A147685 Squares and centered square numbers interleaved.

0, 1, 1, 5, 4, 13, 9, 25, 16, 41, 25, 61, 36, 85, 49, 113, 64, 145, 81, 181, 100, 221, 121, 265, 144, 313, 169, 365, 196, 421, 225, 481, 256, 545, 289, 613, 324, 685, 361, 761, 400, 841, 441, 925, 484, 1013, 529, 1105, 576, 1201, 625, 1301, 676, 1405, 729, 1513

Offset: 0

Author

R. J. Mathar, Nov 10 2008

Keywords

• easy
• nonn

Comments

This could be called the inverse Motzkin transform of A109188 since the substitution x -> x/(1+x+x^2) in the independent variable of the g.f. A109188(x) yields this sequence here.

Examples

			G.f. = x + x^2 + 5*x^3 + 4*x^4 + 13*x^5 + 9*x^6 + 25*x^7 + 16*x^8 + 41*x^9 + ...
		

Links

• Vincenzo Librandi, 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. A000290, A001844, A109188, A120328.

Programs

Maple

A147685:=n->n^2*(1+(-1)^n)/8+(n^2+1)*(1-(-1)^n)/4: seq(A147685(n), n=0..70); # Wesley Ivan Hurt, Sep 06 2015

Mathematica

CoefficientList[Series[x (1 + x + x^2) (1 + x^2)/((1 - x)^3 (1 + x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 07 2014 *)

PARI

{a(n) = if( n%2, (n^2 + 1) / 2, n^2 / 4)}; /* Michael Somos, Aug 07 2014 */

Formula

a(2*n) = A000290(n), a(2*n+1) = A001844(n).

O.g.f.: x*(1+x+x^2)*(1+x^2)/((1-x)^3*(1+x)^3).

a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6), n>5.

Euler transform of length 4 sequence [ 1, 4, -1, -1]. - Michael Somos, Aug 07 2014

a(2n+1) = a(2n) + a(2n+2) for all n in Z. - Michael Somos, Aug 07 2014

A120328(n-1) = 3*n^2 + 2 = a(2*n + 1) - a(2*n)+ a(2*n - 1) for all n in Z. - Michael Somos, Aug 07 2014

a(n) = n^2*(1+(-1)^n)/8+(n^2+1)*(1-(-1)^n)/4. - Wesley Ivan Hurt, Sep 06 2015