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%I #37 Jan 25 2023 12:13:41
%S 1,1,1,1,2,1,2,3,3,1,4,6,6,4,1,8,13,13,10,5,1,17,28,30,24,15,6,1,37,
%T 62,69,59,40,21,7,1,82,140,160,144,105,62,28,8,1,185,320,375,350,271,
%U 174,91,36,9,1,423,740,885,852,690,474,273,128,45,10,1,978,1728,2102,2077
%N Triangle read by rows: T(n,k) is the number of left factors of Motzkin paths without peaks, having length n and endpoint height k.
%C Column 0 is A004148 (RNA secondary structure numbers).
%C This triangle appears identical to A191579 (apart from offsets). - _Philippe Deléham_, Jan 26 2014
%D Cameron, Naiomi, and Everett Sullivan. "Peakless Motzkin paths with marked level steps at fixed height." Discrete Mathematics 344.1 (2021): 112154.
%D He, Tian-Xiao. "A-sequences, Z-sequence, and B-sequences of Riordan matrices." Discrete Mathematics 343.3 (2020): 111718.
%H Alois P. Heinz, <a href="/A097724/b097724.txt">Rows n = 0..140, flattened</a>
%H Naiomi T. Cameron and Asamoah Nkwanta, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL8/Cameron/cameron46.html">On Some (Pseudo) Involutions in the Riordan Group</a>, Journal of Integer Sequences, Vol. 8 (2005), Article 05.3.7.
%H A. Nkwanta, A. Tefera, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Nkwanta/nkwanta4.html">Curious Relations and Identities Involving the Catalan Generating Function and Numbers</a>, Journal of Integer Sequences, 16 (2013), #13.9.5.
%H A. Panayotopoulos and P. Vlamos, <a href="http://dx.doi.org/10.1007/978-3-642-33412-2_49">Cutting Degree of Meanders</a>, Artificial Intelligence Applications and Innovations, IFIP Advances in Information and Communication Technology, Volume 382, 2012, pp 480-489; DOI 10.1007/978-3-642-33412-2_49. - From _N. J. A. Sloane_, Dec 29 2012
%F T(n,k) = (k+1)*Sum_{j=ceiling((n-k+1)/2)..n-k} (C(j,n-k-j)*C(j+k,n+1-j)/j) for 0 <= k < n; T(n,n)=1.
%F G.f.: G/(1-tzG), where G = (1 - z + z^2 - sqrt(1 - 2z - z^2 - 2z^3 + z^4))/(2z^2) is the g.f. for the sequence A004148.
%F T(n,k) = T(n-1,k-1) + Sum_{j>=0} T(n-1-j,k+j), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - _Philippe Deléham_, Jan 26 2014
%F Sum_{j=0..n-1} cos(2*Pi*k/3 + Pi/6)*T(n,k) = cos(Pi*n/2)*sqrt(3)/2 - cos(2*Pi*n/3 + Pi/6). - _Leonid Bedratyuk_, Dec 06 2017
%e Triangle starts:
%e 1;
%e 1, 1;
%e 1, 2, 1;
%e 2, 3, 3, 1;
%e 4, 6, 6, 4, 1;
%e Row n has n+1 terms.
%e T(3,2)=3 because we have HUU, UHU and UUH, where U=(1,1) and H=(1,0).
%p T:=proc(n,k) if k=n then 1 else (k+1)*sum(binomial(j,n-k-j)*binomial(j+k,n+1-j)/j,j=ceil((n-k+1)/2)..n-k) fi end: seq(seq(T(n,k),k=0..n),n=0..12); T:=proc(n,k) if k=n then 1 else (k+1)*sum(binomial(j,n-k-j)*binomial(j+k,n+1-j)/j,j=ceil((n-k+1)/2)..n-k) fi end: TT:=(n,k)->T(n-1,k-1): matrix(10,10,TT); # gives the sequence as a matrix
%t T[n_, k_] := T[n, k] = If[k==n, 1, (k+1)*Sum[Binomial[j, n-k-j]*Binomial[j +k, n+1-j]/j, {j, Ceiling[(n-k+1)/2], n-k}]]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Feb 22 2017, translated from Maple *)
%Y Cf. A004148, A191579, A091964 (row sums).
%K nonn,tabl
%O 0,5
%A _Emeric Deutsch_, Sep 11 2004
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