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A225058
a(4*n) = n-1. a(2*n+1) = a(4*n+2) = 2*n+1.
1
-1, 1, 1, 3, 0, 5, 3, 7, 1, 9, 5, 11, 2, 13, 7, 15, 3, 17, 9, 19, 4, 21, 11, 23, 5, 25, 13, 27, 6, 29, 15, 31, 7, 33, 17, 35, 8, 37, 19, 39, 9, 41, 21, 43, 10, 45, 23, 47, 11, 49, 25, 51, 12, 53, 27, 55, 13, 57, 29, 59, 14, 61, 31, 63, 15, 65, 33, 67, 16, 69
OFFSET
0,4
COMMENTS
Consider the family of sequences with recurrence a(n) = 2*a(n-4)-a(n-8) where a(0) and a(4) move up in steps of 1. This here is characterized by a(0)=-1, a(4)=0:
-2, 1, 1, 3, -1, 5, 3, 7, 0, 9, 5, 11,...
-1, 1, 1, 3, 0, 5, 3, 7, 1, 9, 5, 11,... = a(n)
0, 1, 1, 3, 1, 5, 3, 7, 2, 9, 5, 11,... = A060819
1, 1, 1, 3, 2, 5, 3, 7, 3, 9, 5, 11,... = b(n)
2, 1, 1, 3, 3, 5, 3, 7, 4, 9, 5, 11,... .
a(n+4)+b(n) = A145979(n).
a(n+4)*b(n) = A061037(n+2).
a(n+4)-b(n) = repeat -1, 4, 2, 4 with period of length 4.
FORMULA
a(n) = 2*a(n-4) - a(n-8).
a(n+4) - a(n) = A176895(n).
G.f.: (-1+x+x^2+3*x^3+3*x^5+x^6+x^7+2*x^4)/((-1+x)^2*(1+x)^2*(x^2+1)^2). - R. J. Mathar, Apr 28 2013
MATHEMATICA
a[n_] := 1/16*(11*n-(-1)^n*(5*n+4)-2*(n+4)*Re[I^n]-4); Table[a[n], {n, 0, 47}] (* Jean-François Alcover, Apr 30 2013 *)
LinearRecurrence[{0, 0, 0, 2, 0, 0, 0, -1}, {-1, 1, 1, 3, 0, 5, 3, 7}, 80] (* Harvey P. Dale, Jul 14 2019 *)
PROG
(PARI) x='x+O('x^50); Vec((-1+x+x^2+3*x^3+3*x^5+x^6+x^7+2*x^4)/((-1+x)^2*(1+x)^2*(x^2+1)^2)) \\ G. C. Greubel, Sep 20 2018
(Magma) m:=50; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((-1+x+x^2+3*x^3+3*x^5+x^6+x^7+2*x^4)/((-1+x)^2*(1+x)^2*(x^2+1)^2))); // G. C. Greubel, Sep 20 2018
CROSSREFS
Sequence in context: A284233 A326990 A037284 * A002123 A276408 A225744
KEYWORD
sign,easy
AUTHOR
Paul Curtz, Apr 26 2013
STATUS
approved