A215415 a(2*n) = n, a(4*n+1) = 2*n-1, a(4*n+3) = 2*n+3.

0, -1, 1, 3, 2, 1, 3, 5, 4, 3, 5, 7, 6, 5, 7, 9, 8, 7, 9, 11, 10, 9, 11, 13, 12, 11, 13, 15, 14, 13, 15, 17, 16, 15, 17, 19, 18, 17, 19, 21, 20, 19, 21, 23, 22, 21, 23, 25, 24, 23, 25, 27, 26, 25, 27, 29, 28, 27, 29, 31, 30, 29, 31, 33, 32, 31, 33, 35, 34, 33, 35, 37

Offset

0,4

Comments

a(n) and higher order differences in further rows:

0, -1, 1, 3, 2, 1,

-1, 2, 2, -1, -1, -2, A134430(n).

3, 0, -3, 0, 3, 0,

-3, -3, 3, 3, -3, -3,

0, 6, 0, -6, 0, 6,

6, -6, -6, 6, 6, -6.

a(n) is the binomial transform of 0, -1, 3, -3, 0, 6, -12, 12, 0, -24, 48, -48, 0, 96..., essentially negated A134813.

By definition, all differences a(n+k)-a(n) are periodic sequences with period length 4. For k=1, 3 and 4 these are A134430, A021307 and A007395, for example.

Links

Colin Barker, Table of n, a(n) for n = 0..1000

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

Formula

a(2*n) = n, a(2*n+1) = A097062(n+1).

a(n) = (A214297(n+1) - A214297(n-1))/2.

a(3*n) =3*A004525(n).

a(n) = +2*a(n-1) -2*a(n-2) +2*a(n-3) -a(n-4).

G.f. -x*(1-3*x+x^2) / ( (x^2+1)*(x-1)^2 ). - R. J. Mathar, Aug 11 2012

a(n) = ((-3*I)*((-I)^n-I^n)+2*n)/4. - Colin Barker, Oct 19 2015

Mathematica

Flatten[Table[{2n, 2n - 1, 2n + 1, 2n + 3}, {n, 0, 19}]] (* Alonso del Arte, Aug 09 2012 *)

Prog

(PARI) a(n) = ((-3*I)*((-I)^n-I^n)+2*n)/4 \\ Colin Barker, Oct 19 2015
(PARI) concat(0, Vec(-x*(1-3*x+x^2)/((x^2+1)*(x-1)^2) + O(x^100))) \\ Colin Barker, Oct 19 2015

Crossrefs

Quadrisections: A005843, A060747, A005408, A144396.

Sequence in context: A318317 A129690 A156665 * A244477 A035572 A325531

Adjacent sequences: A215412 A215413 A215414 * A215416 A215417 A215418

Keyword

sign,easy,less

Author

Paul Curtz, Aug 09 2012

Status

approved