A123596 Squares alternating with triangular numbers.

0, 0, 1, 1, 4, 3, 9, 6, 16, 10, 25, 15, 36, 21, 49, 28, 64, 36, 81, 45, 100, 55, 121, 66, 144, 78, 169, 91, 196, 105, 225, 120, 256, 136, 289, 153, 324, 171, 361, 190, 400, 210, 441, 231, 484, 253, 529, 276, 576, 300, 625, 325, 676, 351, 729, 378, 784, 406, 841, 435

Offset: 0

Author

Peter Hansen (babyskbaby(AT)web.de), Nov 14 2006

Keywords

• nonn

Comments

Rearrangement of A054686.

For n >= 2, a(n) is the number of distinct positive slopes of least squares regression lines fitted to n points (j,y_j), 1 <= j <= n, where all y_j are 0 or 1. The total number of distinct slopes is 2*a(n)+1 (a(n) positive, a(n) negative, and the zero slope). - Pontus von Brömssen, Mar 10 2024

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Crossrefs

Cf. A002620, A054686, A124093.

Programs

Magma

[(3*n^2-1+(n^2+1)*(-1)^n)/16: n in [0..10]]; // G. C. Greubel, Oct 26 2017

Mathematica

CoefficientList[Series[x^2*(1+x+x^2)/((1-x)^3*(1+x)^3), {x, 0, 50}], x] (* or *) Table[(3*n^2-1+(n^2+1)*(-1)^n)/16, {n,0,50}] (* G. C. Greubel, Oct 26 2017 *)
With[{nn=30},Riffle[Range[0,nn]^2,Accumulate[Range[0,nn]]]] (* or *) LinearRecurrence[{0,3,0,-3,0,1},{0,0,1,1,4,3},60] (* Harvey P. Dale, Feb 11 2020 *)

PARI

{a(n) = if(n%2,(n^2-1)/8,n^2/4)} \\ Michael Somos, Nov 18 2006

Formula

a(2*n) = n^2, a(2*n+1) = (n^2+n)/2.

From R. J. Mathar, Feb 12 2010: (Start)

a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).

G.f.: x^2*(1+x+x^2)/((1-x)^3*(1+x)^3). (End)

a(n) = (3*n^2-1+(n^2+1)*(-1)^n)/16. - Luce ETIENNE, May 30 2015

Extensions

Edited by Michael Somos, and several other correspondents, Nov 14 2005