cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A001634 a(n) = a(n-2) + a(n-3) + a(n-4), with initial values a(0) = 0, a(1) = 2, a(2) = 3, a(3) = 6.

Original entry on oeis.org

0, 2, 3, 6, 5, 11, 14, 22, 30, 47, 66, 99, 143, 212, 308, 454, 663, 974, 1425, 2091, 3062, 4490, 6578, 9643, 14130, 20711, 30351, 44484, 65192, 95546, 140027, 205222, 300765, 440795, 646014, 946782, 1387574, 2033591, 2980370, 4367947, 6401535, 9381908
Offset: 0

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References

  • E.-B. Escott, Reply to Query 1484, L'Intermédiaire des Mathématiciens, 8 (1901), 63-64.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    a001634 n = a001634_list !! n
    a001634_list = 0 : 2 : 3 : 6 : zipWith (+) a001634_list
       (zipWith (+) (tail a001634_list) (drop 2 a001634_list))
    -- Reinhard Zumkeller, Mar 23 2012
    
  • Maple
    A001634:=-z*(2+3*z+4*z**2)/(1+z)/(z**3+z-1); # Simon Plouffe in his 1992 dissertation
    a:= n-> (Matrix([[0,4,-1,-1]]). Matrix(4, (i,j)-> if (i=j-1) then 1 elif j=1 then [0,1,1,1][i] else 0 fi)^n)[1,1]: seq(a(n), n=0..40); # Alois P. Heinz, Aug 01 2008
  • Mathematica
    LinearRecurrence[{0, 1, 1, 1}, {0, 2, 3, 6}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)
    CoefficientList[Series[x (2+3x+4x^2)/(1-x^2-x^3-x^4),{x,0,50}],x] (* Harvey P. Dale, Mar 26 2023 *)
  • Maxima
    a(n):=(sum(sum(binomial(j,n-2*k-j-1)*binomial(k+1,j),j,0,k+1)/(k+1),k,0,(n-1)/2))*(n+1); /* Vladimir Kruchinin, Mar 22 2016 */
  • PARI
    a(n)=if(n<0,0,polcoeff(x*(2+3*x+4*x^2)/(1-x^2-x^3-x^4)+x*O(x^n),n))
    

Formula

G.f.: x(2 + 3x + 4x^2)/(1 - x^2 - x^3 - x^4).
a(n) = Sum_{k=0..(n-1)/2}(Sum_{j=0..k+1}(binomial(j,n-2*k-j-1)*binomial(k+1,j))/(k+1))*(n+1). - Vladimir Kruchinin, Mar 22 2016