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G.f. (1/x)*Series_Reversion( x*(1-x)*(3 - 2*x + sqrt(1+4*x-4*x^2))/4 ).
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%I #5 Mar 31 2022 09:36:28

%S 1,1,3,7,28,79,350,1075,5020,16180,78023,259417,1278340,4343642,

%T 21740636,75065787,380161308,1328887420,6792111260,23975385148,

%U 123448657904,439228736887,2275311657814,8148868193557,42427160829508

%N G.f. (1/x)*Series_Reversion( x*(1-x)*(3 - 2*x + sqrt(1+4*x-4*x^2))/4 ).

%C Essentially an unsigned version of A351771 (after dropping the initial term).

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

%F The g.f. A(x) satisfies:

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

%F (2) (A(x) - A(-x))/2 = x*(A(x)^2 + A(-x)^2)/2;

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

%F (4) 1 - x*(A(x) - A(-x))/2 = x/Series_Reversion( x - x*(C(x) + C(-x))/2 ), where C(x) = (1 - sqrt(1-4*x))/2 is the Catalan function (A000108);

%F (5a) (1/A(x) + 1/A(-x))/2 = ( 1 - x*(A(x) - A(-x))/2 )^2;

%F (5b) (1/A(x) - 1/A(-x))/2 = (-x)/(1 - 2*x*(A(x) - A(-x))/2);

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

%F (6b) (1/A(x)^2 - 1/A(-x)^2)/2 = -2*x*(1 - x*(A(x) - A(-x))/2)^2/( 1 - x*(A(x) - A(-x)) ).

%F Let B(x) = Series_Reversion( x*(1-x^2)/(1+x^2)^3 ), then

%F (7) A(x) = (1 - sqrt(1 - 4*x - 4*x*B(x)^2))/(2*x);

%F (8) A(x) - x*A(x)^2 = A(-x) + x*A(-x)^2 = 1 + B(x)^2;

%F (9) 1 - x*(A(x) - A(-x))/2 = 1/(1 + B(x)^2);

%F (10) 1/A(x) = 1/(1 + B(x)^2)^2 - x*(1 + B(x)^2)/(1 - B(x)^2);

%F (10a) (1/A(x) + 1/A(-x))/2 = 1/(1 + B(x)^2)^2;

%F (10b) (1/A(x) - 1/A(-x))/2 = (-x)*(1 + B(x)^2)/(1 - B(x)^2);

%F (11) 1/A(x)^2 = 1/(1 + B(x)^2)^3 - 2*x/(1 - B(x)^4);

%F (11a) (1/A(x)^2 + 1/A(-x)^2)/2 = 1/(1 + B(x)^2)^3;

%F (11b) (1/A(x)^2 - 1/A(-x)^2)/2 = -2*x/(1 - B(x)^4).

%e G.f.: A(x) = 1 + x + 3*x^2 + 7*x^3 + 28*x^4 + 79*x^5 + 350*x^6 + 1075*x^7 + 5020*x^8 + 16180*x^9 + 78023*x^10 + 259417*x^11 + ...

%e such that A(x) = (1/x)*Series_Reversion(x*G(x)) and A(x*G(x)) = 1/G(x),

%e where G(x) = (1-x)*(3 - 2*x + sqrt(1+4*x-4*x^2))/4, which starts

%e G(x) = 1 - x - x^2 + 3*x^3 - 8*x^4 + 26*x^5 - 92*x^6 + 344*x^7 - 1336*x^8 + 5336*x^9 - 21776*x^10 + ...

%e Let B(x) = Series_Reversion( x*(1-x^2)/(1+x^2)^3 ),

%e B(x) = x + 4*x^3 + 39*x^5 + 496*x^7 + 7180*x^9 + 112236*x^11 + 1846082*x^13 + 31485120*x^15 + ...,

%e then A(x) = 1 + x*A(x)^2 + B(x)^2, where

%e B(x)^2 = x^2 + 8*x^4 + 94*x^6 + 1304*x^8 + 19849*x^10 + 320600*x^12 + 5396108*x^14 + 93615864*x^16 + ...

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

%o polcoeff(A,n)}

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

%Y Cf. A351771, A352383.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Mar 29 2022