some help is needed Re: SEQ+# A127157 FROM Emeric Deutsch (fwd)
Emeric Deutsch
deutsch at duke.poly.edu
Wed Feb 28 05:09:17 CET 2007
Dear Seqfans,
The bivariate g.f. G(t,z) for the number of ordered trees with
n edges and having 2k nodes of odd degree (not outdegree) satisfies
(1) z^2*G^3-z(z+2)G^2+(1+2z)G-t^2*z-1=0
I have found this in my notes dating back to 1997; I have to recall
the derivation (probably, a minor problem). However, according to my
notes, I have found the formula
(2) [t^k z^n] = 2*binomial(3k-1,2k)*binomial(n-1+k,3k-2)/(3k-1)
only by inspection (consistent with the g.f. at least up to n=50).
Any idea for deriving (2)? Probably a tricky change of variables
will make (1) "Lagrangeable".
Below you'll find the submitted sequence.
Thanks for any input.
Emeric
---------- Forwarded message ----------
Date: Tue, 27 Feb 2007 22:52:40 -0500
From: The On-Line Encyclopedia of Integer Sequences <oeis at research.att.com>
Reply-To: deutsch at duke.poly.edu
To: njas at research.att.com
Cc: deutsch at duke.poly.edu
Subject: SEQ+# A127157 FROM Emeric Deutsch
The following is a copy of the email message that was sent to njas
containing the sequence you submitted.
All greater than and less than signs have been replaced by their html
equivalents. They will be changed back when the message is processed.
This copy is just for your records. No reply is expected.
Subject: PRE-NUMBERED NEW SEQUENCE A127157 FROM Emeric Deutsch
%I A127157
%S A127157 1, 2, 3, 2, 4, 10, 5, 30, 7, 6, 70, 56, 7, 140, 252, 30, 8, 252, 840, 330, 9, 420, 2310, 1980, 143, 10, 660, 5544, 8580, 2002, 11, 990, 12012, 30030, 15015, 728, 12, 1430, 24024, 90090, 80080, 12376, 13, 2002, 45045, 240240, 340340, 111384, 3876, 14, 2730, 80080, 583440, 1225224, 705432, 77520, 15, 3640, 136136, 1312740, 3879876, 3527160, 813960, 21318
%N A127157 Triangle read by rows: T(n,k) is the number of ordered trees with n edges and 2k nodes of odd degree (not outdegree; 1<=k<=ceil(n/2)).
%C A127157 Neil: this is a funny-shaped triangle. END
Row n has ceil(n/2) terms.
Row sums are the Catalan numbers (A000108).
T(n,1)=n;
T(n,2)=2*binom(n+1,4)=2*A000332(n+1);
T(n,3)=7*binom(n+2,7)=7*A000580(n+2);
T(n,4)=30*binom(n+3,10)=30*A001287(n+3);
T(n,5)=143*binom(n+4,13)=143*A010966(n+4);
T(2n-1,n)=A006013(n-1).
%F A127157 T(n,k)=2*binomial(3k-1,2k)*binomial(n-1+k,3k-2)/(3k-1) (formula obtained only by inspection).
G.f.=G-1, where G=G(t,z) satisfies z^2*G^3-z(z+2)G^2+(1+2z)G-t^2*z-1=0.
%e A127157 Triangle starts:
1;
2;
3,2;
4,10;
5,30,7;
6,70,56;
%p A127157 T:=(n,k)->2*binomial(3*k-1,2*k)*binomial(n-1+k,3*k-2)/(3*k-1): for n from 1 to 15 do seq(T(n,k),k=1..ceil(n/2)) od;
%Y A127157 A000108,A000332,A000580,A001287,A010966,A006013
%O A127157 1
%K A127157 ,nonn,
%A A127157 Emeric Deutsch (deutsch at duke.poly.edu), Feb 27 2007
RH
RA 192.20.225.32
RU
RI
Mitch Harris wrote:
>...
>
>the real root to x^3 - x^2 -1 = 0 is
>
> (1/(3 * 2^(1/3))) * (2^(1/3) + cbrt(29 + 3sqrt(93)) + cbrt(29 - 3sqrt(93)))
>...
Thanks to everyone who replied. But after all that I think I am just
going to use the much simpler, if only implicit, "x is the root of x^3
-x^2 -1 = 0."
Am I correct, however, in assuming that limit{n->infinity} r(n)*r(n+1) =
(2/x^3) + (x^3/2) - 2 ?
(r(n) was the nth {rational} term of the continued fraction
[r(1);r(2),...,r(n)] =
A076725(n)/A076725(n-1)^2, for every positive integer n.)
Thanks,
Leroy Quet
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