A199874 G.f. satisfies A(x) = (1 + x*A(x)^2)*(1 + x^2*A(x)^2).
1, 1, 3, 10, 37, 147, 611, 2625, 11564, 51953, 237123, 1096420, 5125063, 24178427, 114974387, 550511901, 2651896733, 12843003108, 62494595022, 305400429548, 1498184696271, 7375179807191, 36421312544431, 180383163330765, 895756907248150, 4459095182031675, 22247684478181317
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 3*x^2 + 10*x^3 + 37*x^4 + 147*x^5 + 611*x^6 +... where A( x/(1+x^2) - x^2 ) = (1+x^2)/(1-x-x^3). Related expansions: A(x)^2 = 1 + 2*x + 7*x^2 + 26*x^3 + 103*x^4 + 428*x^5 + 1838*x^6 +... A(x)^4 = 1 + 4*x + 18*x^2 + 80*x^3 + 359*x^4 + 1632*x^5 + 7506*x^6 +... where A(x) = 1 + x*(1+x)*A(x)^2 + x^3*A(x)^4. The logarithm of the g.f. equals the series: log(A(x)) = (1 + x)*x*A(x) + (1 + 2^2*x + x^2)*x^2*A(x)^2/2 + (1 + 3^2*x + 3^2*x^2 + x^3)*x^3*A(x)^3/3 + (1 + 4^2*x + 6^2*x^2 + 4^2*x^3 + x^4)*x^4*A(x)^4/4 + (1 + 5^2*x + 10^2*x^2 + 10^2*x^3 + 5^2*x^4 + x^5)*x^5*A(x)^5/5 + (1 + 6^2*x + 15^2*x^2 + 20^2*x^3 + 15^2*x^4 + 6^2*x^5 + x^6)*x^6*A(x)^6/6 +... more explicitly, log(A(x)) = x + 5*x^2/2 + 22*x^3/3 + 101*x^4/4 + 481*x^5/5 + 2330*x^6/6 +...
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*AGF^2)*(1+x^2*AGF^2)-AGF,x,j]==0,koef][[1]]; aa[[j]]=koef/.sol[[1]],{j,2,nmax}];Flatten[{1,aa}] (* Vaclav Kotesovec, Aug 18 2013 *) -
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)*(x*A+x*O(x^n))^m/m))); polcoeff(A, n, x)} -
PARI
{a(n)=polcoeff((1/x)*serreverse(x/(1+x^2+x*O(x^n))-x^2),n)} -
PARI
{a(n)=polcoeff(((1+x^2)/(1-x-x^3+x*O(x^n)))^(n+1)/(n+1), n)} -
PARI
{a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (1-x)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j)*x^m*A^m/m))); polcoeff(A, n, x)}
Formula
G.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion( x/(1+x^2) - x^2 ).
(2) A( x*(1-x-x^3)/(1+x^2) ) = (1+x^2)/(1-x-x^3).
(3) a(n) = [x^n] ((1+x^2)/(1-x-x^3))^(n+1) / (n+1).
(4) A(x) = exp( Sum_{n>=1} (Sum_{k=0..n} C(n,k)^2 * x^k) * x^n*A(x)^n/n ).
(5) A(x) = exp( Sum_{n>=1} (1-x)^(2*n+1)*(Sum_{k>=0} C(n+k,k)^2*x^k) * x^n*A(x)^n/n ).
Recurrence: 31*(n-1)*n*(n+1)*(85396*n^4 - 902916*n^3 + 3471647*n^2 - 5767203*n + 3503250)*a(n) = 2*(n-1)*n*(6319304*n^5 - 69975436*n^4 + 290875210*n^3 - 559740413*n^2 + 484175751*n - 138985722)*a(n-1) + 2*(n-1)*(2903464*n^6 - 36506072*n^5 + 179801738*n^4 - 439606930*n^3 + 553204983*n^2 - 328951215*n + 67014378)*a(n-2) + 2*(2*n - 5)*(1964108*n^6 - 24695284*n^5 + 123902749*n^4 - 317652203*n^3 + 438313617*n^2 - 307740825*n + 85471038)*a(n-3) - 32*(n-3)*(2*n - 7)*(85396*n^5 - 860218*n^4 + 3249611*n^3 - 5747414*n^2 + 4753791*n - 1471338)*a(n-4) + 8*(n-4)*(n-3)*(2*n - 9)*(85396*n^4 - 561332*n^3 + 1275275*n^2 - 1191073*n + 390174)*a(n-5). - Vaclav Kotesovec, Aug 18 2013
a(n) ~ c*d^n/n^(3/2), where d=5.28245622984... is the root of the equation -16 + 64*d - 92*d^2 - 68*d^3 - 148*d^4 + 31*d^5 = 0 and c = 0.49559010377906722118329... - Vaclav Kotesovec, Aug 18 2013
a(n) = Sum_{k=0..floor(n/2)} binomial(2*n-2*k+1,k) * binomial(2*n-2*k+1,n-2*k) / (2*n-2*k+1). - Seiichi Manyama, Jul 18 2023