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A001608 Perrin sequence (or Ondrej Such sequence): a(n) = a(n-2) + a(n-3) with a(0) = 3, a(1) = 0, a(2) = 2.

Original entry on oeis.org

3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17, 22, 29, 39, 51, 68, 90, 119, 158, 209, 277, 367, 486, 644, 853, 1130, 1497, 1983, 2627, 3480, 4610, 6107, 8090, 10717, 14197, 18807, 24914, 33004, 43721, 57918, 76725, 101639, 134643, 178364, 236282, 313007, 414646, 549289
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

Has been called the skiponacci sequence or skiponacci numbers. - N. J. A. Sloane, May 24 2013
For n >= 3, also the numbers of maximal independent vertex sets, maximal matchings, minimal edge covers, and minimal vertex covers in the n-cycle graph C_n. - Eric W. Weisstein, Mar 30 2017 and Aug 03 2017
With the terms indexed as shown, has property that p prime => p divides a(p). The smallest composite n such that n divides a(n) is 521^2. For quotients a(p)/p, where p is prime, see A014981.
Asymptotically, a(n) ~ r^n, with r=1.3247179572447... the inverse of the real root of 1-x^2-x^3=0 (see A060006). If n>9 then a(n)=round(r^n). - Ralf Stephan, Dec 13 2002
The recursion can be used to compute a(-n). The result is -A078712(n). - T. D. Noe, Oct 10 2006
For n>=3, a(n) is the number of maximal independent sets in a cycle of order n. - Vincent Vatter, Oct 24 2006
Pisano period lengths are given in A104217. - R. J. Mathar, Aug 10 2012
From Roman Witula, Feb 01 2013: (Start)
Let r1, r2 and r3 denote the roots of x^3 - x - 1. Then the following identity holds:
a(k*n) + (a(k))^n - (a(k) - r1^k)^n - (a(k) - r2^k)^n - (a(k) - r3^k)^n
= 0 for n = 0, 1, 2,
= 6 for n = 3,
= 12*a(k) for n = 4,
= 10*[2*(a(k))^2 - a(-k)] for n = 5,
= 30*a(k)*[(a(k))^2 - a(-k)] for n = 6,
= 7*[6*(a(k))^4 - 9*a(-k)*(a(k))^2 + 2*(a(-k))^2 - a(k)] for n = 7,
= 56*a(k)*[((a(k))^2 - a(-k))^2 - a(k)/2] for n = 8,
where a(-k) = -A078712(k) and the formula (5.40) from the paper of Witula and Slota is used. (End)
The parity sequence of a(n) is periodic with period 7 and has the form (1,0,0,1,0,1,1). Hence we get that a(n) and a(2*n) are congruent modulo 2. Similarly we deduce that a(n) and a(3*n) are congruent modulo 3. Is it true that a(n) and a(p*n) are congruent modulo p for every prime p? - Roman Witula, Feb 09 2013
The trinomial x^3 - x - 1 divides the polynomial x^(3*n) - a(n)*x^(2*n) + ((a(n)^2 - a(2*n))/2)*x^n - 1 for every n>=1. For example, for n=3 we obtain the factorization x^9 - 3*x^6 + 2*x^3 - 1 = (x^3 - x - 1)*(x^6 + x^4 - 2*x^3 + x^2 - x + 1). Sketch of the proof: Let p,s,t be roots of the Perrin polynomial x^3 - x - 1. Then we have (a(n))^2 = (p^n + s^n + t^n)^2 = a(2*n) + 2*a(n)*x^n -2*x^n + 2/x^n for every x = p,s,t, i.e., x^(3*n) - a(n)*x^(2*n) + ((a(n)^2 - a(2*n))/2)*x^n - 1 = 0 for every x = p,s,t, which finishes the proof. By discussion of the power(a(n))^3 = (p^n + s^n + t^n)^3 it can be deduced that the trinomial x^3 - x - 1 divides the polynomial 2*x^(4*n) - a(n)*x^(3*n) - a(2*n)*x^(2*n) + ((a(n)^3 - a(3*n) - 3)/3)*x^n - a(n) = 0. Co-author of these divisibility relations is also my young student Szymon Gorczyca (13 years old as of 2013). - Roman Witula, Feb 09 2013
The sum of powers of the real root and complex roots of x^3-x-1=0 as expressed as powers of the plastic number r, (see A060006). Let r0=1, r1=r, r2=1+r^(-1) and c0=2, c1=-r and c3 = r^(-5) then a(n) = r(n-2)+r(n-3) + c(n-2)+c(n-3). Example: a(5) = 1 + r^(-1) + 1 + r + 2 - r + r^(-5) = 4 + r^(-1) + r^(-5) = 5. - Richard Turk, Jul 14 2016
Also the number of minimal total dominating sets in the n-sun graph. - Eric W. Weisstein, Apr 27 2018
Named after the French engineer François Olivier Raoul Perrin (1841-1910). - Amiram Eldar, Jun 05 2021
a(p) = p*A127687(p) for p prime. - Robert FERREOL, Apr 09 2024

Examples

			From _Roman Witula_, Feb 01 2013: (Start)
We note that if a + b + c = 0 then:
1) a^3 + b^3 + c^3 = 3*a*b*c,
2) a^4 + b^4 + c^4 = 2*((a^2 + b^2 + c^2)/2)^2,
3) (a^5 + b^5 + c^5)/5 = (a^3 + b^3 + c^3)/3 * (a^2 +
    b^2  + c^2)/2,
4) (a^7 + b^7 + c^7)/7 = (a^5 + b^5 + c^5)/5 * (a^2 + b^2 + c^2)/2 = 2*(a^3 + b^3 + c^3)/3 * (a^4 + b^4 + c^4)/4,
5) (a^7 + b^7 + c^7)/7 * (a^3 + b^3 + c^3)/3 = ((a^5 + b^5 + c^5)/5)^2.
Hence, by the Binet formula for a(n) we obtain the relations: a(3) = 3, a(4) = 2*(a(2)/2)^2 = 2, a(5)/5 = a(3)/3 * a(2)/2, i.e., a(5) = 5, and similarly that a(7) = 7. (End)

References

  • Olivier Bordellès, Thèmes d'Arithmétique, Ellipses, 2006, Exercice 4.11, p. 127.0
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.2.
  • Dmitry Fomin, On the properties of a certain recursive sequence, Mathematics and Informatics Quarterly, Vol. 3 (1993), pp. 50-53.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 70.
  • Manfred Schroeder, Number Theory in Science and Communication, 3rd ed., Springer, 1997.
  • A. G. Shannon, P. G. Anderson and A. F. Horadam, Properties of Cordonnier, Perrin and Van der Laan numbers, International Journal of Mathematical Education in Science and Technology, Volume 37:7 (2006), 825-831. See Q_n.
  • 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).

Links

Crossrefs

Closely related to A182097.
Cf. A000931, bisection A109377.
Cf. A013998 (Unrestricted Perrin pseudoprimes).
Cf. A018187 (Restricted Perrin pseudoprimes).

Programs

  • GAP
    a:=[3,0,2];; for n in [4..20] do a[n]:=a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Jul 12 2018
  • Haskell
    a001608 n = a000931_list !! n
a001608_list = 3 : 0 : 2 : zipWith (+) a001608_list (tail a001608_list)
-- Reinhard Zumkeller, Feb 10 2011
  • Magma
    I:=[3,0,2]; [n le 3 select I[n] else Self(n-2) +Self(n-3): n in [1..50]]; // G. C. Greubel, Mar 18 2019
  • Maple
    A001608 :=proc(n) option remember; if n=0 then 3 elif n=1 then 0 elif n=2 then 2 else procname(n-2)+procname(n-3); fi; end proc;
[seq(A001608(n),n=0..50)]; # N. J. A. Sloane, May 24 2013
  • Mathematica
    LinearRecurrence[{0, 1, 1}, {3, 0, 2}, 50] (* Harvey P. Dale, Jun 26 2011 *)
per = Solve[x^3 - x - 1 == 0, x]; f[n_] := Floor @ Re[N[ per[[1, -1, -1]]^n + per[[2, -1, -1]]^n + per[[3, -1, -1]]^n]]; Array[f, 46, 0] (* Robert G. Wilson v, Jun 29 2010 *)
a[n_] := n*Sum[Binomial[k, n-2*k]/k, {k, 1, n/2}]; a[0]=3; Table[a[n] , {n, 0, 45}] (* Jean-François Alcover, Oct 04 2012, after Vladimir Kruchinin *)
CoefficientList[Series[(3 - x^2)/(1 - x^2 - x^3), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 03 2015 *)
Table[RootSum[-1 - # + #^3 &, #^n &], {n, 0, 20}] (* Eric W. Weisstein, Mar 30 2017 *)
RootSum[-1 - # + #^3 &, #^Range[0, 20] &] (* Eric W. Weisstein, Dec 30 2017 *)
  • PARI
    a(n)=if(n<0,0,polsym(x^3-x-1,n)[n+1])
  • PARI
    A001608_list(n) = polsym(x^3-x-1,n) \\ Joerg Arndt, Mar 10 2019
  • Python
    A001608_list, a, b, c = [3, 0, 2], 3, 0, 2
for _ in range(100):
    a, b, c = b, c, a+b
    A001608_list.append(c) # Chai Wah Wu, Jan 27 2015
  • Sage
    ((3-x^2)/(1-x^2-x^3)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Mar 18 2019

Formula

G.f.: (3 - x^2)/(1 - x^2 - x^3). - Simon Plouffe in his 1992 dissertation
a(n) = r1^n + r2^n + r3^n where r1, r2, r3 are three roots of x^3-x-1=0.
a(n-1) + a(n) + a(n+1) = a(n+4), a(n) - a(n-1) = a(n-5). - Jon Perry, Jun 05 2003
From Gary W. Adamson, Feb 01 2004: (Start)
a(n) = trace(M^n) where M is the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 1 0], the companion matrix of the characteristic polynomial of this sequence, P = X^3 - X - 1.
M^n * [3, 0, 2] = [a(n), a(n+1), a(n+2)]; e.g., M^7 * [3, 0, 2] = [7, 10, 12].
a(n) = 2*A000931(n+3) + A000931(n). (End)
a(n) = 3*p(n) - p(n-2) = 2*p(n) + p(n-3), with p(n) := A000931(n+3), n >= 0. - Wolfdieter Lang, Jun 21 2010
From Francesco Daddi, Aug 03 2011: (Start)
a(0) + a(1) + a(2) + ... + a(n) = a(n+5) - 2.
a(0) + a(2) + a(4) + ... + a(2*n) = a(2*n+3).
a(1) + a(3) + a(5) + ... + a(2*n+1) = a(2*n+4) - 2. (End)
From Francesco Daddi, Aug 04 2011: (Start)
a(0) + a(3) + a(6) + a(9) + ... + a(3*n) = a(3*n+2) + 1.
a(0) + a(5) + a(10) + a(15) + ... + a(5*n) = a(5*n+1)+3.
a(0) + a(7) + a(14) + a(21) + ... + a(7*n) = (a(7*n) + a(7*n+1) + 3)/2. (End)
a(n) = n*Sum_{k=1..floor(n/2)} binomial(k,n-2*k)/k, n > 0, a(0)=3. - Vladimir Kruchinin, Oct 21 2011
(a(n)^3)/2 + a(3n) - 3*a(n)*a(2n)/2 - 3 = 0. - Richard Turk, Apr 26 2017
2*a(4n) - 2*a(n) - 2*a(n)*a(3n) - a(2n)^2 + a(n)^2*a(2n) = 0. - Richard Turk, May 02 2017
a(n)^4 + 6*a(4n) - 4*a(3n)*a(n) - 3*a(2n)^2 - 12a(n) = 0. - Richard Turk, May 02 2017
a(n+5)^2 + a(n+1)^2 - a(n)^2 = a(2*(n+5)) + a(2*(n+1)) - a(2*n). - Aleksander Bosek, Mar 04 2019
From Aleksander Bosek, Mar 18 2019: (Start)
a(n+12) = a(n) + 2*a(n+4) + a(n+11);
a(n+16) = a(n) + 4*a(n+9) + a(n+13);
a(n+18) = a(n) + 2*a(n+6) + 5*a(n+12);
a(n+21) = a(n) + 2*a(n+12) + 6*a(n+14);
a(n+27) = a(n) + 3*a(n+9) + 4*a(n+22). (End)
a(n) = Sum_{j=0..floor((n-g)/(2*g))} 2*n/(n-2*(g-2)*j-(g-2)) * Hypergeometric2F1([-(n-2g*j-g)/2, -(2j+1)], [1], 1), g = 3 and n an odd integer. - Richard Turk, Oct 14 2019
E.g.f.: exp(r1*x) + exp(r2*x) + exp(r3*x), where r1, r2, r3 are three roots of x^3 - x - 1 = 0. - Fabian Pereyra, Nov 02 2024

Extensions

Additional comments from Mike Baker, Oct 11 2005
Definition edited by Chai Wah Wu, Jan 27 2015
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

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