A246581 G.f.: x^((k^2 + k)/2) / (Product_{i=1..k} (1 - x^i) * Product_{r>=1} (1 + x^r)) with k = 2.

0, 0, 0, 1, 0, 1, -1, 1, -2, 1, -3, 2, -4, 3, -5, 5, -6, 7, -8, 10, -10, 13, -13, 17, -17, 21, -22, 27, -28, 33, -36, 41, -45, 50, -56, 62, -69, 75, -85, 92, -103, 111, -125, 135, -150, 162, -180, 195, -215, 232, -256, 278, -303, 329, -359, 390, -423

Offset: 0

Author

N. J. A. Sloane, Aug 31 2014

Keywords

• sign

Comments

Empirical: 2*(-1)^n*a(n+1) is equal to the number of partitions mu of n such that the diagram of mu and the diagram of the transpose of mu have exactly n-1 cells in common (see below example). - John M. Campbell, Feb 01 2016

Examples

			From _John M. Campbell_, Feb 01 2016: (Start)
For example, letting n=9, there are 2*(-1)^n*a(n+1) = (-2)*(-3) = 6 partitions mu of n=9 such that the diagram of mu and the diagram of the transpose of mu have exactly n-1 cells in common: (5,2,1,1), (4,3,2), (4,3,1,1), (4,2,2,1), (4,2,1,1,1), (3,3,2,1). For example, the diagram of (3,3,2,1) is
   ooo
   ooo
   oo
   o
and the diagram of the transpose of (3,3,2,1) is
   oooo
   ooo
   oo
and these diagrams share exactly (n-1)=8 cells in common, when the diagrams are positioned so that the upper-left corners of both diagrams coincide. (End)
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..5000
• J. Fulman, Random matrix theory over finite fields, Bull. Amer. Math. Soc., 39 (No. 1, 2002), 51-85, MR1864086 (2002i:60012). See top of page 70, Eq. 3, with k=2.

Crossrefs

For k=0 and 1 we get A081362, A027349 (apart from signs).

Cf. A218907, A246582, A246583.

Programs

Maple

fSp:=proc(k) local a,i,r;
a:=x^((k^2+k)/2)/mul(1-x^i,i=1..k);
a:=a/mul(1+x^r,r=1..101);
series(a,x,101);
seriestolist(%);
end;
fSp(2);

Mathematica

k = 2; CoefficientList[Series[x^((k^2 + k)/2)/(Product[1 - x^i, {i, k}] Product[1 + x^r, {r, 1000}]), {x, 0, 56}], x] (* Michael De Vlieger, Feb 01 2016 *)

Formula

G.f.: x^3/((1-x)*(1-x^2)) * Product_{k>=1} 1/(1+x^k). - Vaclav Kotesovec, Mar 12 2016

a(n) ~ (-1)^(n+1) * 3^(1/4) * exp(sqrt(n/6)*Pi) / (2^(9/4)*Pi*n^(1/4)). - Vaclav Kotesovec, Mar 12 2016