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 A105422 Triangle read by rows: T(n,k) is the number of compositions of n having exactly k parts equal to 1. 8
 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, 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, 48, 35, 8, 9, 0, 1, 55, 139, 199, 209, 170, 125, 63, 44, 9, 10, 0, 1, 89, 240, 366, 404, 360 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,8 COMMENTS 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. - Emeric Deutsch, Sep 21 2008 The cyclic version of this array is A320341(n,k), which counts the (unmarked) cyclic compositions of n with exactly k parts equal to 1, with a minor exception for k=0. The sequence (A320341(n, k=0) - 1: n >= 1) counts the (unmarked) cyclic compositions of n with no parts equal to 1. - Petros Hadjicostas, Jan 08 2019 Also the convolution triangle of Fibonacci(n-2). # Peter Luschny, Oct 07 2022 LINKS Alois P. Heinz, Rows n = 0..140, flattened 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). [Emeric Deutsch, Sep 21 2008] Jia Huang, Partially Palindromic Compositions, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See pp. 4, 10-11. Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4. FORMULA G.f.: (1-z)/(1 - z - z^2 - tz + tz^2). 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. - Philippe Deléham, Nov 18 2009 If the triangle's columns are transposed into rows, then T(n,k) can be interpreted as the number of compositions of n+k having exactly k 1's. Then g.f.: ((1-x)/(1-x-x^2))^(k-1) and T(n,k) = T(n-2,k) + T(n-1,k) - T(n-1, k-1) + T(n, k-1). Also, Sum_{j=1..n} T(n-j, j) = 2^(n-1), the number of compositions of n. - Gregory L. Simay, Jun 28 2019 EXAMPLE 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). Triangle begins: 00: 1; 01: 0, 1; 02: 1, 0, 1; 03: 1, 2, 0, 1; 04: 2, 2, 3, 0, 1; 05: 3, 5, 3, 4, 0, 1; 06: 5, 8, 9, 4, 5, 0, 1; 07: 8, 15, 15, 14, 5, 6, 0, 1; 08: 13, 26, 31, 24, 20, 6, 7, 0, 1; 09: 21, 46, 57, 54, 35, 27, 7, 8, 0, 1; 10: 34, 80, 108, 104, 85, 48, 35, 8, 9, 0, 1; 11: 55, 139, 199, 209, 170, 125, 63, 44, 9, 10, 0, 1; 12: 89, 240, 366, 404, 360, 258, 175, 80, 54, 10, 11, 0, 1; 13: 144, 413, 666, 780, 725, 573, 371, 236, 99, 65, 11, 12, 0, 1; ... MAPLE 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 # second Maple program: T:= proc(n) option remember; local j; if n=0 then 1 else []; for j to n do zip((x, y)-> x+y, %, [`if`(j=1, 0, [][]), T(n-j)], 0) od; %[] fi end: seq(T(n), n=0..20); # Alois P. Heinz, Nov 05 2012 # Uses function PMatrix from A357368. Adds a row above and a column to the left. PMatrix(10, n -> combinat:-fibonacci(n-2)); # Peter Luschny, Oct 07 2022 MATHEMATICA nn = 10; a = x/(1 - x) - x + y x; CoefficientList[CoefficientList[Series[1/(1 - a), {x, 0, nn}], x], y] // Flatten (* Geoffrey Critzer, Dec 23 2011 *) CROSSREFS Column 0 yields A000045 (the Fibonacci numbers). Column 1 yields A006367. Column 2 yields A105423. Row sums yield A011782. Cyclic version is A320341. Sequence in context: A029275 A058739 A128627 * A166291 A162986 A319203 Adjacent sequences: A105419 A105420 A105421 * A105423 A105424 A105425 KEYWORD nonn,tabl AUTHOR Emeric Deutsch, Apr 07 2005 STATUS approved

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