A251091 a(n) = n^2 / gcd(n+2, 4).

0, 1, 1, 9, 8, 25, 9, 49, 32, 81, 25, 121, 72, 169, 49, 225, 128, 289, 81, 361, 200, 441, 121, 529, 288, 625, 169, 729, 392, 841, 225, 961, 512, 1089, 289, 1225, 648, 1369, 361, 1521, 800, 1681, 441, 1849, 968, 2025, 529, 2209, 1152, 2401, 625, 2601, 1352

Offset: 0

Author

Paul Curtz, May 08 2015

Keywords

• nonn
• easy
• mult

Comments

A061038(n), which appears in 4*a(n) formula, is a permutation of n^2.

Origin. In December 2010, I wrote in my 192-page Exercise Book no. 5, page 41, the array (difference table of the first row):

1 0, 1/3, 1, 9/5, 8/3, 25/7, 9/2, 49/9, ...

-1, 1/3, 2/3, 4/5, 13/15, 19/21, 13/14, 17/18, 43/45, ...

Numerators are listed in A176126, denominators are in A064038, and denominator - numerator = 2, 2, 1, 1,... (A014695).

4/3, 1/3, 2/15, 1/15, 4/105, 1/42, 1/63, 1/90, 4/495, ...

-1, -1/5, -1/15, -1/35, -1/70, -1/126, -1/210, -1/330, -1/495, ...

where the denominators of the second row are listed in A000332.

Also for those of the inverse binomial transform

1, -1, 4/3, -1, 4/5, -2/3, 4/7, -1/2, 4/9, -2/5, 4/11, -1/3, ... ?

a(n) is the (n+1)-th term of the numerators of the first row.

Examples

			a(0) = 0/2, a(1) = 1/1, a(2) = 4/4, a(3) = 9/1.
		

Links

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

Crossrefs

Cf. A000290, A000332, A016754, A026741, A060819, A061038, A109008, A139098, A168077, A176895, A181900.

Programs

Magma

[(1-(1/16)*(1+(-1)^n)*(5-(-1)^(n div 2)) )*n^2: n in [0..60]]; // Vincenzo Librandi, Jun 12 2015

Maple

seq(seq((4*i+j-1)^2/[2,1,4,1][j],j=1..4),i=0..30); # Robert Israel, May 14 2015

Mathematica

f[n_] := Switch[ Mod[n, 4], 0, n^2/2, 1, n^2, 2, n^2/4, 3, n^2]; Array[f, 50, 0] (* or *) Table[(4 i + j - 1)^2/{2, 1, 4, 1}[[j]], {i, 0, 12}, {j, 4}] // Flatten (* after Robert Israel *) (* or *) LinearRecurrence[{0, 0, 0, 3, 0, 0, 0, -3, 0, 0, 0, 1}, {0, 1, 1, 9, 8, 25, 9, 49, 32, 81, 25, 121}, 53] (* or *) CoefficientList[ Series[-((x (1 + x (1 + x (9 + x (8 + x (22 + x (6 + x (22 + x (8 + x (9 + x + x^2))))))))))/(-1 + x^4)^3), {x, 0, 52}], x] (* Robert G. Wilson v, May 19 2015 *)

PARI

concat(0, Vec(-x*(x^10 + x^9 + 9*x^8 + 8*x^7 + 22*x^6 + 6*x^5 + 22*x^4 + 8*x^3 + 9*x^2 + x + 1) / ((x-1)^3*(x+1)^3*(x^2+1)^3) + O(x^100))) \\ Colin Barker, May 14 2015

Formula

a(n) = n^2/(period 4: repeat 2, 1, 4, 1).

a(4n) = 8*n^2, a(2n+1) = a(4n+2) = (2*n+1)^2.

a(n+4) = a(n) + 8*A060819(n).

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12), n>11.

4*a(n) = (period 4: repeat 2, 1, 4, 1) * A061038(n).

G.f.: -x*(x^10+x^9+9*x^8+8*x^7+22*x^6+6*x^5+22*x^4+8*x^3+9*x^2+x+1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - Colin Barker, May 14 2015

a(2n) = A181900(n), a(2n+1) = A016754(n). [Bruno Berselli, May 14 2015]

a(n) = ( 1 - (1/16)*(1+(-1)^n)*(5-(-1)^(n/2)) )*n^2. - Bruno Berselli, May 14 2015

Sum_{n>=1} 1/a(n) = 13*Pi^2/48. - Amiram Eldar, Aug 12 2022

Extensions

Missing term (1521) inserted in the sequence by Colin Barker, May 14 2015

Definition uses a formula by Jean-François Alcover, Jul 01 2015

Keyword:mult added by Andrew Howroyd, Aug 06 2018