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A195822 Triangle read by rows in which row n lists the Dyson's ranks of all partitions of n that do not contain 1 as a part, in nonincreasing order.

%I A195822 #33 Mar 20 2019 04:24:02
%S A195822 0,1,2,3,0,4,1,5,2,1,-1,6,3,2,0,7,4,3,2,1,0,-2,8,5,4,3,2,1,0,-1,9,6,5,
%T A195822 4,3,3,2,1,1,0,-1,-3,10,7,6,5,4,4,3,2,2,1,1,0,-1,-2,11,8,7,6,5,5,4,4,
%U A195822 3,3,2,2,2,1,1,0,0,-1,-1,-2,-4
%N A195822 Triangle read by rows in which row n lists the Dyson's ranks of all partitions of n that do not contain 1 as a part, in nonincreasing order.
%C A195822 The sum of row n is equal to A000041(n-1), if n >= 2. Proof: Dyson defined the rank of a partition as the largest part minus the number of parts. On the other hand the total number of parts in all partitions of n equals the sum of largest parts of all partitions of n (Cf. A006128), hence the sum of the ranks of all partitions of n is equal to zero. Let p(n) be the number of partitions of n. If we now add an part equal to 1 in each partition of n we obtain the partitions of n+1 that contain 1 as a part. The sum of the ranks of these partitions is p(n)*(-1) because the largest parts are the same but now there is an additional part in each partition. On the other hand the sum of the ranks of all partitions of n+1 is equal to zero, hence the sum of the ranks of all partitions of n+1 that do not contain 1 as a part is equal to p(n).
%H A195822 A. O. L. Atkins and F. G. Garvan, <a href="https://arxiv.org/abs/math/0208050">Relations between the ranks and cranks of partitions</a>, arXiv:math/0208050 [math.NT], 2002.
%e A195822 Triangle begins:
%e A195822 0;
%e A195822 1;
%e A195822 2;
%e A195822 3, 0;
%e A195822 4, 1;
%e A195822 5, 2, 1, -1;
%e A195822 6, 3, 2, 0;
%e A195822 7, 4, 3, 2, 1, 0, -2;
%e A195822 8, 5, 4, 3, 2, 1, 0, -1;
%e A195822 9, 6, 5, 4, 3, 3, 2, 1, 1, 0, -1, -3;
%e A195822 10, 7, 6, 5, 4, 4, 3, 2, 2, 1, 1, 0, -1, -2;
%Y A195822 Row n has length A002865(n), n >= 2.
%Y A195822 Cf. A000041, A105805, A135010.
%K A195822 sign,tabf
%O A195822 1,3
%A A195822 _Omar E. Pol_, Nov 06 2011