, 30}], x], x] (* _Vaclav Kotesovec_, Mar 15 2022 *)

%o (PARI) /* Using Series Reversion */

%o {a(n) = my(A = 1 + serreverse( x*(1+x)*(3 + 2*x + sqrt(1-4*x-4*x^2 +x^2*O(x^n)))/4)); polcoeff(A,n)}

%o for(n=0,30,print1(a(n),", "))

%o (PARI) /* From [x^(2*n)] A(x) - A(x)^2 = 0 and [x^(2*n+1)] A(x)^2 - A(x)^3 = 0  */

%o {a(n) = my(A = 1 + x +x^2*O(x^n));

%o for(k=2,n, if(k%2==0,

%o A = A + x^k*polcoeff(A^1 - A^2,k),

%o A = A + x^k*polcoeff(A^2 - A^3,k)));

%o polcoeff(A,n)}

%o for(n=0,30,print1(a(n),", "))

%Y Cf. A352383, A352384, A000108.

%K sign

%O 0,4

%A _Paul D. Hanna_, Mar 14 2022