A124418 Triangle read by rows: T(n,k) is the number of partitions of the set {1,2,...,n} having exactly k blocks that contain both odd and even entries (0<=k<=floor(n/2)).
1, 1, 1, 1, 2, 3, 4, 9, 2, 10, 30, 12, 25, 100, 72, 6, 75, 370, 372, 60, 225, 1369, 1922, 600, 24, 780, 5587, 9920, 4500, 360, 2704, 22801, 51200, 33750, 5400, 120, 10556, 101774, 273920, 234000, 55800, 2520, 41209, 454276, 1465472, 1622400, 576600, 52920, 720
Offset: 0
Examples
T(4,1) = 9 because we have 1234, 134|2, 1|234, 124|3, 14|2|3, 1|2|34, 123|4, 1|23|4 and 12|3|4. Triangle starts: 1; 1; 1, 1; 2, 3; 4, 9, 2; 10, 30, 12; 25, 100, 72, 6; ...
Links
- Alois P. Heinz, Rows n = 0..200, flattened
Crossrefs
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,s=1},Q[n])) od: for n from 0 to 13 do seq(coeff(P[n],x,j),j=0..floor(n/2)) od; # yields sequence in triangular form # second Maple program: with(combinat): T:= proc(n, k) local g, u; g:= floor(n/2); u:=ceil(n/2); add(binomial(g, i)*stirling2(i, k)*bell(g-i), i=k..g)* add(binomial(u, i)*stirling2(i, k)*bell(u-i), i=k..u)*k! 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
T[n_, k_] := Module[{g = Floor[n/2], u = Ceiling[n/2]}, Sum[Binomial[g, i] * StirlingS2[i, k]*BellB[g-i], {i, k, g}]*Sum[Binomial[u, i]*StirlingS2[i, k] * BellB[u-i], {i, k, u}]*k!]; Table[Table[T[n, k], {k, 0, Floor[n/2]}], {n, 0, 15}] // Flatten (* Jean-François Alcover, Feb 20 2015, after Alois P. Heinz *)
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PARI
{T(n,k)=if(k<0||k>n,0, k!*(n\2)!*((n+1)\2)!*polcoeff(polcoeff(exp((1+y)*(exp(x+x*O(x^n))-1)),n\2),k) *polcoeff(polcoeff(exp((1+y)*(exp(x+x*O(x^n))-1)),(n+1)\2),k))} \\ Paul D. Hanna, Nov 08 2006
Formula
The generating polynomial of row n is P[n](x)=Q[n](1,1,x), 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.
Conjecture: T(n,k) = k!*A049020([n/2],k)*A049020([(n+1)/2],k) where A049020(n,k)=Sum_{i=0..n} S2(n,i)*C(i,k) and S2(n,k)=(1/k!)*Sum_{j=0..k} (-1)^(k-j)*C(k,j)*j^n (the Stirling numbers of 2nd kind). - Paul D. Hanna, Nov 08 2006
Sum_{k=0..floor(n/2)} = k * A362495(n). - Alois P. Heinz, Jun 05 2023
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