A246861 - OEIS

%I #23 Sep 17 2023 21:29:49

%S 1,1,1,2,7,21,54,141,407,1231,3691,10990,33144,101674,314679,977289,

%T 3047527,9557503,30133759,95390622,302960929,965282651,3085146472,

%U 9888455045,31774215928,102334358736,330298415136,1068242904256,3461372341327,11235251353747,36527859658661

%N G.f. satisfies: A(x) = Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * (x*A(x))^(2*k).

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

%F (1) A(x) = Sum_{n>=0} x^n * [Sum_{k>=0} C(n+k,k)^2 *(x*A(x))^(2*k)] * (1 - x^2*A(x)^2)^(2*n+1).

%F (2) A(x) = Sum_{n>=0} x^(3*n) * A(x)^(2*n) * [Sum_{k>=0} C(n+k,k)^2 * x^k].

%F (3) A(x) = Sum_{n>=0} x^(3*n) * A(x)^(2*n) * [Sum_{k=0..n} C(n,k)^2 * x^k] /(1-x)^(2n+1).

%F (4) A(x) = exp( Sum_{n>=1} (x^n/n) * Sum_{k=0..n} C(2*n,2*k) * (x*A(x))^(2*k) ).

%F (5) A(x) = exp( Sum_{n>=1} (x^n/n) * ((1 + x*A(x))^(2*n) + (1 - x*A(x))^(2*n))/2 ).

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

%F (7) A(x) = x*A(x) + x^3*A(x)^3 + sqrt(1 + 4*x^4*A(x)^4).

%F (8) A(x) = 1/x * Series_Reversion( x / (x + x^3 + sqrt(1 + 4*x^4)) ).

%e G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 7*x^4 + 21*x^5 + 54*x^6 + 141*x^7 + ...

%e where the g.f. A = A(x) equals the binomial series:

%e A(x) = 1 + x*(1 + x^2*A^2) + x^2*(1 + 2^2*x^2*A^2 + x^4*A^4)

%e + x^3*(1 + 3^2*x^2*A^2 + 3^2*x^4*A^4 + x^6*A^6)

%e + x^4*(1 + 4^2*x^2*A^2 + 6^2*x^4*A^4 + 4^2*x^6*A^6 + x^8*A^8)

%e + x^5*(1 + 5^2*x^2*A^2 + 10^2*x^4*A^4 + 10^2*x^6*A^6 + 5^2*x^8*A^8 + x^10*A^10) + ...

%e Let A = g.f. A(x), then the g.f. satisfies:

%e log(A(x)) = x*(1 + x^2*A^2) + x^2*(1 + 6*x^2*A^2 + x^4*A^4)/2

%e + x^3*(1 + 15*x^2*A^2 + 15*x^4*A^4 + x^6*A^6)/3

%e + x^4*(1 + 28*x^2*A^2 + 70*x^4*A^4 + 28*x^6*A^6 + x^8*A^8)/4

%e + x^5*(1 + 45*x^2*A^2 + 210*x^4*A^4 + 210*x^6*A^6 + 45*x^8*A^8 + x^10*A^10)/5 + ...

%e RELATED SERIES:

%e A(x)^2 = 1 + 2*x + 3*x^2 + 6*x^3 + 19*x^4 + 60*x^5 + 168*x^6 + ...

%e A(x)^3 = 1 + 3*x + 6*x^2 + 13*x^3 + 39*x^4 + 126*x^5 + 376*x^6 + ...

%e A(x)^4 = 1 + 4*x + 10*x^2 + 24*x^3 + 71*x^4 + 232*x^5 + 726*x^6 + ...

%e A(x)^6 = 1 + 6*x + 21*x^2 + 62*x^3 + 192*x^4 + 642*x^5 + 2145*x^6 + ...

%e where 1 = (1-x)^2*A(x)^2 - 2*x^3*(1+x)*A(x)^4 + x^6*A(x)^6.

%e Let G(x) = x/Series_Reversion(x*A(x)), then G(x*A(x)) = A(x), where

%e G(x) = x + x^3 + 1 + 2*x^4 - 2*x^8 + 4*x^12 - 10*x^16 + 28*x^20 - 84*x^24 + ...

%e G(x) = x + x^3 + sqrt(1 + 4*x^4).

%o (PARI) /* By definition: */

%o {a(n)=local(A=1); for(i=1,n, A=sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2*(x*A)^(2*k)) +x*O(x^n))); polcoeff(A, n)}

%o for(n=0, 35, print1(a(n), ", "))

%o (PARI) /* (1) From a binomial series identity: */

%o {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, x^m * sum(k=0, n, binomial(m+k, k)^2*(x*A +x*O(x^n))^(2*k)) * (1-x^2*A^2)^(2*m+1) )); polcoeff(A, n)}

%o for(n=0, 35, print1(a(n), ", "))

%o (PARI) /* (2) From a binomial series identity: */

%o {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n\2, x^(3*m)*(A +x*O(x^n))^(2*m)*sum(k=0, n, binomial(m+k, k)^2*x^k))); polcoeff(A, n)}

%o for(n=0, 35, print1(a(n), ", "))

%o (PARI) /* (3) From a binomial series identity: */

%o {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n\2, x^(3*m)*A^(2*m) * sum(k=0, m, binomial(m, k)^2*x^k) / (1-x +x*O(x^n))^(2*m+1) )); polcoeff(A, n)}

%o for(n=0, 35, print1(a(n), ", "))

%o (PARI) /* (4) From exponential series formula: */

%o {a(n)=local(A=1); for(i=1,n, A=exp(sum(m=1, n, ((1+x*A)^(2*m) + (1-x*A)^(2*m))/2 * x^m/m) +x*O(x^n))); polcoeff(A, n)}

%o for(n=0, 35, print1(a(n), ", "))

%o (PARI) /* (6) From functional equation: */

%o {a(n)=local(A=1); for(i=1,n, A =  1 / sqrt((1 - x*(1 - x*A)^2) * (1 - x*(1 + x*A)^2)) +x*O(x^n)); polcoeff(A, n)}

%o for(n=0, 40, print1(a(n), ", "))

%o (PARI) /* (7) From functional equation: */

%o {a(n)=local(A=1); for(i=1,n, A = x*A + x^3*A^3 + sqrt(1 + 4*x^4*A^4 +x*O(x^n)) ); polcoeff(A, n)}

%o for(n=0, 40, print1(a(n), ", "))

%o (PARI) /* (8) From explicit formula: */

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

%o for(n=0, 40, print1(a(n), ", "))

%Y Cf. A246840, A181665.

%K nonn

%O 0,4

%A _Paul D. Hanna_, Sep 05 2014