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Irregular triangle read by rows: T(n,k), n >= 1, -(n-1) <= k <= n-1, = number of partitions of n with rank k.
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%I #40 May 21 2023 09:46:46

%S 1,1,0,1,1,0,1,0,1,1,0,1,1,1,0,1,1,0,1,1,1,1,1,0,1,1,0,1,1,2,1,2,1,1,

%T 0,1,1,0,1,1,2,1,3,1,2,1,1,0,1,1,0,1,1,2,2,3,2,3,2,2,1,1,0,1,1,0,1,1,

%U 2,2,3,3,4,3,3,2,2,1,1,0,1,1,0,1,1,2,2,4,3,5,4,5,3,4,2,2,1,1,0,1,1,0,1,1,2

%N Irregular triangle read by rows: T(n,k), n >= 1, -(n-1) <= k <= n-1, = number of partitions of n with rank k.

%C The rank of a partition is the largest part minus the number of parts.

%C The rows are symmetric: for every partition of rank r there is its conjugate with rank -r. [_Joerg Arndt_, Oct 07 2012]

%H Reinhard Zumkeller, <a href="/A063995/b063995.txt">Rows n = 1..72 of triangle, flattened</a>

%H G. E. Andrews, <a href="http://www.math.psu.edu/vstein/alg/antheory/preprint/andrews/17.pdf">The number of smallest parts in the partitions of n</a>. [Also Selected Works, p. 603, see N(m,n).] - _N. J. A. Sloane_, Dec 16 2013

%H A. O. L. Atkin and P. Swinnerton-Dyer, <a href="https://doi.org/10.1112/plms/s3-4.1.84">Some properties of partitions</a>, Proc. London Math. Soc. (3) 4, (1954). 84-106. Math. Rev. 15,685d.

%H Alexander Berkovich and Frank G. Garvan, <a href="https://doi.org/10.1006/jcta.2002.3281">Some observations on Dyson's new symmetries of partitions</a>, Journal of Combinatorial Theory, Series A 100.1 (2002): 61-93.

%H Freeman J. Dyson, <a href="https://doi.org/10.1016/S0021-9800(69)80006-2">A new symmetry of partitions</a>, Journal of Combinatorial Theory 7.1 (1969): 56-61. See Table 1.

%H Freeman J. Dyson, <a href="https://doi.org/10.1016/0097-3165(89)90043-5">Mappings and symmetries of partitions</a>, J. Combin. Theory Ser. A 51 (1989), 169-180.

%e The partition 5 = 4+1 has largest summand 4 and 2 summands, hence has rank 4-2 = 2.

%e Triangle begins:

%e [ 1] 1,

%e [ 2] 1, 0, 1,

%e [ 3] 1, 0, 1, 0, 1,

%e [ 4] 1, 0, 1, 1, 1, 0, 1,

%e [ 5] 1, 0, 1, 1, 1, 1, 1, 0, 1,

%e [ 6] 1, 0, 1, 1, 2, 1, 2, 1, 1, 0, 1,

%e [ 7] 1, 0, 1, 1, 2, 1, 3, 1, 2, 1, 1, 0, 1,

%e [ 8] 1, 0, 1, 1, 2, 2, 3, 2, 3, 2, 2, 1, 1, 0, 1,

%e [ 9] 1, 0, 1, 1, 2, 2, 3, 3, 4, 3, 3, 2, 2, 1, 1, 0, 1,

%e [10] 1, 0, 1, 1, 2, 2, 4, 3, 5, 4, 5, 3, 4, 2, 2, 1, 1, 0, 1,

%e [11] 1, 0, 1, 1, 2, ...

%e Row 20 is:

%e T(20, k) = 1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 12, 14, 20, 22, 30, 33, 40, 42, 48, 45, 48, 42, 40, 33, 30, 22, 20, 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1; -19 <= k <= 19.

%e Another view of the table of p(n,m) = number of partitions of n with rank m, taken from Dyson (1969):

%e n\m -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

%e -----------------------------------------------------

%e 0 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,

%e 1 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,

%e 2 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0,

%e 3 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0,

%e 4 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0,

%e 5 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0,

%e 6 0, 1, 0, 1, 1, 2, 1, 2, 1, 1, 0, 1, 0,

%e 7 1, 0, 1, 1, 2, 1, 3, 1, 2, 1, 1, 0, 1,

%e ...

%e The central triangle is the present sequence, the right-hand triangle is A105806. - _N. J. A. Sloane_, Jan 23 2020

%t Table[ Count[ (First[ # ]-Length[ # ]& /@ IntegerPartitions[ k ]), # ]& /@ Range[ -k+1, k-1 ], {k, 16} ]

%o (Haskell)

%o import Data.List (sort, group)

%o a063995 n k = a063995_tabf !! (n-1) !! (n-1+k)

%o a063995_row n = a063995_tabf !! (n-1)

%o a063995_tabf = [[1], [1, 0, 1]] ++ (map

%o (\rs -> [1, 0] ++ (init $ tail $ rs) ++ [0, 1]) $ drop 2 $ map

%o (map length . group . sort . map rank) $ tail pss) where

%o rank ps = maximum ps - length ps

%o pss = [] : map (\u -> [u] : [v : ps | v <- [1..u],

%o ps <- pss !! (u - v), v <= head ps]) [1..]

%o -- _Reinhard Zumkeller_, Jul 24 2013

%Y For the number of partitions of n with rank 0 (balanced partitions) see A047993.

%Y Cf. A105806 (right half of triangle), A005408 (row lengths), A000041 (row sums), A047993 (central terms).

%K nonn,nice,tabf

%O 1,30

%A _N. J. A. Sloane_, Sep 19 2001

%E More terms from _Vladeta Jovovic_ and _Wouter Meeussen_, Sep 19 2001