A062781 Number of arithmetic progressions of four terms and any mean which can be extracted from the set of the first n positive integers.

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, 100, 108, 117, 126, 135, 145, 155, 165, 176, 187, 198, 210, 222, 234, 247, 260, 273, 287, 301, 315, 330, 345, 360, 376, 392

Offset: 1

Author

Santi Spadaro, Jul 18 2001

Keywords

• nonn
• easy

Comments

This sequence seems to be a shifted version of the Somos sequence A058937.

Equal to the partial sums of A002264 (cf. A130518) but with initial index 1 instead of 0. - Hieronymus Fischer, Jun 01 2007

Apart from offset, the same as A130518. - R. J. Mathar, Jun 13 2008

Apart from offset, the same as A001840. - Michael Somos, Sep 18 2010

Links

• Muniru A Asiru, Table of n, a(n) for n = 1..750
• Michael Somos, Somos Polynomials
• Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Crossrefs

Cf. A058937, A001840.

Cf. A002620, A130519, A130520.

Programs

Maple

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

Mathematica

RecurrenceTable[{a[0]==0, a[n]==Floor[n/3] + a[n-1]}, a, {n, 49}] (* Jon Maiga, Nov 25 2018 *)

Sage

[floor(binomial(n,2)/3) for n in range(0,50)] # Zerinvary Lajos, Dec 01 2009

Formula

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.

From Hieronymus Fischer, Jun 01 2007: (Start)

a(n) = (1/2)*floor((n-1)/3)*(2*n - 3 - 3*floor((n-1)/3)).

G.f.: x^4/((1 - x^3)*(1 - x)^2). (End)

a(n) = floor((n-1)/3) + a(n-1). - Jon Maiga, Nov 25 2018

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