A024495 a(n) = C(n,2) + C(n,5) + ... + C(n, 3*floor(n/3)+2).
0, 0, 1, 3, 6, 11, 21, 42, 85, 171, 342, 683, 1365, 2730, 5461, 10923, 21846, 43691, 87381, 174762, 349525, 699051, 1398102, 2796203, 5592405, 11184810, 22369621, 44739243, 89478486, 178956971, 357913941, 715827882, 1431655765, 2863311531, 5726623062
Offset: 0
References
- A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.
- D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, 2nd. ed., Problem 38, p. 70.
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..3000
- Paul Barry, A note on Krawtchouk Polynomials and Riordan Arrays, JIS 11 (2008) 08.2.2, example 9.
- Antoine-Augustin Cournot, Solution d'un problème d'analyse combinatoire, Bulletin des Sciences Mathématiques, Physiques et Chimiques, item 34, volume 11, 1829, pages 93-97. Also at Google Books. Page 97 case p=3 formula y^(2) = a(n).
- Christian Ramus, Solution générale d'un problème d'analyse combinatoire, Journal für die Reine und Angewandte Mathematik (Crelle's journal), volume 11, 1834, pages 353-355. Page 353 case p=3 formula y^(2) = a(n).
- Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
- Eric Weisstein's World of Mathematics, Plane division by lines
- Index entries for linear recurrences with constant coefficients, signature (3,-3,2).
Crossrefs
Programs
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Magma
R
:=PowerSeriesRing(Integers(), 30); [0,0] cat Coefficients(R!( x^2/((1-x)^3-x^3) )); // G. C. Greubel, Apr 11 2023 -
Maple
a:= proc(n) option remember; `if`(n=0, 0, 2*a(n-1)+ [-1, 0, 1, 1, 0, -1, -1][1+(n mod 6)]) end: seq(a(n), n=0..33); # Paul Weisenhorn, May 17 2020
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Mathematica
LinearRecurrence[{3,-3,2},{0,0,1},40] (* Harvey P. Dale, Sep 20 2016 *)
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PARI
a(n) = sum(k=0,n\3,binomial(n,3*k+2)) /* Michael Somos, Feb 14 2006 */
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PARI
a(n)=if(n<0, 0, ([1,0,1;1,1,0;0,1,1]^n)[3,1]) /* Michael Somos, Feb 14 2006 */
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SageMath
def A024495(n): return (2^n - chebyshev_U(n, 1/2) - chebyshev_U(n-1, 1/2))/3 [A024495(n) for n in range(41)] # G. C. Greubel, Apr 11 2023
Formula
a(n) = ( 2^n + 2*cos((n-4)*Pi/3) )/3 = (2^n - A057079(n))/3.
a(n) = 2*a(n-1) + A010892(n-2) = a(n-1) + A024494(n-1). With initial zero, binomial transform of A011655 which is effectively A010892 unsigned. - Henry Bottomley, Jun 04 2001
a(2) = 1, a(3) = 3, a(n+2) = a(n+1) - a(n) + 2^n. - Benoit Cloitre, Sep 04 2002
a(n) = Sum_{k=0..n} 2^k*2*sin(Pi*(n-k)/3 + Pi/3)/sqrt(3) (offset 0). - Paul Barry, May 18 2004
G.f.: x^2/((1-x)^3 - x^3) = x^2 / ( (1-2*x)*(1-x+x^2) ).
a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3). - Paul Curtz, Nov 18 2007
From Paul Curtz, May 29 2011: (Start)
a(n) + a(n+3) = 3*2^n = A007283(n).
a(n+6) - a(n) = 21*2^n = A175805(n).
a(n) + a(n+9) = 171*2^n.
a(n+12) - a(n) = 1365*2^n. (End)
Start with x(0)=1, y(0)=0, z(0)=0 and set x(n+1) = x(n) + z(n), y(n+1) = y(n) + x(n), z(n+1) = z(n) + y(n). Then a(n) = z(n). - Stanislav Sykora, Jun 10 2012
G.f.: -x^2/( x^3 - 1 + 3*x/Q(0) ) where Q(k) = 1 + k*(x+1) + 3*x - x*(k+1)*(k+4)/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Mar 15 2013
a(n) = 1/18*(-4*(-1)^floor((n - 1)/3) - 6*(-1)^floor(n/3) - 3*(-1)^floor((n + 1)/3) + (-1)^(1 + floor((n + 2)/3)) + 3*2^(n + 1)). - John M. Campbell, Dec 23 2016
a(n) = (1/63)*(-40 + 21*2^n - 42*floor(n/6) + 32*floor((n+3)/6) + 16*floor((n+ 4)/6) - 24*floor((n+5)/6) - 22*floor((n+7)/6) + 21*floor((n+8)/6) + 10*floor((n+9)/6) + 5*floor((n+10)/6) + 3*floor((n+11)/6) + floor((n+ 13)/6)). - John M. Campbell, Dec 24 2016
a(n+m) = a(n)*A024493(m) + A131708(n)*A131708(m) + A024493(n)*a(m). - Vladimir Shevelev, Aug 01 2017
From Kevin Ryde, Sep 24 2020: (Start)
a(n) = (1/3)*2^n - (1/3)*cos((1/3)*Pi*n) - (1/sqrt(3))*sin((1/3)*Pi*n). [Cournot]
a(n) + A111927(n) + A131708(n) = 2^n - 1. [Cournot, page 96 last formula, but misprint should be 2^x - 1 rather than 2^p - 1] (End)
E.g.f.: (exp(2*x) - exp(x/2)*(cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2)))/3. - Stefano Spezia, Feb 06 2025
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