A182399 G.f. A(x) satisfies: A(A(x)) - A(A(x))^2 = x + x^2.
1, 1, 1, 3, 7, 21, 61, 187, 583, 1837, 5885, 19027, 62167, 204917, 680621, 2275211, 7648519, 25852573, 87812093, 299349795, 1023570647, 3515918501, 12140103149, 41894710427, 143835281351, 501071173901, 1808088546557, 6212411239539, 17720665594455
Offset: 1
Keywords
Examples
G.f.: A(x) = x + x^2 + x^3 + 3*x^4 + 7*x^5 + 21*x^6 + 61*x^7 + 187*x^8 +... Related expansions: A(A(x)) = x + 2*x^2 + 4*x^3 + 12*x^4 + 40*x^5 + 144*x^6 + 544*x^7 + 2128*x^8 +... A(A(x))^2 = x^2 + 4*x^3 + 12*x^4 + 40*x^5 + 144*x^6 + 544*x^7 + 2128*x^8 +... where A(A(x)) - A(A(x))^2 = x + x^2. Let C(x) satisfy C(x-x^2) = x, where C(x) begins: C(x) = x + x^2 + 2*x^3 + 5*x^4 + 14*x^5 + 42*x^6 +...+ A000108(n-1)*x^n +... then A(-C(-x)) = x + x^3 + 4*x^5 + 21*x^7 + 122*x^9 + 758*x^11 + 4958*x^13 +...+ (-1)^(n-1)*A179270(2*n-1)*x^(2*n-1) +...
Links
- Paul D. Hanna, Table of n, a(n) for n = 1..256
Programs
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Maxima
T(n, m):= if n=m then 1 else ((sum((binomial(k+m,n-k-m)*binomial(2*k+m-1,k+m-1))/(k+m),k,0,n-m))*m -sum(T(n, i) *T(i, m), i, m+1, n-1))/2; makelist(T(n, 1), n, 1, 10); /* Vladimir Kruchinin, Apr 28 2012 */
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PARI
{a(n)=local(A=x+x^2,G);for(i=1,n,G=subst(A,x,A+x*O(x^n));A=A+(x+x^2-G+G^2)/2);polcoeff(A,n)} for(n=1,30,print1(a(n),", ")) -
PARI
/* Faster vectorized version: */ {MM=100;A=[1];B=x;C=(1-sqrt(1-4*(x+x^2+x*O(x^MM))))/2; for(n=1,oo,A=concat(A,0);B=x*Ser(A); A[n]=Vec((B+subst(C+x*O(x^n),x,serreverse(B)))/2)[n]; print1(A[n],", "))} -
PARI
/* PARI/GP Version of Vladimir Kruchinin's formula: */ {T(n, m)=if(n==m,1, if(n>m, (sum(k=0,n-m,(binomial(k+m,n-k-m)*binomial(2*k+m-1,k+m-1))/(k+m))*m - sum(i=m+1,n-1,T(n, i) *T(i, m)))/2 ))} {a(n)=T(n,1)}
Formula
G.f. satisfies: A(-A(-x)) = x.
G.f. satisfies: A(A(x)) = (1 - sqrt(1-4*(x+x^2)))/2 is the g.f. of A025227; thus, A(A(x)) = C(x+x^2) where C(x-x^2) = x.
G.f. satisfies: A(-C(-x)) = -I*G(I*x) where C(x-x^2) = x and G(x) is the g.f. of A179270 such that the inverse of function G(x) + I*G(x)^2 equals the complex conjugate: G(x) - I*G(x)^2.
a(n) = T(n,1), with T(n, m) = (sum((binomial(k+m,n-k-m)*binomial(2*k+m-1,k+m-1))/(k+m),k,0,n-m)*m -sum(T(n, i) *T(i, m), i, m+1, n-1))/2, n>m, T(n,n) = 1. - Vladimir Kruchinin, Apr 28 2012
Comments