A103923 Triangle of partitions of n with parts of sizes 1,2,...,m, each of two different kinds, m>=1.

1, 1, 1, 2, 2, 1, 3, 4, 2, 1, 5, 7, 5, 2, 1, 7, 12, 9, 5, 2, 1, 11, 19, 17, 10, 5, 2, 1, 15, 30, 28, 19, 10, 5, 2, 1, 22, 45, 47, 33, 20, 10, 5, 2, 1, 30, 67, 73, 57, 35, 20, 10, 5, 2, 1, 42, 97, 114, 92, 62, 36, 20, 10, 5, 2, 1, 56, 139, 170, 147, 102, 64, 36, 20, 10, 5, 2, 1, 77, 195

Offset: 0

Author

Wolfdieter Lang, Mar 24 2005

Keywords

• nonn
• easy
• tabl

Comments

The corresponding Fine-Riordan triangle is A008951.

This is the array p_2(n,m) of Gupta et al. written as a triangle. p_2(n,m) is defined on p. x of this reference as the number of partitions of n into parts consisting of two varieties of each of the integers 1 to m and one variety of each larger integer. Therefore a(n,m) gives these numbers for the partitions of n-m.

a(n,m)= sum over partitions of n+t(m)-m of binomial(q(partition),m), with t(m):=A000217(m) and q the number of distinct parts of a given partition. m>=0.

a(n,m)= number of partitions of 2*n-m with exactly m odd parts.

a(n,m)= sum over partitions of n+t(m)-m of product(k[j],j=1..m), with t(m):=A000217(m) and k[j]=number of parts of size j (exponent of j in a given partition of n), if m>=1. If m=0 then a(n,0)=p(n):=A000041(n) (number of partitions of n). 0 is counted as a part for n=0 and only for this n.

Examples

			Triangle starts:
[1];
[1,1];
[2,2,1];
[3,4,2,1];
[5,7,5,2,1];
...
a(4,2)=5 from the partitions of 4-2=2 with two varieties of parts 1 and of 2, namely (2),(2'),(1^2),(1'^2) and (1,1').
a(4,2)=5 from the partitions of 4+t(2)-2=5 which have products of the exponents of parts 1 and 2: 0*0,1*0,0*1,2*1,1*2,5*0 and sum to 4.
a(4,2)=5 from the partitions of 4+t(2)-2=5 which have number of distinct parts (q values) 1,2,2,2,2,2,1. The corresponding binomial(q,2) values are 0,1,1,1,1,0 and sum to 4.
a(4,2)=5 from the partitions of 2*4-2=6 with exactly two odd parts, namely (1,5), (3^2), (1^2,4), (1,2,3) and (1^2,2^2), which are 5 in number.
		

References

• H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958 (reprinted 1962), pp. 90-121.
• J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199.

Links

• Alois P. Heinz, Rows n = 0..140, flattened
• J. Huang, A. Senger, P. Wear, T. Wu, Partition statistics equidistributed with the number of hook difference one cells, 2013. See Remark 5.7. - _N. J. A. Sloane_, May 20 2014
• W. Lang: First 16 rows.

Crossrefs

The column sequences (without leading 0's) are, for m=0..10: A000041, A000070, A000097, A000098, A000710, A103924-A103929.

Cf. A000712, A124577.

Programs

Maple

with(numtheory):
b:= proc(n, k) option remember; `if`(n=0, 1, add(add(d*
      `if`(d<=k, 2, 1), d=divisors(j)) *b(n-j, k), j=1..n)/n)
    end:
A:= (n, k)-> b(n, k) -`if`(k=0, 0, b(n, k-1)):
seq(seq(A(n, k), k=0..n), n=0..14);  # Alois P. Heinz, Sep 14 2014

Mathematica

a[n_, 0] := a[n, 0] = PartitionsP[n]; a[n_, m_] /; n= m >= 0 := a[n, m] = a[n-1, m-1] + a[n-m, m]; Table[a[n, m], {n, 0, 14}, {m, 0, n}] // Flatten (* Jean-François Alcover, Dec 09 2014 *)
Flatten@Table[Length@IntegerPartitions[n-m, All, Range@n~Join~Range@m],  {n, 0, 12}, {m, 0, n}] (* Robert Price, Jul 29 2020 *)

Formula

a(n, m) = a(n-1, m-1) + a(n-m, m), n>=m>=0, with a(n, 0)= A000041(n) (partition numbers), a(n, m)=0 if n

a(n, m) = sum(a(n-1-j*m, m-1), j=0..floor((n-m)/m)), m>=1, input a(n, 0)= A000041(n).

G.f. column m: product(1/(1-x^j), j=1..m)*P(x), with P(x)= product(1/(1-x^j), j=1..infty), the o.g.f. for the partition numbers A000041.

G.f. column m>=1: (product(1/(1-x^k), k=1..m)^2)*product(1/(1-x^j), j=(m+1)..infty). For m=0 put the first product equal to 1.

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