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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).

Original entry on oeis.org

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

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Author

Emeric Deutsch, Oct 03 2015

Keywords

Comments

Sum of entries in row n is A000041(n) (the number of partitions of n).
Number of entries in row n is A003040(n).
Sum_{k>=1} k*T(n,k) = A000085(n) = number of standard tableaux of size n.
In the Maple program (too slow and too complex) (i) pp(n) yields the prime number preceding n and 1 if n=2; (ii) B(n) yields the partition corresponding to the Heinz number n; (iii) a(n) yields the number of standard tableaux of the Ferrers diagram of the partition B(n); (iv) Q(n) yields the generating polynomial of the partitions of n with respect to the number of standard tableaux of their Ferrers diagrams. For example, Q(4) = 2x + x^2 + 2x^3; indeed, the partitions [4], [1,1,1,1] have each 1 standard tableau, the partition [2,2] has 2 standard tableaux, and the partitions [3,1], [2,1,1] each have 3 standard tableaux.
The Heinz number of a partition p = [p_1, p_2, ..., p_r] is defined as Product(p_j-th prime, j=1...r) (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 1, 4] we get 2*2*2*7 = 56.

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;
		

Crossrefs

Programs

  • 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