A059250 Square array read by antidiagonals: T(k,n) = binomial(n-1, k) + Sum_{i=0..k} binomial(n, i), k >= 1, n >= 0.
1, 1, 2, 1, 2, 4, 1, 2, 4, 6, 1, 2, 4, 8, 8, 1, 2, 4, 8, 14, 10, 1, 2, 4, 8, 16, 22, 12, 1, 2, 4, 8, 16, 30, 32, 14, 1, 2, 4, 8, 16, 32, 52, 44, 16, 1, 2, 4, 8, 16, 32, 62, 84, 58, 18, 1, 2, 4, 8, 16, 32, 64, 114, 128, 74, 20, 1, 2, 4, 8, 16, 32, 64, 126, 198, 186, 92, 22, 1, 2, 4, 8, 16, 32, 64
Offset: 1
Examples
Array begins 1, 2, 4, 6, 8, 10, 12, ... 1, 2, 4, 8, 14, 22, ... 1, 2, 4, 8, 16, ...
References
- L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 73, Problem 4.
Links
- G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
- K. E. Morrison, From bocce to positivity: some probabilistic linear algebra, arXiv:1405.2994 [math.PR], 2014; Math. Mag., 86 (2013) 110-119.
- L. Schläfli, Theorie der vielfachen Kontinuität, 1901. (See p. 41)
- J. G. Wendel, A problem in geometric probability, Math. Scand., 11 (1962) 109-111.
Crossrefs
Programs
-
Mathematica
getvalue[n_, k_] := If[n==0, 1, Binomial[n-1, k]+Sum[Binomial[n, i],{i, 0,k}]]; lexicographicLattice[{dim_, maxHeight_}] := Flatten[Array[Sort@Flatten[(Permutations[#1] &) /@ IntegerPartitions[#1 + dim - 1, {dim}], 1] &, maxHeight], 1]; pairs=lexicographicLattice[{2, 13}]-1; Table[getvalue[First[pairs[[j]]], Last[pairs[[j]]]+1], {j, 1, Length[pairs]}] (* Frank M Jackson, Mar 16 2013 *)
Formula
T(k,n) = 2 * Sum_{i=0..k-1} binomial(n-1, i), k >= 1, n >= 1. - Kent E. Morrison, Nov 11 2017
Extensions
Corrected and edited by N. J. A. Sloane, Aug 31 2011, following a suggestion from Frank M Jackson
Comments