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A060819
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a(n) = n / gcd(n,4).
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23
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1, 1, 3, 1, 5, 3, 7, 2, 9, 5, 11, 3, 13, 7, 15, 4, 17, 9, 19, 5, 21, 11, 23, 6, 25, 13, 27, 7, 29, 15, 31, 8, 33, 17, 35, 9, 37, 19, 39, 10, 41, 21, 43, 11, 45, 23, 47, 12, 49, 25, 51, 13, 53, 27, 55, 14, 57, 29, 59, 15, 61, 31, 63, 16, 65, 33, 67, 17, 69, 35, 71, 18, 73, 37, 75, 19
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| a(n) = A167192(n+4,4). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Oct 30 2009]
a(n) = Numerator(sum(1/((k+1)*(k+2)),k=1..n)). This summation has a closed form of 1/2-1/(n+2) and denominator of A145979(n). [From Gary Detlefs, Sep 16 2011]
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LINKS
| Harry J. Smith, Table of n, a(n) for n=1,...,1000
Index to sequences with linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).
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FORMULA
| G.f.: (x^7+x^6+3x^5+x^4+3x^3+x^2+x)/(1-x^4)^2.
a(n)= +2*a(n-4) -a(n-8).
a(n) = n/16*(11-5*(-1)^n-i^n-(-i)^n). - Ralf Stephan (ralf(AT)ark.in-berlin.de), Mar 15 2003
a(2n+1) = a(4n+2) = 2n+1, a(4n+4) = n+1. - Ralf Stephan, Jun 10 2005
Multiplicative with a(2^e) = 2^max(0, e-2), a(p^e) = p^e, p >= 3 (from Mitch Harris, Jun 29 2005)
a(n) = A109045(n)/4. Dirichlet g.f. zeta(s-1)*(1-1/2^s-1/2^(2s)). - R. J. Mathar, Apr 18 2011
a(n+4) - a(n) = A176895(n) (Period 4: repeat 1,4,2,4.) - Paul Curtz, Apr 05 2011
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MAPLE
| seq(numer(1/2-1/(n+2)), n= 1..25). [From Gary Detlefs, Sep 16 2011]
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MATHEMATICA
| f[n_] := n/GCD[n, 4]; Array[f, 76]
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PROG
| (Other) sage: [lcm(n, 4)/4for n in xrange(1, 77)] # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 07 2009]
(PARI) { for (n=1, 1000, write("b060819.txt", n, " ", n / gcd(n, 4)); ) } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Jul 12 2009]
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CROSSREFS
| Cf. A026741, A051176, A060791, A060789. A061037, A061038.
Sequence in context: A162742 A081432 A136655 * A089654 A097062 A200498
Adjacent sequences: A060816 A060817 A060818 * A060820 A060821 A060822
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KEYWORD
| nonn,easy,mult
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AUTHOR
| Len Smiley (smiley(AT)math.uaa.alaska.edu), Apr 30 2001
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EXTENSIONS
| More terms from Larry Reeves (larryr(AT)acm.org), May 07 2001
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