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%I #21 Dec 07 2019 03:50:59
%S 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,
%T 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,
%U 5,5,6,5,6,6,5,6,5,5,12,15,7,6,8,7,8,8,8,7,8,6,7,15
%N 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.
%C Compare with A284592.
%H Alois P. Heinz, <a href="/A284593/b284593.txt">Antidiagonals n = 0..200, flattened</a>
%H H. S. Wilf, <a href="https://www.math.upenn.edu/~wilf/PIMS/PIMSLectures.pdf">Lectures on Integer Partitions</a>
%F 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).
%F Antidiagonal sums are A032302.
%e Square array begins
%e n\k| 0 1 2 3 4 5 6 7 8 9 10 11 12 13
%e - - - - - - - - - - - - - - - - - - - - - - - - - - - -
%e 0 | 1 1 1 2 2 3 4 5 6 8 10 12 15 18: A000009
%e 1 | 1 0 1 1 1 2 2 3 3 5 5 7 8 10: A096765
%e 2 | 1 1 0 1 2 2 2 3 4 5 6 7 9 11: A015744
%e 3 | 2 1 1 2 2 3 4 6 6 8 9 12 15 18
%e 4 | 2 1 2 2 2 3 5 5 7 9 10 14 15 19
%e 5 | 3 2 2 3 3 6 6 8 9 12 16 19 22 28
%e 6 | 4 2 2 4 5 6 8 9 11 16 18 22 27 33
%e 7 | 5 3 3 6 5 8 9 14 16 20 23 29 34 41
%e ...
%e 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).
%p # A284593 as a square array
%p ser := taylor(taylor(mul(1 + x^j + y^j, j = 1..10), x, 11), y, 11):
%p convert(ser, polynom):
%p s := convert(%, polynom):
%p with(PolynomialTools):
%p for n from 0 to 10 do CoefficientList(coeff(s, y, n), x) end do;
%p # second Maple program:
%p b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
%p b(n, i-1)+expand((x^i+1)*b(n-i, min(n-i, i-1)))))
%p end:
%p T:= (n, k)-> coeff(b(n+k$2), x, k):
%p seq(seq(T(n, d-n), n=0..d), d=0..14); # _Alois P. Heinz_, Aug 24 2019
%t nmax = 12; M = CoefficientList[#, y][[;; nmax+1]]& /@ (Product[1 + x^j + y^j, {j, 1, nmax}] + O[x]^(nmax+1) // CoefficientList[#, x]& // Expand);
%t T[n_, k_] := M[[n+1, k+1]];
%t Table[T[n-k, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Dec 07 2019 *)
%Y Cf. A000009 (row 0), A096765 (row 1), A015744 (row 2), A032032 (antidiagonal sums).
%Y Cf. A284592, A322210.
%Y Main diagonal gives 2*A108796 (for n>0).
%K nonn,tabl,easy
%O 0,7
%A _Peter Bala_, Mar 30 2017
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