A227970 Triangular arithmetic on half-squares: b(n)*(b(n) - 1)/2 where b(n) = floor(n^2/2).

0, 0, 1, 6, 28, 66, 153, 276, 496, 780, 1225, 1770, 2556, 3486, 4753, 6216, 8128, 10296, 13041, 16110, 19900, 24090, 29161, 34716, 41328, 48516, 56953, 66066, 76636, 87990, 101025, 114960, 130816, 147696, 166753, 186966, 209628, 233586, 260281, 288420, 319600, 352380, 388521, 426426

Offset: 0

Author

Richard R. Forberg, Aug 01 2013

Keywords

• nonn
• easy
• changed

Comments

Analogous to A083374 for the squares A000290.

A "mirrored" repeating pattern of cycle length 20 exists in the last digit.

Links

• Index entries for linear recurrences with constant coefficients, signature (2,2,-6,0,6,-2,-2,1).

Crossrefs

Cf. A007590, A083374.

Programs

Maple

A227970:=n->binomial(floor(n^2/2),2); seq(A227970(k),k=0..50); # Wesley Ivan Hurt, Oct 05 2013

Mathematica

Table[Binomial[Floor[n^2/2], 2], {n, 0, 50}] (* Wesley Ivan Hurt, Sep 27 2013 *)

Formula

Let b(n) = floor(n^2/2), for n => 0, then a(n) = b(n)*(b(n)- 1)/2.

G.f.: -x^2*(1+4*x+14*x^2+4*x^3+x^4) / ( (1+x)^3*(x-1)^5 ). - R. J. Mathar, Aug 14 2013

a(n) = binomial(floor(n^2/2), 2). - Wesley Ivan Hurt, Sep 27 2013