A097090 G.f. A(x) satisfies: A(A(x)) = x*(1+2*x)^2.
1, 2, -2, 6, -22, 80, -228, 18, 6694, -65604, 396804, -1336332, -2510244, 64799884, -302351144, -1410221598, 23754923526, 16833211660, -2598949277964, 14767224078420, 229529725999500, -3478598282993328, -13287810766972728, 667271251276705140, -1630867775606147844
Offset: 1
Keywords
Examples
G.f.: A(x) = x + 2*x^2 - 2*x^3 + 6*x^4 - 22*x^5 + 80*x^6 - 228*x^7 + 18*x^8 +... where A(A(x)) = x + 4*x^2 + 4*x^3. Illustrate Vladimir Kruchinin's formula by the triangular matrix T: 1; 2, 1; -2, 4, 1; 6, 0, 6, 1; -22, 4, 6, 8, 1; 80, -16, 2, 16, 10, 1; -228, 48, -6, 8, 30, 12, 1; 18, -12, 0, 0, 30, 48, 14, 1; 6694, -1460, 232, -32, 10, 76, 70, 16, 1; ... in which the g.f. of column k = A(x)^k and A(x) is the g.f. of this sequence. The matrix square, T^2, of the above triangle begins: 1; 4, 1; 4, 8, 1; 0, 24, 12, 1; 0, 32, 60, 16, 1; 0, 16, 160, 112, 20, 1; 0, 0, 240, 448, 180, 24, 1; 0, 0, 192, 1120, 960, 264, 28, 1; 0, 0, 64, 1792, 3360, 1760, 364, 32, 1; ... in which the g.f. column k = (x + 4*x^2 + 4*x^3)^k = sum(n>=k, binomial(2*k,n-k)*2^(n-k)*x^n).
Links
- Paul D. Hanna, Table of n, a(n) for n = 1..130
- Dmitry Kruchinin, Vladimir Kruchinin, Method for solving an iterative functional equation $A^{2^n}(x)=F(x)$, arXiv:1302.1986
Crossrefs
Cf. A107099.
Programs
-
Mathematica
T[n_, m_] := T[n, m] = If[n == m, 1, (1/2)*(Binomial[2 m, n - m]*2^(n - m) - Sum[T[n, i]*T[i, m], {i, m+1, n-1}])]; a[n_] := T[n, 1] // Numerator; Table[a[n], {n, 1, 25}] (* Jean-François Alcover, Oct 29 2015, after Vladimir Kruchinin *) -
PARI
{a(n)=local(A,B,F);F=x*(1+2*x+x*O(x^n))^2;A=F; for(i=0,n,B=serreverse(A);A=(A+subst(B,x,F))/2);polcoeff(A,n,x)} -
PARI
/* Vladimir Kruchinin's formula: a(n) = T(n,1) where: */ {T(n,k)=if(n==k,1,1/2*(binomial(2*k,n-k)*2^(n-k)-sum(j=k+1,n-1,T(n,j)*T(j,k))))} {a(n)=T(n,1)} /* Paul D. Hanna, Nov 10 2011 */
Formula
a(n)=T(n,1), T(n,m)=if n=m then 1 else 1/2*(binomial(2*m,n-m)*2^(n-m)-sum(i=m+1..n-1, T(n,i)*T(i,m))). [From Vladimir Kruchinin, Nov 10 2011]