A106254 Partition table in square format.
1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 3, 3, 2, 1, 1, 3, 4, 3, 2, 1, 1, 4, 5, 5, 3, 2, 1, 1, 4, 7, 6, 5, 3, 2, 1, 1, 5, 8, 9, 7, 5, 3, 2, 1, 1, 5, 10, 11, 10, 7, 5, 3, 2, 1, 1, 6, 12, 15, 13, 11, 7, 5, 3, 2, 1, 1, 6, 14, 18, 18, 14, 11, 7, 5, 3, 2, 1, 1, 7, 16, 23, 23, 20, 15, 11, 7, 5, 3, 2, 1
Offset: 1
Examples
The partitions of 6 are: 1 + 1 + 1 + 1 + 1 + 1; 2 + 1 + 1 + 1 + 1; 3 + 1 + 1 + 1; 2 + 2 + 1 + 1; 4 + 1 + 1; 3 + 2 + 1; 2 + 2 + 2; 5 + 1, 4 + 2, 3 + 3, 6. There are 9 partitions of 6 having summands no larger than 4, so p_4(6) = 9.
References
- Ivan Niven, "Mathematics of Choice, How to Count Without Counting", MAA, 1965, pp. 98-99 (table p. 98).
Links
- Martin Y. Champel, Table of n, a(n) for n = 1..405
- A. Comtet, S. N. Majumdar and S. Ouvry, Integer partitions and exclusion statistics, J. Phys. A: Math. Theor. vol. 40 (2007) pp. 11255-11269.
- A. Comtet et al., Integer Partitions and Exclusion Statistics, arXiv:0705.2640 [cond-mat.stat-mech], 2007, eq. (12).
Crossrefs
Essentially the same as A008284, except for missing one diagonal thereof (which would be zero row of this array).
Programs
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Mathematica
T[n_, k_] := T[n, k] = If[n == 0 || k == 1, 1, T[n, k-1] + If[k > n, 0, T[n-k, k]]]; Table[T[n-k+1, k], {n, 1, 13}, {k, 1, n}] // Flatten (* Jean-François Alcover, Aug 08 2018, after Alois P. Heinz *)
Formula
Antidiagonals of table of values of p_k(n) (the number of partitions of n with no summand greater than k).
T(n,m) = sum_{i=1..m} A008284(n,i). T(n,m) = A026820(n,m) if m<=n and T(n,m)=T(n,n) if m>=n. G.f. column m: 1/(1-x)/(1-x^2)/.../(1-x^m) = sum_(n=1,2,3..) T(n,m) x^n [Comtet]. - R. J. Mathar, Aug 31 2007
Extensions
Edited, corrected and extended by Franklin T. Adams-Watters, Jan 12 2006
Comments