A152072 Triangle read by rows: T(n,k) = the largest product of a partition of n into k positive integers (1 <= k <= n).
1, 2, 1, 3, 2, 1, 4, 4, 2, 1, 5, 6, 4, 2, 1, 6, 9, 8, 4, 2, 1, 7, 12, 12, 8, 4, 2, 1, 8, 16, 18, 16, 8, 4, 2, 1, 9, 20, 27, 24, 16, 8, 4, 2, 1, 10, 25, 36, 36, 32, 16, 8, 4, 2, 1, 11, 30, 48, 54, 48, 32, 16, 8, 4, 2, 1, 12, 36, 64, 81, 72, 64, 32, 16, 8, 4, 2, 1
Offset: 1
Examples
Triangle begins: 1 2,1 3,2,1 4,4,2,1 5,6,4,2,1 6,9,8,4,2,1 7,12,12,8,4,2,1 8,16,18,16,8,4,2,1 9,20,27,24,16,8,4,2,1 10,25,36,36,32,16,8,4,2,1 ... T(7,3)=12 since there are 12 ways to selected class representatives from the mod 3 partitioning of {1,..,7} = {1,4,7} U {2,5} U {3,6}. - _Dennis P. Walsh_, Nov 27 2012
References
- Cees H. Elzinga, M. Studer, Normalization of Distance and Similarity in Sequence Analysis in G. Ritschard & M. Studer (eds), Proceedings of the International Conference on Sequence Analysis and Related Methods, Lausanne, June 8-10, 2016, pp 445-468.
- K. Mahler and J. Popken, Over een Maximumprobleem uit de Rekenkunde (in Dutch), (On a Maximum Problem in Arithmetic), Nieuw Archief voor Wiskunde (3) 1 (1953), 1-15.
- David W. Wilson, Posting to Sequence Fans mailing List, Mar 11 2009
Links
- David W. Wilson, Table of n, a(n) for n = 1..10011
- Zhiwei Lin, H. Wang, C. H. Elzinga, Concordance and the Smallest Covering Set of Preference Orderings, arXiv preprint arXiv:1609.04722 [cs.AI], 2016.
Crossrefs
Programs
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Maple
T:= (n,k)-> mul(floor((n+i)/k), i=0..k-1): seq(seq(T(n, k), k=1..n), n=1..12);
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Mathematica
T[n_, k_] := Product[ Floor[(n + i)/k], {i, 0, k - 1}]; Flatten@ Table[ T[n, k], {n, 12}, {k, n}] (* Robert G. Wilson v, Jul 08 2016 *)
Formula
T(n,k) = PROD(0 <= i < k; [(n+i)/k]).
T(n,n-d) = 2^d = A000079(d) (d <= n/2).
MAX(1 <= k <= n, T(n,k)) = A000792(n).
T(n,k) = (ceiling(n/k))^(n mod k)*(floor(n/k))^(k-n mod k). - Dennis P. Walsh, Nov 27 2012
Sum_{k = 1..n} T(n,k) = A152074(n). - David W. Wilson, Jul 07 2016
Comments