A014107 a(n) = n*(2*n-3).
0, -1, 2, 9, 20, 35, 54, 77, 104, 135, 170, 209, 252, 299, 350, 405, 464, 527, 594, 665, 740, 819, 902, 989, 1080, 1175, 1274, 1377, 1484, 1595, 1710, 1829, 1952, 2079, 2210, 2345, 2484, 2627, 2774, 2925, 3080, 3239, 3402, 3569, 3740, 3915, 4094, 4277
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..5000
- Emily Barnard and Nathan Reading, Coxeter-biCatalan combinatorics, arXiv:1605.03524 [math.CO], 2016. See p. 51.
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Crossrefs
Programs
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Maple
A014107:=n->n*(2*n-3); seq(A014107(n), n=0..100); # Wesley Ivan Hurt, Nov 19 2013
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Mathematica
Table[2n^2 - 3n, {n, 0, 49}] (* Alonso del Arte, Dec 15 2012 *) LinearRecurrence[{3,-3,1},{0,-1,2},50] (* Harvey P. Dale, Sep 18 2018 *) -
PARI
a(n)=n*(2*n-3)
Formula
a(n) = A100345(n, n - 3) for n > 2.
a(n) = A014106(-n) for all n in Z. - Michael Somos, Nov 06 2005
From Michael Somos, Nov 06 2005: (Start)
G.f.: x*(-1 + 5*x)/(1 - x)^3.
E.g.f: x*(-1 + 2*x)*exp(x). (End)
a(n) = 2*a(n-1) - a(n-2) + 4, n > 1; a(0) = 0, a(1) = -1, a(2) = 2. - Zerinvary Lajos, Feb 18 2008
a(n) = a(n-1) + 4*n - 5 with a(0) = 0. - Vincenzo Librandi, Nov 20 2010
a(n) = (2*n-1)*(n-1) - 1. Also, with an initial offset of -1, a(n) = (2*n-1)*(n+1) = 2*n^2 + n - 1. - Alonso del Arte, Dec 15 2012
(a(n) + 1)^2 + (a(n) + 2)^2 + ... + (a(n) + n)^2 = (a(n) + n + 1)^2 + (a(n) + n + 2)^2 + ... + (a(n) + 2n - 1)^2 starting with a(1) = -1. - Jeffreylee R. Snow, Sep 17 2013
a(n) = A014105(n-1) - 1 for all n in Z. - Michael Somos, Nov 23 2021
From Amiram Eldar, Feb 20 2022: (Start)
Sum_{n>=1} 1/a(n) = -2*(1 - log(2))/3.
Sum_{n>=1} (-1)^n/a(n) = Pi/6 + log(2)/3 + 2/3. (End)
Comments