A124422 Triangle read by rows: T(n,k) is the number of partitions of the set {1,2,...,n}, having exactly k blocks consisting only even entries (0<=k<=floor(n/2)).
1, 1, 1, 1, 3, 2, 5, 8, 2, 22, 25, 5, 52, 101, 45, 5, 283, 423, 156, 15, 855, 1889, 1143, 238, 15, 5451, 9726, 5002, 916, 52, 19921, 48382, 35805, 10540, 1275, 52, 144074, 292223, 187515, 49155, 5400, 203, 614866, 1609551, 1379753, 512710, 89425, 7089, 203
Offset: 0
Examples
T(4,1) = 8 because we have 134|2, 13|24, 14|2|3, 1|24|3, 1|2|34, 123|4, 1|23|4 and 12|3|4. Triangle starts: 1; 1; 1, 1; 3, 2; 5, 8, 2; 22, 25, 5; 52, 101, 45, 5; ...
Links
- Alois P. Heinz, Rows n = 0..200, flattened
Programs
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Maple
Q[0]:=1: for n from 1 to 13 do if n mod 2 = 1 then Q[n]:=expand(t*diff(Q[n-1],t)+x*diff(Q[n-1],s)+x*diff(Q[n-1],x)+t*Q[n-1]) else Q[n]:=expand(x*diff(Q[n-1],t)+s*diff(Q[n-1],s)+x*diff(Q[n-1],x)+s*Q[n-1]) fi od: for n from 0 to 13 do P[n]:=sort(subs({t=1,x=1},Q[n])) od: for n from 0 to 13 do seq(coeff(P[n],s,j),j=0..floor(n/2)) od; # yields sequence in triangular form # second Maple program: T:= proc(n, k) local g, u; g:= floor(n/2); u:=ceil(n/2); add(Stirling2(i, k)*binomial(g, i)* add(Stirling2(u, j)*j^(g-i), j=0..u), i=k..g) end: seq(seq(T(n,k), k=0..floor(n/2)), n=0..15); # Alois P. Heinz, Oct 23 2013
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Mathematica
Unprotect[Power]; 0^0 = 1; T[n_, k_] := Module[{g = Floor[n/2], u = Ceiling[n/2]}, Sum[StirlingS2[i, k]*Binomial[g, i]*Sum[StirlingS2[u, j]*j^(g-i), {j, 0, u}], {i, k, g}]]; Table[Table[T[n, k], {k, 0, Floor[n/2]}], {n, 0, 15}] // Flatten (* Jean-François Alcover, May 22 2015, after Alois P. Heinz *)
Formula
The generating polynomial of row n is P[n](s)=Q[n](1,s,1), where the polynomials Q[n]=Q[n](t,s,x) are defined by Q[0]=1; Q[n]=t*dQ[n-1]/dt + x*dQ[n-1]/ds + x*dQ[n-1]/dx + t*Q[n-1] if n is odd and Q[n]=x*dQ[n-1]/dt + s*dQ[n-1]/ds + x*dQ[n-1]/dx + s*Q[n-1] if n is even.
Comments