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 A180184 Irregular triangle read by rows: T(n,k) is the number of compositions of n with k parts, all >= 4, for n >= 4 and 1 <= k <= floor(n/4). 1
 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 1, 6, 3, 1, 7, 6, 1, 8, 10, 1, 9, 15, 1, 1, 10, 21, 4, 1, 11, 28, 10, 1, 12, 36, 20, 1, 13, 45, 35, 1, 1, 14, 55, 56, 5, 1, 15, 66, 84, 15, 1, 16, 78, 120, 35, 1, 17, 91, 165, 70, 1, 1, 18, 105, 220, 126, 6, 1, 19, 120, 286, 210, 21, 1, 20 (list; graph; refs; listen; history; text; internal format)
 OFFSET 4,8 COMMENTS Sum of the entries in row n is A003269(n-3). Row n contains floor(n/4) entries. From Petros Hadjicostas, Apr 15 2020: (Start) From equations (3) and (23) in Mathar (2016), we have Sum_{m,k >= 0} T_{4 x m}(4,k)*z^m*t^k = 1/(1 - z - z^4*t), where T_{4xm}(4,k) counts the ways of tiling the 4 x m rectangle with k non-overlapping squares of shape 4 x 4 and with 4*m - 16*k unit squares (see Definition 1 in his paper for more details). By expanding 1/(1 - (z + z^4*t)) as an infinite geometric series, using the binomial theorem, and changing indices of summation, we may show that T_{4xm}(4,k) = binomial(m - 3*k, k) = T(m+4, k+1) for m >= 0 and 0 <= k <= floor(m/4) (where the current array T(n,k) should be distinguished from Mathar's T_{4 x m}(4,k)). Indeed, we have a bijection between the above tilings of the 4 x m rectangle with k non-overlapping squares of shape 4 x 4 and 4*m - 16*k unit squares and the compositions of m+4 with k+1 parts, all >= 4. To construct the bijection, let us agree that an a x b rectangle has height a and base b. Given such a composition, a_1 + a_2 + ... + a_{k+1} = m + 4 (with a_i >= 4), paste together a 4 x 4 square followed by a_1 - 4 columns of 4 unit squares, another 4 x 4 square followed by a_2 - 4 columns of 4 units squares, and so on, and finally a 4 x 4 square followed by a_{k+1} - 4 columns of 4 unit squares. Remove the first 4 x 4 square, and we get a tiling of the 4 x m rectangle with k 4 x 4 squares and 4*Sum_{i=1..(k+1)} (a_i - 4) = 4*(m+4) - 16*(k+1) = 4*m - 16*k unit squares. The above process can be reversed to complete the bijection, but we omit the details. (End) LINKS R. J. Mathar, Tiling n x m rectangles with 1 x 1 and s x s squares, arXiv:1609.03964 [math.CO] (2016), Section 4.3. FORMULA T(n, k) = binomial(n-3*k-1, k-1). T(n, k) = A228572(2*n-4, 2*k-1) + A228572(2*n-7, 2*k-2) - A228572(2*n-3, 2*k-1) for n >= 4 and 1 <= k <= floor(n/4). - Johannes W. Meijer, Aug 26 2013 [Range of k adjusted by Petros Hadjicostas, Apr 15 2020 to start at 1 rather than 0] G.f.: Sum_{n,k} T(n,k)*z^n*t^k = z^4*t/(1-z-t*z^4). - R. J. Mathar, Aug 24 2016 [Adjusted by Petros Hadjicostas, Apr 14 2020 to agree with the offset] EXAMPLE Triangle T(n,k) (with n >= 4 and 1 <= k <= floor(n/4)) starts as follows:   1;   1;   1;   1;   1,  1;   1,  2;   1,  3;   1,  4;   1,  5,  1;   1,  6,  3;   1,  7,  6;   1,  8, 10;   1,  9, 15,  1;   1, 10, 21,  4;   1, 11, 28, 10;   1, 12, 36, 20;   ... T(14,3) = 6 because we have the following compositions (ordered partitions) of 14 with 3 parts, all >= 4: [5,5,4], [4,6,4], [5,4,5], [6,4,4], [4,5,5], [4,4,6]. MAPLE for n from 4 to 27 do seq(binomial(n-3*k-1, k-1), k = 1 .. floor((1/4)*n)) end do; T := (n, k) -> binomial(n-3*k-1, k-1): seq(seq(T(n, k), k=1..floor(n/4)), n=4..26); # Johannes W. Meijer, Aug 26 2013 MATHEMATICA Flatten[Table[Binomial[n-3k-1, k-1], {n, 4, 30}, {k, Floor[n/4]}]] (* Harvey P. Dale, Feb 05 2013 *) CROSSREFS Cf. A003269, A102547, A228572. Sequence in context: A324933 A085343 A049077 * A330752 A222266 A077609 Adjacent sequences:  A180181 A180182 A180183 * A180185 A180186 A180187 KEYWORD nonn,tabf,easy AUTHOR Emeric Deutsch, Aug 15 2010 STATUS approved

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Last modified June 6 16:34 EDT 2020. Contains 334828 sequences. (Running on oeis4.)