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Triangle read by rows: row n (n>=0) gives the number of partitions of (n,0), (n-1,1), (n-2,2), ..., (0,n) respectively into sums of distinct pairs.
13

%I #92 Aug 17 2020 01:59:37

%S 1,1,1,1,2,1,2,3,3,2,2,5,5,5,2,3,7,9,9,7,3,4,10,14,17,14,10,4,5,14,21,

%T 27,27,21,14,5,6,19,31,42,46,42,31,19,6,8,25,44,64,74,74,64,44,25,8,

%U 10,33,61,93,116,123,116,93,61,33,10

%N Triangle read by rows: row n (n>=0) gives the number of partitions of (n,0), (n-1,1), (n-2,2), ..., (0,n) respectively into sums of distinct pairs.

%C By analogy with ordinary partitions into distinct parts (A000009). The empty partition gives T(0,0)=1 by definition. A054225 and A201376 give pair partitions with repeats allowed.

%C Also number of partitions into pairs which are not both even.

%C In the paper by S. M. Luthra: "Partitions of bipartite numbers when the summands are unequal", the square table on page 370 contains an errors. In the formula (6, p. 372) for fixed m there should be factor 1/m!. The correct asymptotic formula is q(m, n) ~ (sqrt(12*n)/Pi)^m * exp(Pi*sqrt(n/3)) / (4*3^(1/4)*m!*n^(3/4)). The same error is also in article by F. C. Auluck (see A054225). - _Vaclav Kotesovec_, Feb 02 2016

%H Alois P. Heinz, <a href="/A054242/b054242.txt">Rows n = 0..75, flattened</a>

%H S. M. Luthra, <a href="http://www.insa.nic.in/writereaddata/UpLoadedFiles/PINSA/Vol23A_1957_5_Art04.pdf">Partitions of bipartite numbers when the summands are unequal</a>, Proceedings of the Indian National Science Academy, vol.23, 1957, issue 5A, p. 370-376. [broken link]

%H Reinhard Zumkeller, <a href="/A054225/a054225_1.lhs.txt">Haskell programs for A054225, A054242, A201376, A201377</a>

%F G.f.: (1/2)*Product(1+x^i*y^j), i, j>=0.

%e The second row (n=1) is 1,1 since (1,0) and (0,1) each have a single partition.

%e The third row (n=2) is 1, 2, 1 from (2,0), (1,1) or (1,0)+(0,1), (0,2).

%e In the fourth row, T(1,3)=5 from (1,3), (0,3)+(1,0), (0,2)+(1,1), (0,2)+(0,1)+(1,0), (0,1)+(1,2).

%e The triangle begins:

%e 1;

%e 1, 1;

%e 1, 2, 1;

%e 2, 3, 3, 2;

%e 2, 5, 5, 5, 2;

%e 3, 7, 9, 9, 7, 3;

%e 4, 10, 14, 17, 14, 10, 4;

%e 5, 14, 21, 27, 27, 21, 14, 5;

%e 6, 19, 31, 42, 46, 42, 31, 19, 6;

%e 8, 25, 44, 64, 74, 74, 64, 44, 25, 8;

%e ...

%t max = 10; f[x_, y_] := Product[1 + x^n*y^k, {n, 0, max}, {k, 0, max}]/2; se = Series[f[x, y], {x, 0, max}, {y, 0, max}] ; coes = CoefficientList[ se, {x, y}]; t[n_, k_] := coes[[n-k+1, k+1]]; Flatten[ Table[ t[n, k], {n, 0, max}, {k, 0, n}]] (* _Jean-François Alcover_, Dec 06 2011 *)

%o (Haskell) see Zumkeller link.

%Y See A201377 for the same triangle formatted in a different way.

%Y The outer diagonals are T(n,0) = T(n,n) = A000009(n).

%Y Cf. A054225.

%Y T(2*n,n) = A219554(n). Row sums give A219555. - _Alois P. Heinz_, Nov 22 2012

%Y Columns 0-5: A000009, A036469, A268345, A268346, A268347, A268348.

%K easy,nonn,tabl,nice

%O 0,5

%A _Marc LeBrun_, Feb 08 2000 and Jul 01 2003

%E Entry revised by _N. J. A. Sloane_, Nov 30 2011, to incorporate corrections provided by _Reinhard Zumkeller_, who also contributed the alternative version A201377.