Search: id:A088218
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%I A088218
%S A088218 1,1,3,10,35,126,462,1716,6435,24310,92378,352716,1352078,5200300,
%T A088218 20058300,77558760,300540195,1166803110,4537567650,17672631900,
%U A088218 68923264410,269128937220,1052049481860,4116715363800,16123801841550
%N A088218 Total number of leaves in all rooted ordered trees with n edges.
%C A088218 Number of ordered partitions of n into n parts, allowing zeros (cf. A097070) 
               is binomial(2*n-1,n) = a(n) = essentially A001700. - Vladeta Jovovic 
               (vladeta(AT)eunet.rs), Sep 15 2004
%C A088218 a(n) = A110556(n)*(-1)^n, central terms in triangle A110555. - Reinhard 
               Zumkeller (reinhard.zumkeller(AT)gmail.com), Jul 27 2005
%C A088218 Hankel transform is A000027; example: Det([1,1,3,10;1,3,10,35;3,10,35,
               126;10,35,126,462])=4 . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), 
               Apr 13 2007
%C A088218 a(n) is the number of functions f:[n]->[n] such that for all x,y in [n] 
               if x<y then f(x)<=f(y). So 2*a(n)-n=A045992(n) [From Geoffrey Critzer 
               (critzer.geoffrey(AT)usd443.org), Apr 02 2009]
%C A088218 Hankel transform of the aeration of this sequence is A000027 doubled: 
               1,1,2,2,3,3,... [From Paul Barry (pbarry(AT)wit.ie), Sep 26 2009]
%D A088218 Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal 
               Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
%F A088218 a(n)=(0^n+C(2n, n))/2. - Paul Barry (pbarry(AT)wit.ie), May 21 2004
%F A088218 a(n) is the coefficient of x^n in 1/(1-x)^n and also the sum of the first 
               n coefficients of 1/(1-x)^n. Given B(x) with the property that the 
               coefficient of x^n in B(x)^n equals the sum of the first n coefficients 
               of B(x)^n, then B(x)=B(0)/(1-x).
%F A088218 G.f.: 1/(2-C(x)) where C(x) is g.f. for Catalan numbers A000108.
%F A088218 G.f.: (1+1/sqrt(1-4x))/2. a(n)=binomial(2n-1,n).
%F A088218 a(n)=sum{k=0..n, binomial(2n, k)cos((n-k)*pi)}; a(n)=sum{k=0..n, binomial(n, 
               (n-k)/2)(1+(-1)^(n-k))cos(k*pi/2)/2} (with interpolated zeros); a(n)=sum{k=0..floor(n/
               2), binomial(n, k)cos((n-2k)pi/2)} (with interpolated zeros); - Paul 
               Barry (pbarry(AT)wit.ie), Nov 02 2004
%F A088218 a(n)=Sum_{k, 0<=k<=n}A094527(n,k)*(-1)^k . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), 
               Mar 14 2007
%e A088218 The five rooted ordered trees with 3 edges have 10 leaves.
%e A088218 ..x........................
%e A088218 ..o..x.x..x......x.........
%e A088218 ..o...o...o.x..x.o..x.x.x..
%e A088218 ..r...r....r....r.....r....
%p A088218 seq(abs(binomial(-n,-2*n)), n=0..24); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Oct 03 2007
%o A088218 (PARI) a(n)=sum(i=0,n,binomial(n+i-2,i))
%o A088218 (PARI) a(n)=if(n<0,0,polcoeff((1+1/sqrt(1-4*x+x*O(x^n)))/2,n))
%o A088218 (PARI) a(n)=if(n<0,0,polcoeff(1/(1-x+x*O(x^n))^n,n))
%o A088218 (PARI) a(n)=if(n<0, 0, binomial(2*n-1,n))
%o A088218 (PARI) {a(n)=if(n<1, n==0, polcoeff( subst((1-x)/(1-2*x),x,serreverse(x-x^2+x*O(x^n))), 
               n))}
%Y A088218 A001700(n)=a(n+1). a(n)=A024718(n)-A024718(n-1).
%Y A088218 Sequence in context: A099908 A167403 A001700 this_sequence A110556 A072266 
               A085282
%Y A088218 Adjacent sequences: A088215 A088216 A088217 this_sequence A088219 A088220 
               A088221
%K A088218 nonn
%O A088218 0,3
%A A088218 Michael Somos, Sep 24 2003
%E A088218 Essentially the same as A001700, which has much more information.

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