Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.
%I #16 Jul 17 2017 03:53:30
%S 1,1,2,4,1,13,2,46,6,184,18,1,805,69,3,3840,288,12,19775,1324,47,1,
%T 109180,6578,213,4,642382,35136,1032,20,4007712,200398,5390,96,1,
%U 26399764,1214136,30027,505,5,182939900,7778856,177744,2792,30,1329327991,52501052,1112969,16362,170,1
%N Triangle read by rows: T(n,k) is the number of partitions of the set {1,2,...,n} having k adjacent blocks of size 3, i.e., blocks of the form (i,i+1,i+2) (0 <= k <= floor(n/3)).
%C Number of entries in row n is 1 + floor(n/3).
%C Sum of entries in row n = A000110(n) (the Bell numbers).
%C T(n,0) = A184177(n).
%C Sum_{k>=0}k*T(n,k) = A052889(n-2).
%H Alois P. Heinz, <a href="/A184176/b184176.txt">Rows n = 0..250, flattened</a>
%F T(n,k) = Sum_{j=k..floor(n/3)}(-1)^(k+j) * C(j,k) * C(n-2j,j) * Bell(n-3j).
%e T(4,1) = 2 because we have 123-4 and 1-234.
%e Triangle starts:
%e 1;
%e 1;
%e 2;
%e 4, 1;
%e 13, 2;
%e 46, 6;
%e 184, 18, 1;
%p with(combinat): q := 3: a := proc (n, k) options operator, arrow: sum((-1)^(k+j)*binomial(j, k)*binomial(n+j-j*q, j)*bell(n-j*q), j = k .. floor(n/q)) end proc: for n from 0 to 15 do seq(a(n, k), k = 0 .. floor(n/q)) end do; # yields sequence in triangular form
%t q = 3; a[n_, k_] := Sum[(-1)^(k+j)*Binomial[j, k]*Binomial[n+j-j*q, j]* BellB[n-j*q], {j, k, Floor[n/q]}]; Table[a[n, k], {n, 0, 15}, {k, 0, Floor[n/q]}] // Flatten (* _Jean-François Alcover_, Feb 22 2017, translated from Maple *)
%Y Cf. A000110, A052889, A184174, A184175, A184177.
%K nonn,tabf
%O 0,3
%A _Emeric Deutsch_, Feb 09 2011