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%I #12 Oct 12 2015 10:31:57
%S 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,
%T 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,
%U 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
%N 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).
%C Sum of entries in row n is A000041(n) (the number of partitions of n).
%C Number of entries in row n is A003040(n).
%C Sum_{k>=1} k*T(n,k) = A000085(n) = number of standard tableaux of size n.
%C 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.
%C 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.
%H Alois P. Heinz, <a href="/A258124/b258124.txt">Rows n = 1..14, flattened</a>
%H FindStat - Combinatorial Statistic Finder, <a href="http://www.findstat.org/StatisticsDatabase/St000003/">The number of standard Young tableaux of the partition</a>
%e 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].
%e Triangle starts:
%e 1;
%e 2;
%e 2,1;
%e 2,1,2;
%e 2,0,0,2,2,1;
%e 2,0,0,0,4,0,0,0,2,2,0,0,0,0,0,1;
%p 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
%Y Cf. A215366, A000041, A003040, A000085.
%K nonn,tabf
%O 1,2
%A _Emeric Deutsch_, Oct 03 2015
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