

A215415


a(2*n) = n, a(4*n+1) = 2*n1, a(4*n+3) = 2*n+3.


1



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
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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(n1))/2.
a(3*n) =3*A004525(n).
a(n) = +2*a(n1) 2*a(n2) +2*a(n3) a(n4).
G.f. x*(13*x+x^2) / ( (x^2+1)*(x1)^2 ).  R. J. Mathar, Aug 11 2012
a(n) = ((3*I)*((I)^nI^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)^nI^n)+2*n)/4 \\ Colin Barker, Oct 19 2015
(PARI) concat(0, Vec(x*(13*x+x^2)/((x^2+1)*(x1)^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



