A215576 G.f. satisfies A(x) = (1 + x^2)*(1 + x*A(x)^2).
1, 1, 3, 8, 24, 80, 278, 997, 3670, 13782, 52588, 203314, 794726, 3135540, 12470444, 49942305, 201233170, 815205699, 3318291966, 13565162636, 55669063762, 229257178198, 947142023262, 3924380904498, 16303716754884, 67899954924360, 283425070356740, 1185551594834910
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 3*x^2 + 8*x^3 + 24*x^4 + 80*x^5 + 278*x^6 + 997*x^7 +... Related expansions: A(x)^2 = 1 + 2*x + 7*x^2 + 22*x^3 + 73*x^4 + 256*x^5 + 924*x^6 + 3414*x^7 +... where A(x) = 1+x^2 + x*(1+x^2)*A(x)^2. The logarithm of the g.f. A = A(x) equals the series: log(A(x)) = (1 + x/A^2)*A*x + (1 + 2^2*x/A^2 + x^2/A^4)*A^2*x^2/2 + (1 + 3^2*x/A^2 + 3^2*x^2/A^4 + x^3/A^6)*A^3*x^3/3 + (1 + 4^2*x/A^2 + 6^2*x^2/A^4 + 4^2*x^3/A^6 + x^4/A^8)*A^4*x^4/4 + (1 + 5^2*x/A^2 + 10^2*x^2/A^4 + 10^2*x^3/A^6 + 5^2*x^4/A^8 + x^5/A^10)*A^5*x^5/5 + (1 + 6^2*x/A^2 + 15^2*x^2/A^4 + 20^2*x^3/A^6 + 15^2*x^4/A^8 + 6^2*x^5/A^10 + x^6/A^12)*A^6*x^6/6 +... more explicitly, log(A(x)) = x + 5*x^2/2 + 16*x^3/3 + 57*x^4/4 + 231*x^5/5 + 938*x^6/6 + 3830*x^7/7 + 15833*x^8/8 +...
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..1000
Crossrefs
Programs
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Mathematica
nmax=20;aa=ConstantArray[0,nmax]; aa[[1]]=1;Do[AGF=1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j; sol=Solve[Coefficient[(1+x^2)*(1+x*AGF^2)-AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}];Flatten[{1,aa}] (* Vaclav Kotesovec, Aug 19 2013 *) CoefficientList[Series[(1 - Sqrt[1 - 4*x*(1 + x^2)^2]) / (2*x*(1 + x^2)), {x, 0, 30}], x] (* Vaclav Kotesovec, Oct 11 2018 *) Table[Sum[Binomial[2*n - 4*i + 1, i] * Binomial[2*n - 4*i + 1, n - 2*i]/(2*n - 4*i + 1), {i, 0, Floor[n/2]}], {n, 0, 30}] (* Vaclav Kotesovec, Oct 11 2018, after Vladimir Kruchinin *)
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PARI
{a(n)=polcoeff((1 - sqrt(1 - 4*x*(1+x^2 +x*O(x^n))^2)) / (2*x*(1+x^2 +x*O(x^n))),n)} for(n=0,31,print1(a(n),", "))
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PARI
{a(n)=local(A=1+x); for(i=1, n, A=(1+x*A^2)*(1+x^2)+x*O(x^n)); polcoeff(A, n)}
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PARI
{a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*x^j/A^(2*j))*(x*A+x*O(x^n))^m/m))); polcoeff(A, n, x)}
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PARI
{a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (1-x/A^2)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j/A^(2*j))*x^m*A^m/m))); polcoeff(A, n, x)}
Formula
G.f. satisfies:
(1) A(x) = (1 - sqrt(1 - 4*x*(1+x^2)^2)) / (2*x*(1+x^2)).
(2) A(x) = exp( Sum_{n>=1} (x^n/n) * A(x)^n * (Sum_{k=0..n} C(n,k)^2 * x^k / A(x)^(2*k)) ).
(3) A(x) = exp( Sum_{n>=1} (1-x/A(x)^2)^(2*n+1) * (Sum_{k>=0} C(n+k,k)^2*x^k/A(x)^(2*k)) * x^n*A(x)^n/n ).
(4) A(x) = x / Series_Reversion( x*G(x) ) where G(x) is the g.f. of A200717.
(5) A(x) = G(x/A(x)) where G(x) = A(x*G(x)) is the g.f. of A200717.
Recurrence: (n+1)*a(n) = 2*(2*n-1)*a(n-1) - (n+1)*a(n-2) + 6*(2*n-5)*a(n-3) + 6*(2*n-9)*a(n-5) + 2*(2*n-13)*a(n-7). - Vaclav Kotesovec, Aug 19 2013
a(n) ~ c*d^n/n^(3/2), where d = 4.41997678... is the root of the equation -4-8*d^2-4*d^4+d^5=0 and c = sqrt(d*(8 + 16*d^2 + 8*d^4 + 3*d^5 + d^7) / (Pi*(1 + d^2)^3))/4 = 0.648259186485429075561822659694489853... - Vaclav Kotesovec, Aug 19 2013, updated Oct 11 2018
a(n) = Sum_{i=0..floor(n/2)} C(2*n-4*i+1,i)*C(2*n-4*i+1,n-2*i)/(2*n-4*i+1). - Vladimir Kruchinin, Oct 11 2018
Comments