A258124 Triangle read by rows: T(n,k) is the number of partitions of n having k standard tableaux in their Ferrers diagrams (n>=1, k>=1).
1, 2, 2, 1, 2, 1, 2, 2, 0, 0, 2, 2, 1, 2, 0, 0, 0, 4, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 4, 2, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1
Offset: 1
Examples
T(6,5) = 4 because there are 4 partitions of 6 having 5 standard tableaux in their Ferrers diagrams: [5,1], [3,3], [2,2,2], and [2,1,1,1,1]. Triangle starts: 1; 2; 2,1; 2,1,2; 2,0,0,2,2,1; 2,0,0,0,4,0,0,0,2,2,0,0,0,0,0,1;
Links
- Alois P. Heinz, Rows n = 1..14, flattened
- FindStat - Combinatorial Statistic Finder, The number of standard Young tableaux of the partition
Programs
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Maple
with(numtheory): pp := proc (n) if n = 2 then 1 else prevprime(n) end if end proc: B := proc (n) local nn, j, m: nn := op(2, ifactors(n)): for j to nops(nn) do m[j] := op(j, nn) end do: [seq(seq(pi(op(1, m[i])), q = 1 .. op(2, m[i])), i = 1 .. nops(nn))] end proc: a := proc (n) local pp, FS: pp := proc (n) if n = 1 then 1 elif n = 2 then 1 else prevprime(n) end if end proc: FS := factorset(n): if n = 1 then 1 else add(a(n*pp(FS[j])/FS[j]), j = 1 .. nops(FS)) end if end proc; Q := proc (n) local R, i: R := 0: for i from ithprime(n) to 2^n do if sum(B(i)[j], j = 1 .. nops(B(i))) = n then R := R+x^a(i) else end if end do: sort(R) end proc: T := proc (n, k) options operator, arrow: coeff(Q(n), x, k) end proc: seq(seq(T(n, k), k = 1 .. degree(Q(n))), n = 1 .. 8); # yields sequence in triangular form
Comments