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Partition triangle A008284 read from right to left.
89

%I #27 Apr 24 2024 13:15:45

%S 1,1,1,1,1,1,1,1,2,1,1,1,2,2,1,1,1,2,3,3,1,1,1,2,3,4,3,1,1,1,2,3,5,5,

%T 4,1,1,1,2,3,5,6,7,4,1,1,1,2,3,5,7,9,8,5,1,1,1,2,3,5,7,10,11,10,5,1,1,

%U 1,2,3,5,7,11,13,15,12,6,1,1,1,2,3,5,7,11,14,18,18,14,6,1,1,1,2,3,5,7,11

%N Partition triangle A008284 read from right to left.

%C a(n,m) is the number of partitions of n with n-(m-1) parts or, equivalently, with greatest part n-(m-1).

%C The columns are the diagonals of triangle A008284. The diagonals are the columns of the partition array p(n,m), n >= 0, m >= 1, with p(n,m) the number of partitions of n in which every part is <= m; p(0,m) := 1. For n >= 1 this array is obtained from table A026820 read as lower triangular array with extension of the rows according to p(n,m)=A000041(n) for m>n.

%D L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 94, 96 and 307.

%D M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 27.

%H Seiichi Manyama, <a href="/A058398/b058398.txt">Rows n = 1..140, flattened</a>

%H Henry Bottomley, <a href="/A008284/a008284.gif">Illustration of initial terms</a>

%H Roser Homs and Anna-Lena Winz, <a href="https://arxiv.org/abs/2309.06871">Deformations of local Artin rings via Hilbert-Burch matrices</a>, arXiv:2309.06871 [math.AC], 2023. See p. 16.

%F a(n, m)= A008284(n, n-(m-1)).

%F a(n, m)= p(m-1, n-m+1), n >= m >= 1 with the p(n, m) array defined in the comment.

%F a(n, m)=0 if n<m or m<=0 or n=0; a(1, 1)=1; a(n, m)= a(n-1, m)+a(m-1, 2*m-n+1).

%F Viewed as a square array by antidiagonals, T(n,k) = 0 if n<0; T(n,1) = 1; otherwise T(n,k) = T(n,k-1) + T(n-k,k). - _Franklin T. Adams-Watters_, Jul 25 2006

%F Let x be a triangular number C(n,2), where n is the integer being partitioned. Then a(x) = a(x+1) = a(x+2) = 1. Also, a(x+3) = 2 for x>3 and a(x-1) = floor(n/2). - _Allan Bickle_, Apr 18 2024

%e Lower triangular matrix:

%e 1;

%e 1,1;

%e 1,1,1;

%e 1,1,2,1;

%e 1,1,2,2,1;

%e 1,1,2,3,3,1;

%e 1,1,2,3,4,3,1;

%e 1,1,2,3,5,5,4,1;

%e ...

%t row[n_] := Table[ IntegerPartitions[n, k] // Length, {k, 0, n}] // Differences // Reverse; Table[row[n], {n, 1, 14}] // Flatten (* _Jean-François Alcover_, Feb 28 2013 *)

%Y Cf. A026820, A008284, A000041.

%K nonn,easy,tabl,nice

%O 1,9

%A _Wolfdieter Lang_, Dec 11 2000