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%I #40 Jan 24 2020 03:23:12
%S 0,1,-1,2,0,-2,3,1,0,-1,-3,4,2,1,0,-1,-2,-4,5,3,2,1,1,0,-1,-1,-2,-3,
%T -5,6,4,3,2,2,1,0,0,0,-1,-2,-2,-3,-4,-6,7,5,4,3,2,3,2,1,1,0,1,0,-1,-1,
%U -2,-1,-2,-3,-3,-4,-5,-7,8,6,5,4,3,4,3,2,1,2,1,0,2,1,0,0,-1,-1,0,-1,-2,-2,-3,-2,-3,-4,-4,-5,-6,-8,9,7,6,5,4,3,5,4,3
%N Irregular triangle read by rows: T(n,k) is the Dyson's rank of the k-th partition of n in Abramowitz-Stegun order.
%C The rank of a partition is the largest part minus the number of parts.
%C Row lengths give A000041, n >= 1.
%C Just for n <= 6, row n is antisymmetric due to conjugation of partitions (see links under A105806): a(n, p(n)-(k-1)) = a(n,k), k = 1..floor(p(n)/2). [Comment corrected by _Franklin T. Adams-Watters_, Jan 17 2006]
%C First differs from A330368 at a(49) = T(7,5). - _Omar E. Pol_, Dec 31 2019
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%H A. M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, pp. 831-2.
%H Freeman J. Dyson, <a href="https://www.jstor.org/stable/2304400">Problems for solution nr. 4261</a>, Am. Math. Month. 54 (1947) 418.
%H Wolfdieter Lang, <a href="/A105805/a105805.txt"> First 16 rows.</a>
%F a(n,k) = A049085(n,k) - A036043(n,k). - _Alford Arnold_, Aug 02 2010
%e Triangle begins:
%e [0];
%e [1, -1];
%e [2, 0, -2];
%e [3, 1, 0, -1, -3];
%e [4, 2, 1, 0, -1, -2, -4];
%e [5, 3, 2, 1, 1, 0, -1, -1, -2, -3, -5];
%e ...
%e Row 3 for partitions of 3 in the mentioned order: 3,(1,2),1^3 with ranks 2,0,-2.
%e From _Wolfdieter Lang_, Jul 18 2013: (Start)
%e Row n = 7 is [6, 4, 3, 2, 2, 1, 0 , 0, 0, -1, -2, -2, -3, -4, -6].
%e This is also antisymmetric, but by accident, because a(7,7) = 0 for the partition (1,3^2), conjugate to (2^2,3) with a(7,8) = 0, and a(7,9) = 0 for (1^3,4) which is self-conjugate.
%e Row n=8 (see the link) is no longer antisymmetric. See the _Franklin T. Adams-Watters_ correction above. (End)
%p # ASPrts is implemented in A119441
%p A105805 := proc(n,k)
%p local pi;
%p pi := ASPrts(n)[k] ;
%p max(op(pi))-nops(pi) ;
%p end proc:
%p for n from 1 do
%p for k from 1 to A000041(n) do
%p printf("%d,",A105805(n,k)) ;
%p end do:
%p printf("\n") ;
%p end do: # _R. J. Mathar_, Jul 17 2013
%Y Cf. A000041, A036043, A049085, A209616 (sum of positive ranks), A330368 (another version).
%K sign,easy,tabf
%O 1,4
%A _Wolfdieter Lang_, Apr 28 2005
%E Name clarified by _Omar E. Pol_, Dec 31 2019
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