A011848 a(n) = floor(binomial(n, 2)/2).
0, 0, 0, 1, 3, 5, 7, 10, 14, 18, 22, 27, 33, 39, 45, 52, 60, 68, 76, 85, 95, 105, 115, 126, 138, 150, 162, 175, 189, 203, 217, 232, 248, 264, 280, 297, 315, 333, 351, 370, 390, 410, 430, 451, 473, 495, 517, 540, 564, 588, 612, 637, 663, 689, 715, 742, 770, 798
Offset: 0
Examples
G.f. = x^3 + 3*x^4 + 5*x^5 + 7*x^6 + 10*x^7 + 14*x^8 + 18*x^9 + 22*x^10 + ... p(0) = p(1) = 1, p(2) = 1 + x, p(3) = 1 + x + x^3, p(4) = 1 + 2*x + 2*x^2 + x^3 + x^5. - _Michael Somos_, Mar 22 2023
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Kyu-Hwan Lee and Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.
- Eric Weisstein's World of Mathematics, Matching Number.
- Eric Weisstein's World of Mathematics, Triangular Graph.
- Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).
Crossrefs
Programs
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GAP
List([0..60],n->Int(Binomial(n,2)/2)); # Muniru A Asiru, Apr 05 2018
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Haskell
a011848 n = if n < 2 then 0 else flip div 2 $ a007318 n 2 -- Reinhard Zumkeller, Mar 04 2015
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Magma
[ Floor(n*(n-1)/4) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 09 2014
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Maple
seq(floor(binomial(n,2)/2), n=0..57); # Zerinvary Lajos, Jan 12 2009
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Mathematica
Table[Floor[n (n - 1)/4], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jun 28 2011 *) CoefficientList[Series[x^3/((1 + x^2) (1 - x)^3), {x, 0, 70}], x] (* Vincenzo Librandi, Jun 21 2013 *) LinearRecurrence[{3, -4, 4, -4, 1}, {0, 0, 1, 3, 5}, {0, 20}] (* Eric W. Weisstein, Jun 02 2017 *) Table[Floor[Binomial[n, 2]/2], {n, 0, 20}] (* Eric W. Weisstein, Jun 02 2017 *) Table[1/4 (-1 + (-1 + n) n + Cos[n Pi/2] + Sin[n Pi/2]), {n, 0, 20}] (* Eric W. Weisstein, Jun 02 2017 *) Floor[Binomial[Range[0, 20], 2]/2] (* Eric W. Weisstein, Apr 03 2018 *)
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PARI
a(n) = binomial(n, 2)\2;
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PARI
vector(100, n, n--; floor(n*(n-1)/4)) \\ Altug Alkan, Sep 30 2015
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Python
def a(n): return n*(n-1)//4 # Christoph B. Kassir, Oct 07 2022
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Sage
[floor(binomial(n,2)/2) for n in range(0,58)] # Zerinvary Lajos, Dec 01 2009
Formula
G.f.: x^3*(1-x^2)/((1-x)^3*(1-x^4)).
G.f.: x^3/((1+x^2)*(1-x)^3). - Jon Perry, Mar 31 2004
a(n) = +3*a(n-1) -4*a(n-2) +4*a(n-3) -3*a(n-4) +a(n-5). - R. J. Mathar, Apr 15 2010
a(n) = floor((n/(1+e^(1/n)))^2). - Richard R. Forberg, Jun 19 2013
a(n) = floor(n*(n-1)/4). - T. D. Noe, Jun 20 2013
a(n) = (1/4) * ( n^2 - n - 1 + (-1)^floor(n/2) ). - Ralf Stephan, Aug 11 2013
a(4*n) = A033991(n). a(4*n+1) = A007742(n). a(4*n+2) = A033954(n). a(4*n+3) = A001107(n+1). - Bob Selcoe, Sep 26 2015
E.g.f.: (sin(x) + cos(x) + (x^2 - 1)*exp(x))/4. - Ilya Gutkovskiy, Nov 18 2016
A156859(n) = a(2*n+2). - Michael Somos, Nov 18 2016
Euler transform of length 4 sequence [ 3, -1, 0, 1]. - Michael Somos, Nov 18 2016
From Amiram Eldar, Mar 18 2022: (Start)
Sum_{n>=3} 1/a(n) = 40/9 - 2*Pi/3.
Sum_{n>=3} (-1)^(n+1)/a(n) = 32/9 - 4*log(2). (End)
0 = a(n+2)*(a(n)*(a(n) -6*a(n+1) +4*a(n+2)) +a(n+1)*(8*a(n+1) -10*a(n+2)) + 3*a(n+2)^2) +a(n+3)*(a(n)*(+a(n) -2*a(n+1)) +a(n+2)*(2*a(n+1) -a(n+2))) for all n in Z. - Michael Somos, Mar 22 2023
2*a(n) + 2*a(n-2) = (n-1)*(n-2). - R. J. Mathar, Feb 12 2024
Comments