%I A105422
%S A105422 1,0,1,1,0,1,1,2,0,1,2,2,3,0,1,3,5,3,4,0,1,5,8,9,4,5,0,1,8,15,15,14,5,
6,
%T A105422 0,1,13,26,31,24,20,6,7,0,1,21,46,57,54,35,27,7,8,0,1,34,80,108,104,85,
%U A105422 48,35,8,9,0,1,55,139,199,209,170,125,63,44,9,10,0,1,89,240,366,404,360
%N A105422 Triangle read by rows: T(n,k) is the number of compositions of n having
exactly k parts equal to 1.
%C A105422 T(n,k) is also the number of length n bit strings beginning with 0 having
k singletons. Example: T(4,2)=3 because we have 0010, 0100 and 0110.
[From Emeric Deutsch (deutsch(AT)duke.poly.edu), Sep 21 2008]
%D A105422 D. Baccherini, D. Merlini and R. Sprugnoli, Level generating trees and
proper Riordan arrays, Applicable Analysis and Discrete Mathematics,
2, 2008, 69-91 (see p. 83). [From Emeric Deutsch (deutsch(AT)duke.poly.edu),
Sep 21 2008]
%F A105422 G.f.=(1-z)/(1-z-z^2-tz+tz^2).
%F A105422 T(n,k)=T(n-1,k)+T(n-2,k)+T(n-1,k-1)-T(n-2,k-1), T(0,0)=1, T(1,0)=0. [From
Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 18 2009]
%e A105422 T(6,2)=9 because we have (1,1,4),(1,4,1),(4,1,1),(1,1,2,2),(1,2,1,2),
(1,2,2,1),(2,1,1,2),(2,1,2,1) and (2,2,1,1).
%e A105422 Triangle begins:
%e A105422 1;
%e A105422 0,1;
%e A105422 1,0,1;
%e A105422 1,2,0,1;
%e A105422 2,2,3,0,1;
%p A105422 G:=(1-z)/(1-z-z^2-t*z+t*z^2): Gser:=simplify(series(G,z=0,15)): P[0]:=1:
for n from 1 to 14 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 13
do seq(coeff(t*P[n],t^k),k=1..n+1) od; # yields sequence in triangular
form
%Y A105422 Column 0 yields A000045 (the Fibonacci numbers). Column 1 yields A006367.
Column 2 yields A105423. Row sums yield A011782.
%Y A105422 Sequence in context: A029275 A058739 A128627 this_sequence A166291 A162986
A128584
%Y A105422 Adjacent sequences: A105419 A105420 A105421 this_sequence A105423 A105424
A105425
%K A105422 nonn,tabl,new
%O A105422 0,8
%A A105422 Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 07 2005
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