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.

A001949 Solutions of a fifth-order probability difference equation.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 4, 8, 16, 32, 63, 124, 244, 480, 944, 1856, 3649, 7174, 14104, 27728, 54512, 107168, 210687, 414200, 814296, 1600864, 3147216, 6187264, 12163841, 23913482, 47012668, 92424472, 181701728, 357216192, 702268543, 1380623604, 2714234540
Offset: 0

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This sequence is the case r = 5 in the solution to an r-th order probability difference equation that can be found in Eqs. (4) and (3) on p. 356 of Dunkel (1925). (Equation (3) follows equation (4) in the paper!) For r = 2, we get a shifted version of A000071. For r = 3, we get a shifted version of A008937. For r = 4, we get a shifted version of A107066. For r = 6, we get a shifted version of A172316. See also the table in A172119. - Petros Hadjicostas, Jun 15 2019

References

  • 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

Column k = 1 of A141020 (with a different offset) and second main diagonal of A141021 (with no zeros).
Column k = 5 of A172119.
Partial sums of A001591.

Programs

  • Maple
    A001949:=1/(z-1)/(z**5+z**4+z**3+z**2+z-1); # Simon Plouffe in his 1992 dissertation
  • Mathematica
    t={0,0,0,0,0};Do[AppendTo[t,t[[-5]]+t[[-4]]+t[[-3]]+t[[-2]]+t[[-1]]+1],{n,40}];t (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
    LinearRecurrence[{2,0,0,0,0,-1},{0,0,0,0,0,1},40] (* Harvey P. Dale, Jan 17 2015 *)
  • Maxima
    a(n):=sum(sum((-1)^j*binomial(n-5*j-5,k-1)*binomial(n-k-5*j-4,j),j,0,(n-k-4)/5),k,1,n-4); /* Vladimir Kruchinin, Oct 19 2011 */
    
  • PARI
    x='x+O('x^99); concat(vector(5), Vec(x^5/((x-1)*(x^5+x^4+x^3+x^2+x-1)))) \\ Altug Alkan, Oct 04 2017

Formula

For n >= 6, a(n+1) = 2*a(n) - a(n-5).
G.f.: x^5 / ( (x-1)*(x^5 + x^4 + x^3 + x^2 + x - 1) ).
a(n) = Sum_{k=1..n-4} Sum_{j=0..floor((n-k-4)/5)} (-1)^j*binomial(n-5*j-5, k-1)*binomial(n-k-5*j-4, j). - Vladimir Kruchinin, Oct 19 2011
4*a(n) = A000322(n+1) - 1. - R. J. Mathar, Aug 16 2017
From Petros Hadjicostas, Jun 15 2019: (Start)
a(n) = 1 + a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) for n >= 5. (See Eq. (4) and the Theorem with r = 5 on p. 356 of Dunkel (1925).)
a(n) = T(n - 5, 5) for n >= 5, where T(n, k) = Sum_{j = 0..floor(n/(k+1))} (-1)^j * binomial(n - k*j, n - (k+1)*j) * 2^(n - (k+1)*j) for 0 <= k <= n. This is Richard Choulet's formula in A172119.
(End)

Extensions

Name edited by Petros Hadjicostas, Jun 15 2019