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%I #26 May 28 2019 05:06:27
%S 1,1,0,1,1,0,1,3,1,0,1,6,7,1,0,1,10,25,15,1,0,1,15,65,90,31,1,0,1,21,
%T 140,350,301,63,0,0,1,28,266,1050,1701,966,119,0,0,1,36,462,2646,6951,
%U 7770,2989,210,0,0,1,45,750,5880,22827,42525,33985,8925,336,0,0
%N The triangle in A151359 read by rows upwards.
%C Conjectured: The i-th element of row j is the number of different equivalence relationships, within a set of (j-1) element, having (j-i) equivalence classes. For example: row 5 = [1, 6, 7, 1, 0] means that, in a set of 4 elements, there exists 7 equivalence relationships having 3 different equivalence classes. - _Philippe Beaudoin_, Nov 09 2013
%H Moa Apagodu, David Applegate, N. J. A. Sloane, and Doron Zeilberger, <a href="http://arxiv.org/abs/1701.08394">Analysis of the Gift Exchange Problem</a>, arXiv:1701.08394, 2017.
%H David Applegate and N. J. A. Sloane, <a href="http://arxiv.org/abs/0907.0513">The Gift Exchange Problem</a> (arXiv:0907.0513, 2009) (see Table 7 E5(n,k) page 16).
%e Triangle begins:
%e 1
%e 1 0
%e 1 1 0
%e 1 3 1 0
%e 1 6 7 1 0
%e 1 10 25 15 1 0
%e 1 15 65 90 31 1 0
%e 1 21 140 350 301 63 0 0
%e 1 28 266 1050 1701 966 119 0 0
%t Unprotect[Power]; 0^0 = 1; a[n_ /; 1 <= n <= 6] = 1; a[_] = 0; t[n_, k_] := t[n, k] = If[k == 0, a[0]^n, Sum[Binomial[n - 1, j - 1] a[j] t[n - j, k - 1], {j, 0, n - k + 1}]]; Table[Table[t[n - 1, k], {k, n - 1, 0, -1}], {n, 0, 10}] // Flatten (* _Jean-François Alcover_, Jan 20 2016, using _Peter Luschny_'s Bell transform *)
%Y Cf. A148092 (row sums), A151511 (row-reversed).
%K nonn,tabl,easy
%O 0,8
%A _N. J. A. Sloane_, May 14 2009
%E Row 9 added by _Michel Marcus_, Feb 13 2014
%E Row 10 from _R. J. Mathar_, May 28 2019
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