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A180184
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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).
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3
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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
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OFFSET
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4,8
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COMMENTS
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Sum of the entries in row n is A003269(n-3).
Row n contains floor(n/4) entries.
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)
T(n+7,k+1) is the number of k-subsets of {1..n} with values at least 4 apart. For example, T(17,4) = 4 corresponds to the subsets {1,5,9},{1,5,10},{1,6,10},{2,6,10} of {1..10} (A102547 gives the number of k-subsets of {1..n} with values at least 3 apart and A011973 with values at least 2 apart). - Enrique Navarrete, Jan 29 2022
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LINKS
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FORMULA
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T(n, k) = binomial(n-3*k-1, k-1).
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]
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EXAMPLE
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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].
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MAPLE
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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
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MATHEMATICA
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Flatten[Table[Binomial[n-3k-1, k-1], {n, 4, 30}, {k, Floor[n/4]}]] (* Harvey P. Dale, Feb 05 2013 *)
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CROSSREFS
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KEYWORD
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nonn,tabf,easy
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AUTHOR
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STATUS
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approved
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