login
Imbalance of the sum of largest parts of all partitions of n.
2

%I #14 Oct 16 2018 11:23:54

%S 0,-2,1,-5,3,-12,7,-25,17,-47,36,-88,69,-155,133,-262,240,-439,415,

%T -717,705,-1142,1165,-1803,1874,-2797,2975,-4276,4632,-6478,7094,

%U -9698,10741,-14355,16059,-21079,23719,-30670,34716,-44243,50315,-63372

%N Imbalance of the sum of largest parts of all partitions of n.

%C Consider the three-dimensional structure of the shell model of partitions version "tree". Note that only the larges parts > 1 produce the imbalance. Note that every column where is located a largest part contains largest parts of the same size, thesame as a periodic table (see example). For more information see A135010.

%F a(n) = Sum_{k=2..n} (-1)^(k-1)*A138137(k), n >= 2.

%e For n = 6 the illustration of the shell model with 6 shells shows an imbalance of largest parts (see below):

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

%e Partitions Tree Table 1.0

%e of 6. A194805 A135010

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

%e 6 6 6 . . . . .

%e 3+3 3 3 . . 3 . .

%e 4+2 4 4 . . . 2 .

%e 2+2+2 2 2 . 2 . 2 .

%e 5+1 1 5 5 . . . . 1

%e 3+2+1 1 3 3 . . 2 . 1

%e 4+1+1 4 1 4 . . . 1 1

%e 2+2+1+1 2 1 2 . 2 . 1 1

%e 3+1+1+1 1 3 3 . . 1 1 1

%e 2+1+1+1+1 2 1 2 . 1 1 1 1

%e 1+1+1+1+1+1 1 1 1 1 1 1 1

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

%e The sum of largest parts > 1 on the left hand side is 23 and the sum of largest parts > 1 on the right hand side is 11, so a(6) = -23 + 11 = -12. On the other hand for n = 6 we have that 0 together with the first n-1 terms > 1 of A138137 are 0, 2, 3, 6, 8, 15 so a(6) = 0-2+3-6+8-15 = -12.

%Y Cf. A135010, A138121, A138137, A141285, A194795-A194797, A194805.

%K sign

%O 1,2

%A _Omar E. Pol_, Feb 02 2012