cp's OEIS Frontend

The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), giving a bijective correspondence between positive integers and integer partitions.

The Dyson rank of a nonempty partition is its maximum part minus its length. The rank of an empty partition is undefined.

We define the k-th rank of a partition as the k-th part minus the number of parts >= k. Every partition of n has n ranks. This is a generalization of the Dyson's rank of a partition which is the largest part minus the number of parts. Since the first part of a partition is also the largest part of the same partition so the Dyson's rank of a partition is the case for k = 1.

The sum of the k-th ranks of all partitions of n is equal to zero.

Also T(n,k) = number of partitions of n with negative k-th rank.

It appears that reversed rows converge to A000070, the same as A208482. - Omar E. Pol, Mar 11 2012

From Omar E. Pol, Dec 12 2019: (Start)

1) The k-th part of a partition of n is also the number of parts >= k of its conjugate partition.

2) The k-th rank of a partitions is also the number of parts >= k of its conjugate partition minus the number of parts >= k.

For example: for n = 9 consider the partition [5, 3, 1]. The first part is 5, so the conjugate partition has five parts >= 1. The second part is 3, so the conjugate partition has three parts >= 2. The third part is 1, so the conjugate partition has only one part >= 3. The mentioned conjugate partition is [3, 2, 2, 1, 1]. And conversely, consider the partition [3, 2, 2, 1, 1]. The first part is 3, so the conjugate partition has three parts >= 1. The second part is 2, so the conjugate partition has two parts >= 2. the Third part is 2, so the conjugate partition has two parts >= 3, and so on. In this case the conjugate partition is [5, 3, 1].

3) The difference between the k-th part and the (k+1)-st part of the partition of n is also the number of k's in its conjugate partition. For example: consider the partition [5, 3, 1]. The difference between the first and the second part is 5 - 3 = 2, equals the number of 1's in its conjugate partition. The difference between the second and the third part is 3 - 1 = 2, equals the number of 2's in its conjugate partition. The difference between the third and the fourth (virtual) part is 1 - 0 = 1, equals the number of 3's in its conjugate partition [3, 2, 2, 1, 1]. And conversely, consider the partition [3, 2, 2, 1, 1]. The difference between the first and the second part is 3 - 2 = 1, equals the number of 1's in its conjugate partition. The difference between the second and the third part is 2 - 2 = 0, equals the number of 2's in its conjugate partition. The difference between the third and the fourth part is 2 - 1 = 1, equals the number of 3's in its conjugate partition, and so on.

4) The list of n ranks of a partition of n equals the list of n ranks multiplied by -1 of its conjugate partition. For example the nine ranks of the partition [5, 3, 1] of 9 are [2, 1, -1, -1, -1, -1, 0, 0, 0], and the nine ranks of its conjugate partition [3, 2, 2, 1, 1] are [-2, -1, 1, 1, 1, 1, 0, 0, 0].

For a list of partitions of the positive integers ordered by its k-th ranks see A330370. (End)

The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), giving a bijective correspondence between positive integers and integer partitions.

The Dyson rank of a nonempty partition is its maximum part minus its length. The rank of an empty partition is undefined.

T(n,k) is the number of partitions of n such that (greatest part) = k*(number of parts).

Column k > 1 is asymptotic to k! * Pi^k * exp(sqrt(2*Pi*n/3)) / (2^((k+4)/2) * 3^((k+1)/2) * n^((k+2)/2)). Equivalently, for fixed k > 1, T(n,k) ~ k! * Pi^k * A000041(n) / (6^(k/2) * n^(k/2)). - Vaclav Kotesovec, Oct 17 2024

For the definition of the k-th rank see A208478.

It appears that the sum of the k-th ranks of all partitions of n is equal to zero.

It appears that reversed rows converge to A000070, the same as A208478. - Omar E. Pol, Mar 10 2012

Column k=3 in the triangle A063995.

T(n,0) = A010054(n), T(n,1) = 1-A010054(n) for n>1;

A000009(n) = Sum(T(n,k): 0<=k

A117192(n) = Sum(T(n,k)*(1 - k mod 2): 0<=k

A117193(n) = Sum(T(n,k)*(k mod 2): 0<=k

A117194(n) = Sum(T(n,k)*(1 - k mod 2): 0

Examples

			Triangle starts:
[ 1]   1,
[ 2]   0, 1,
[ 3]   1, 0, 1,
[ 4]   0, 1, 0, 1,
[ 5]   0, 1, 1, 0, 1,
[ 6]   1, 0, 1, 1, 0, 1,
[ 7]   0, 1, 1, 1, 1, 0, 1,
[ 8]   0, 1, 1, 1, 1, 1, 0, 1,
[ 9]   0, 1, 1, 2, 1, 1, 1, 0, 1,
[10]   1, 0, 2, 1, 2, 1, 1, 1, 0, 1,
[11]   0, 1, 1, 2, 2, 2, 1, 1, 1, 0, 1,
[12]   0, 1, 2, 2, 2, 2, 2, 1, 1, 1, 0, 1,
[13]   0, 1, 1, 3, 2, 3, 2, 2, 1, 1, 1, 0, 1,
[14]   0, 1, 2, 2, 4, 2, 3, 2, 2, 1, 1, 1, 0, 1, ...
T(12,0) = #{} = 0,
T(12,1) = #{5+4+2+1} = 1,
T(12,2) = #{6+3+2+1, 5+4+3} = 2,
T(12,3) = #{6+5+1, 6+4+2} = 2,
T(12,4) = #{7+4+1, 7+3+2} = 2,
T(12,5) = #{8+3+1, 7+5} = 2,
T(12,6) = #{9+2+1, 8+4} = 2,
T(12,7) = #{9+3} = 1,
T(12,8) = #{10+2} = 1,
T(12,9) = #{11+1} = 1,
T(12,10) = #{} = 0,
T(12,11) = #{12} = 1.
		

Links

• Alois P. Heinz, Rows n = 1..141, flattened
• Maria Monks, Number theoretic properties of generating functions related to Dyson's rank for partitions into distinct parts, Proceedings of The American Mathematical Society, vol.138, no.02, pp.481-494, 2009.

Crossrefs

Cf. A063995, A105806.

Programs

• Maple

  b:= proc(n, i, k) option remember;
        if n<0 or k<0 then []
      elif n=0 then [0$k, 1]
      elif i<1 then []
      else zip ((x, y)-> x+y, b(n, i-1, k), b(n-i, i-1, k-1), 0)
        fi
      end:
  T:= proc(n) local j, r; r:= [];
        for j from 0 to n do
          r:= zip ((x, y)-> x+y, r, b(n-j, j-1, j-1), 0)
        od; r[]
      end:
  seq (T(n), n=1..20);  # Alois P. Heinz, Aug 29 2011

• Mathematica

  b[n_, i_, k_] := b[n, i, k] = Which[n<0 || k<0, {}, n == 0, Append[Array[0&, k], 1], i<1, {}, True, Plus @@ PadRight[{b[n, i-1, k], b[n-i, i-1, k-1]}]]; T[n_] := Module[{j, r}, r = {}; For[j = 0, j <= n, j++, r = Plus @@ PadRight[{r, b[n-j, j-1, j-1]}]]; r]; Table[T[n], {n, 1, 20}] // Flatten (* Jean-François Alcover, Jan 30 2014, after Alois P. Heinz *)

• PARI

  N=33;  L=1+2*ceil(sqrtint(N));
  q='q+O(q^N);
  gf=sum(n=1,L, q^(n*(n+1)/2) / prod(k=1,n,1-z*q^k) );
  v=Vec(gf);
  { for (n=1,#v,  /* print triangle: */
      p = Pol(v[n], 'z) + 'c0;
      p = polrecip(p);
      rw = Vec(p);  rw[1] -= 'c0;
      print1("[", n, "]   " );
      print( rw );
  ); }
  /* Joerg Arndt, Oct 07 2012 */

Formula

G.f.: sum(n>=1, q^(n*(n+1)/2) / prod(k=1..n, 1-z*q^k) ), see Monks reference. [Joerg Arndt, Oct 07 2012]

			a(6)=1 because the 11 partitions 6,51,42,411,33,321,3111,222,2211,21111,111111 have ranks 5,3,2,1,1,0,-1,-1,-2,-3,-5, respectively.
		

			The sequence of terms together with their prime indices begins:
     4: {1,1}
    12: {1,1,2}
    18: {1,2,2}
    27: {2,2,2}
    40: {1,1,1,3}
    60: {1,1,2,3}
    90: {1,2,2,3}
   100: {1,1,3,3}
   112: {1,1,1,1,4}
   135: {2,2,2,3}
   150: {1,2,3,3}
   168: {1,1,1,2,4}
   225: {2,2,3,3}
   250: {1,3,3,3}
   252: {1,1,2,2,4}
   280: {1,1,1,3,4}
   352: {1,1,1,1,1,5}
   375: {2,3,3,3}
   378: {1,2,2,2,4}
   392: {1,1,1,4,4}
		

			The sequence of terms together with their prime indices begins:
    3: {2}
    4: {1,1}
   10: {1,3}
   12: {1,1,2}
   15: {2,3}
   18: {1,2,2}
   25: {3,3}
   27: {2,2,2}
   28: {1,1,4}
   40: {1,1,1,3}
   42: {1,2,4}
   60: {1,1,2,3}
   63: {2,2,4}
   70: {1,3,4}
   88: {1,1,1,5}
   90: {1,2,2,3}
   98: {1,4,4}
  100: {1,1,3,3}
  105: {2,3,4}
  112: {1,1,1,1,4}