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%I #35 Mar 12 2021 22:24:42
%S 0,0,0,1,2,3,5,7,9,12,15,18,22,26,30,35,40,45,51,57,63,70,77,84,92,
%T 100,108,117,126,135,145,155,165,176,187,198,210,222,234,247,260,273,
%U 287,301,315,330,345,360,376,392
%N Number of arithmetic progressions of four terms and any mean which can be extracted from the set of the first n positive integers.
%C This sequence seems to be a shifted version of the Somos sequence A058937.
%C Equal to the partial sums of A002264 (cf. A130518) but with initial index 1 instead of 0. - _Hieronymus Fischer_, Jun 01 2007
%C Apart from offset, the same as A130518. - _R. J. Mathar_, Jun 13 2008
%C Apart from offset, the same as A001840. - _Michael Somos_, Sep 18 2010
%H Muniru A Asiru, <a href="/A062781/b062781.txt">Table of n, a(n) for n = 1..750</a>
%H Michael Somos, <a href="https://grail.eecs.csuohio.edu/~somos/somospol.html">Somos Polynomials</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,1,-2,1).
%F a(n) = P(n,4), where P(n,k) = n*floor(n/(k - 1)) - (1/2)(k - 1)(floor(n/(k - 1))*(floor(n/(k - 1)) + 1)); recursion: a(n) = a(n-3) + n - 3; a(1) = a(2) = a(3) = 0.
%F From _Hieronymus Fischer_, Jun 01 2007: (Start)
%F a(n) = (1/2)*floor((n-1)/3)*(2*n - 3 - 3*floor((n-1)/3)).
%F G.f.: x^4/((1 - x^3)*(1 - x)^2). (End)
%F a(n) = floor((n-1)/3) + a(n-1). - _Jon Maiga_, Nov 25 2018
%F E.g.f.: ((4 - 6*x + 3*x^2)*exp(x) - 4*exp(-x/2)*cos(sqrt(3)*x/2))/18. - _Franck Maminirina Ramaharo_, Nov 25 2018
%p seq(coeff(series(x^4/((1-x^3)*(1-x)^2),x,n+1), x, n), n = 1 .. 50); # _Muniru A Asiru_, Nov 25 2018
%t RecurrenceTable[{a[0]==0, a[n]==Floor[n/3] + a[n-1]}, a, {n, 49}] (* _Jon Maiga_, Nov 25 2018 *)
%o (Sage) [floor(binomial(n,2)/3) for n in range(0,50)] # _Zerinvary Lajos_, Dec 01 2009
%Y Cf. A058937, A001840.
%Y Cf. A002620, A130519, A130520.
%K nonn,easy
%O 1,5
%A _Santi Spadaro_, Jul 18 2001
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