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A098181
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Two consecutive odd numbers separated by multiples of four, repeated twice, between them, written in increasing order.
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5
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1, 3, 4, 4, 5, 7, 8, 8, 9, 11, 12, 12, 13, 15, 16, 16, 17, 19, 20, 20, 21, 23, 24, 24, 25, 27, 28, 28, 29, 31, 32, 32, 33, 35, 36, 36, 37, 39, 40, 40, 41, 43, 44, 44, 45, 47, 48, 48, 49, 51, 52, 52, 53, 55, 56, 56, 57, 59, 60, 60, 61, 63, 64, 64, 65, 67, 68, 68, 69, 71, 72, 72
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OFFSET
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0,2
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COMMENTS
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Essentially partial sums of A007877.
a(n) is the number of odd coefficients of the q-binomial coefficient [n+2 choose 2]. (Easy to prove.) - Richard Stanley, Oct 12 2016
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LINKS
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FORMULA
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G.f.: (1+x)/((1-x)^2*(1+x^2)).
a(n) = ( (2*n+3) - cos(Pi*n/2) + sin(Pi*n/2) )/2.
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4).
a(n) = floor(C(n+3, 2)/2)-floor(C(n+1, 2)/2). - Paul Barry, Jan 01 2005
a(4*n) = 4*n+1, a(4*n+1) = 4*n+3, a(4*n+2) = a(4*n+3) = 4*n+4. - Philippe Deléham, Apr 06 2007
Euler transform of length 4 sequence [ 3, -2, 0, 1]. - Michael Somos, Sep 11 2014
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EXAMPLE
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G.f. = 1 + 3*x + 4*x^2 + 4*x^3 + 5*x^4 + 7*x^5 + 8*x^6 + 8*x^7 + 9*x^8 + ...
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MAPLE
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A:=seq((2*n+3 - cos(Pi*n/2) + sin(Pi*n/2))/2, n=0..50); \\ Bernard Schott, Jun 07 2019
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MATHEMATICA
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Table[Floor[Binomial[n+3, 2]/2] -Floor[Binomial[n+1, 2]/2], {n, 0, 80}] (* or *) CoefficientList[Series[(1+x)/((1-x)^2*(1+x^2)), {x, 0, 80}], x] (* Michael De Vlieger, Oct 12 2016 *)
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PROG
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(PARI) {a(n) = n\4*4 + [1, 3, 4, 4][n%4+1]}; /* Michael Somos, Sep 11 2014 */
(Magma) R<x>:=PowerSeriesRing(Integers(), 80); Coefficients(R!( (1+x)/((1-x)^2*(1+x^2)) )); // G. C. Greubel, May 22 2019
(Sage) ((1+x)/((1-x)^2*(1+x^2))).series(x, 80).coefficients(x, sparse=False) # G. C. Greubel, May 22 2019
(GAP) a:=[1, 3, 4, 4];; for n in [5..80] do a[n]:=2*a[n-1]-2*a[n-2]+2*a[n-3] -a[n-4]; od; a; # G. C. Greubel, May 22 2019
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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