A119262 Number of B-trees of order infinity with n leaves, where a(n) = Sum_{k=1..floor(n/2)} a(k)*C(n-k-1,n-2*k) for n >= 2, with a(0)=0, a(1)=1.
0, 1, 1, 1, 2, 3, 5, 8, 14, 25, 46, 85, 158, 294, 548, 1022, 1908, 3567, 6683, 12556, 23669, 44781, 85046, 162122, 310157, 595322, 1146057, 2212004, 4278908, 8292738, 16097018, 31286456, 60873574, 118543329, 231009934, 450434739, 878687665
Offset: 0
Keywords
Examples
A(x) = x + x^2 + x^3 + 2*x^4 + 3*x^5 + 5*x^6 + 8*x^7 + 14*x^8 + ... Series form: A(x) = x + x^2/(1-x) + x^4/((1-x)*((1-x)-x^2)) + x^8/((1-x)*((1-x)-x^2)*((1-x)*((1-x)-x^2)-x^4)) + ... + x^(2^n)/D(n,x) + x^(2^(n+1))/[D(n,x)*(D(n,x)-x^(2^n))] + ... Terms also satisfy the series: x = x*(1-x) + x^2*(1-x^2)/(1+x) + x^3*(1-x^3)/(1+x)^2 + 2*x^4*(1-x^4)/(1+x)^3 + 3*x^5*(1-x^5)/(1+x)^4 + 5*x^6*(1-x^6)/(1+x)^5 + 8*x^7*(1-x^7)/(1+x)^6 + 14*x^8*(1-x^8)/(1+x)^7 + 25*x^9*(1-x^9)/(1+x)^8 + ... + a(n)*x^n*(1-x^n)/(1+x)^(n-1) + ... From _Gus Wiseman_, Oct 07 2018: (Start) The a(1) = 1 through a(7) = 8 balanced series-reduced rooted plane trees: o (oo) (ooo) (oooo) (ooooo) (oooooo) (ooooooo) ((oo)(oo)) ((oo)(ooo)) ((oo)(oooo)) ((oo)(ooooo)) ((ooo)(oo)) ((ooo)(ooo)) ((ooo)(oooo)) ((oooo)(oo)) ((oooo)(ooo)) ((oo)(oo)(oo)) ((ooooo)(oo)) ((oo)(oo)(ooo)) ((oo)(ooo)(oo)) ((ooo)(oo)(oo)) (End)
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..1000
- Eric Weisstein's World of Mathematics, B-Tree.
Crossrefs
Programs
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Mathematica
nn=38;f[x_]:=Sum[a[n]x^n,{n,0,nn}];a[0]=0;sol=SolveAlways[0==Series[f[x]-x-f[x^2/(1-x)],{x,0,nn}],x];Table[a[n],{n,0,nn}]/.sol (* Geoffrey Critzer, Mar 28 2013 *)
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PARI
a(n)=if(n==0,0,if(n==1,1,sum(k=1,n\2,a(k)*binomial(n-k-1,n-2*k)))) for(n=1, 20, print1(a(n), ", "))
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PARI
/* From: A(x) = x + A(x^2/(1-x)) */ {a(n)=local(A=x);for(i=1,n,A=x+subst(A,x,x^2/(1-x+x*O(x^n))));polcoeff(A,n)} for(n=1, 20, print1(a(n), ", "))
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PARI
/* From: x = Sum_{n>=1} a(n)*x^n*(1-x^n)/(1+x)^(n-1) */ a(n)=if(n==1, 1, -polcoeff(sum(k=1, n-1, a(k)*x^k*(1-x^k)/(1+x+x*O(x^n))^(k-1)), n)) for(n=1, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Jul 31 2013
Formula
G.f. A(x) satisfies: A(x) = x + A(x^2/(1-x)).
G.f.: Sum_{n>=0} x^(2^n)/D(n,x) where D(0,x)=1, D(n+1,x) = D(n,x)*[D(n,x) - x^(2^n)].
G.f.: x = Sum_{n>=1} a(n) * x^n * (1-x^n) / (1+x)^(n-1). - Paul D. Hanna, Jul 31 2013
Conjecture: Let M_n be an n X n matrix whose elements are m_ij = 0 for i < j - 1, m_ij = -1 for i = j - 1, and m_ij = binomial(i - j, n - i) otherwise. Then a(n + 1) = Det(M_n). - Benedict W. J. Irwin, Apr 19 2017
Comments