A078712 -id:A078712 - cp's OEIS frontend

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.

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

N. J. A. Sloane

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

			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)

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Index entries for linear recurrences with constant coefficients, signature (0,1,1).

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

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

a001608 n = a000931_list !! n
a001608_list = 3 : 0 : 2 : zipWith (+) a001608_list (tail a001608_list)
-- Reinhard Zumkeller, Feb 10 2011

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

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

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])
```

A001608_list(n) = polsym(x^3-x-1,n) \\ Joerg Arndt, Mar 10 2019

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

((3-x^2)/(1-x^2-x^3)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Mar 18 2019

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

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

A001945 a(n+6) = -a(n+5) + a(n+4) + 3a(n+3) + a(n+2) - a(n+1) - a(n). a(n) = sign(n) if abs(n)<=3.

Original entry on oeis.org

0, 1, 1, 1, 5, 1, 7, 8, 5, 19, 11, 23, 35, 27, 64, 61, 85, 137, 133, 229, 275, 344, 529, 599, 875, 1151, 1431, 2071, 2560, 3481, 4697, 5953, 8245, 10649, 14111, 19048, 24605, 33227, 43739, 57591, 77275, 101107, 134848, 178709, 235405, 314089, 413909

Offset: 0

N. J. A. Sloane

It seems likely that this sequence contains infinitely many primes. In the paper by Einsiedler, Everest, Ward the heuristics for the Mersenne sequence are adapted to argue that approximately c*log(N) of the first N terms should be prime, where c is constant. Numerical evidence is provided to support this. - Graham Everest (g.everest(AT)uea.ac.uk), Mar 01 2001
For n>=4 a(n-4) is the resultant of the polynomials x^3-x-1 and x^(n+1)-x^n-1. For n=4 in fact the result is 0 as we see from the identity x^5-x^4-1=(x^3-x-1)(x^2-x+1). The characteristic polynomial of the sequence is x^6+x^5-x^4-3x^3-x^2+x+1 = (x^3-x-1)*(x^3+x^2-1). - Richard Choulet, Aug 14 2007
From Peter Bala, Sep 15 2019: (Start)
This is a linear divisibility sequence of order 6. It is a particular case of a family of divisibility sequences studied by Roettger et al. The o.g.f. has the form x*d/dx(f(x)/(x^3*f(1/x))) where f(x) = x^3 - x - 1.
More generally, if f(x) = 1 + P*x + Q*x^2 + x^3 or f(x) = -1 + P*x + Q*x^2 + x^3, where P and Q are integers, then the rational function x*d/dx(f(x)/(x^3*f(1/x))) is the generating function for a linear divisibility sequence of order 6. Cf. A001351. There are corresponding results when f(x) is a monic quartic polynomial with constant term 1. (End)
Resultant of the (s_3, s_3+n) pair where s_n(X) is X^n-X-1. See Rush link. - Michel Marcus, Sep 30 2019

			G.f. = x + x^2 + x^3 + 5*x^4 + x^5 + 7*x^6 + 8*x^7 + 5*x^8 + 19*x^9 + ...

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David E. Rush, Degree n Relatives of the Golden Ratio and Resultants of the Corresponding Polynomials, Fib. Q. 50(4), 2012, 313-325. See p. 319.
Index entries for linear recurrences with constant coefficients, signature (-1,1,3,1,-1,-1).

Cf. A001608, A078712, A104499, A001351.

import Data.List (zipWith6)
a001945 n = a001945_list !! n
a001945_list = 0 : 1 : 1 : 1 : 5 : 1 : zipWith6
   (\u v w x y z -> - u + v + 3*w + x - y - z)
     (drop 5 a001945_list) (drop 4 a001945_list) (drop 3 a001945_list)
     (drop 2 a001945_list) (drop 1 a001945_list) (drop 0 a001945_list)
-- Reinhard Zumkeller, Jan 11 2012

A001945:=z*(1+2*z+z**2+2*z**3+z**4)/(z**3-z-1)/(z**3+z**2-1); # conjectured by Simon Plouffe in his 1992 dissertation

a[0] = 0; a[1] = a[2] = a[3] = a[5] = 1; a[4] = 5; a[n_] := a[n] = -a[n - 1] + a[n - 2] + 3a[n - 3] + a[n - 4] - a[n - 5] - a[n - 6]; Table[ a[n], {n, 0, 46}] (* Robert G. Wilson v, Mar 10 2005 *)
LinearRecurrence[{-1, 1, 3, 1, -1, -1}, {0, 1, 1, 1, 5, 1}, 50] (* T. D. Noe, Jan 11 2012 *)
a[ n_] := Sign[n] SeriesCoefficient[ x * (1 + 2 x + x^2 + 2 x^3 + x^4) / (1 + x - x^2 - 3 x^3 - x^4 + x^5 + x^6), {x, 0, Abs @ n}]; (* Michael Somos, Apr 25 2014 *)

{a(n) = sign(n) * polcoeff( x * (1 + 2*x + x^2 + 2*x^3 + x^4) / (1 + x - x^2 - 3*x^3 - x^4 + x^5 + x^6) + x * O(x^abs(n)), abs(n))}; /* Michael Somos, Apr 25 2014 */

a(n)=([0,1,0,0,0,0; 0,0,1,0,0,0; 0,0,0,1,0,0; 0,0,0,0,1,0; 0,0,0,0,0,1; -1,-1,1,3,1,-1]^n*[0;1;1;1;5;1])[1,1] \\ Charles R Greathouse IV, Jul 19 2016

L3(n) = polsym(x^3-x-1, n)[n+1]; \\ A001608
a(n) = my(L3n=L3(n)); L3n - matdet([L3n, L3(2*n);1, L3n])/2; \\ Michel Marcus, Sep 30 2019

G.f.: (x^5+2x^4+x^3+2x^2+x)/(x^6+x^5-x^4-3x^3-x^2+x+1). - Ralf Stephan, Dec 15 2002
a(n) ~ r1^n-2*real(r2^n), with r1=1.324717957 the inverse real root of x^3+x^2-1=0 and r2=(0.87744+0.7448617i) one inverse complex root of x^3-x-1=0. With n>9, a(n) = round(r1^n-2*real(r2^n)). - Ralf Stephan, Dec 17 2002
a(n) = A001608(n) + A078712(n). - Ralf Stephan, Dec 27 2002
a(A104499(n+1)) = A204138(n). - Reinhard Zumkeller, Jan 11 2012
a(-n) = -a(n). - Michael Somos, Apr 25 2014
a(n) = (alpha^n - 1)*(beta^n - 1)*(gamma^n - 1) where alpha, beta and gamma are the zeros of x^3 - x - 1. - Peter Bala, Sep 15 2019

More terms from James Sellers, Dec 23 1999

cp's OEIS Frontend

A294028 Composite numbers k such that k | 1 - A078712(k).

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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.

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A001945 a(n+6) = -a(n+5) + a(n+4) + 3a(n+3) + a(n+2) - a(n+1) - a(n). a(n) = sign(n) if abs(n)<=3.

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A117602 Padovan numbers which can be divided by their digital root.

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