A284593 Square array read by antidiagonals: T(n,k) = the number of pairs of partitions of n and k respectively, such that each partition is composed of distinct parts and the pair of partitions have no part in common.

1, 1, 1, 1, 0, 1, 2, 1, 1, 2, 2, 1, 0, 1, 2, 3, 1, 1, 1, 1, 3, 4, 2, 2, 2, 2, 2, 4, 5, 2, 2, 2, 2, 2, 2, 5, 6, 3, 2, 3, 2, 3, 2, 3, 6, 8, 3, 3, 4, 3, 3, 4, 3, 3, 8, 10, 5, 4, 6, 5, 6, 5, 6, 4, 5, 10, 12, 5, 5, 6, 5, 6, 6, 5, 6, 5, 5, 12, 15, 7, 6, 8, 7, 8, 8, 8, 7, 8, 6, 7, 15

Offset: 0

Author

Peter Bala, Mar 30 2017

Keywords

• nonn
• tabl
• easy

Comments

Compare with A284592.

Examples

			Square array begins
  n\k| 0  1  2  3  4  5  6   7   8   9  10  11  12  13
- - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0  | 1  1  1  2  2  3  4   5   6   8  10  12  15  18: A000009
  1  | 1  0  1  1  1  2  2   3   3   5   5   7   8  10: A096765
  2  | 1  1  0  1  2  2  2   3   4   5   6   7   9  11: A015744
  3  | 2  1  1  2  2  3  4   6   6   8   9  12  15  18
  4  | 2  1  2  2  2  3  5   5   7   9  10  14  15  19
  5  | 3  2  2  3  3  6  6   8   9  12  16  19  22  28
  6  | 4  2  2  4  5  6  8   9  11  16  18  22  27  33
  7  | 5  3  3  6  5  8  9  14  16  20  23  29  34  41
  ...
T(3,7) = 6: the six pairs of partitions of 3 and 7 into distinct parts and with no parts in common are (3, 7), (3, 6 + 1), (3, 5 + 2), (3, 4 + 2 + 1), (2 + 1, 7) and (2 + 1, 4 + 3).
		

Links

• Alois P. Heinz, Antidiagonals n = 0..200, flattened
• H. S. Wilf, Lectures on Integer Partitions

Crossrefs

Rows n=0..2 give A000009, A096765, A015744.

Main diagonal gives A365662.

Antidiagonal sums give A032302.

Cf. A284592, A322210.

Programs

Maple

# A284593 as a square array
ser := taylor(taylor(mul(1 + x^j + y^j, j = 1..10), x, 11), y, 11):
convert(ser, polynom):
s := convert(%, polynom):
with(PolynomialTools):
for n from 0 to 10 do CoefficientList(coeff(s, y, n), x) end do;
# second Maple program:
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
      b(n, i-1)+expand((x^i+1)*b(n-i, min(n-i, i-1)))))
    end:
T:= (n, k)-> coeff(b(n+k$2), x, k):
seq(seq(T(n, d-n), n=0..d), d=0..14);  # Alois P. Heinz, Aug 24 2019

Mathematica

nmax = 12; M = CoefficientList[#, y][[;; nmax+1]]& /@ (Product[1 + x^j + y^j, {j, 1, nmax}] + O[x]^(nmax+1) // CoefficientList[#, x]& // Expand);
T[n_, k_] := M[[n+1, k+1]];
Table[T[n-k, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 07 2019 *)

Formula

O.g.f. Product_{j >= 1} (1 + x^j + y^j) = Sum_{n,k >= 0} T(n,k)*x^n*y^k (see Wilf, Example 7).

Antidiagonal sums are A032302.