A050443 a(0)=4, a(1)=0, a(2)=0, a(3)=3; thereafter a(n) = a(n-3) + a(n-4).

4, 0, 0, 3, 4, 0, 3, 7, 4, 3, 10, 11, 7, 13, 21, 18, 20, 34, 39, 38, 54, 73, 77, 92, 127, 150, 169, 219, 277, 319, 388, 496, 596, 707, 884, 1092, 1303, 1591, 1976, 2395, 2894, 3567, 4371, 5289, 6461, 7938, 9660, 11750, 14399, 17598, 21410, 26149, 31997

Offset

0,1

Comments

Related to Perrin sequence. a(p) is divisible by p for primes p.

Wells states that Mihaly Bencze [Beneze] (1998) proved the divisibility property for this sequence: that a(n) is always divisible by n when n is prime. - Gary W. Adamson, Nov 14 2006

As a(n) = trace(M^n) where M = [0,1,0,0; 0,0,1,0; 0,0,0,1; 1,1,0,0], the previous property comes from the fact that trace(M^n) = trace(M) (= 0) mod n for n prime. - Robert FERREOL, Apr 09 2024

References

David Wells, "Prime Numbers, the Most Mysterious Figures in Math", John Wiley & Sons, Inc.; 2005, p. 103.

Links

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe)

Sadjia Abbad and Hacène Belbachir, The r-Fibonacci polynomial and its companion sequences linked with some classical sequences, Integers (2025), Vol. 25, Art. No. A38. See p. 17.

Mihaly Bencze, Dan Saracino, and Allen Stenger, Solution of Problem 10655: A Recurrence Generating Multiples of Primes, American Mathematical Monthly 107 (2000) 281-282.

Johann Cigler, Recurrences for certain sequences of binomial sums in terms of (generalized) Fibonacci and Lucas polynomials, arXiv:2212.02118 [math.NT], 2022.

Gregory T. Minton, Linear recurrence sequences satisfying congruence conditions, Proc. Amer. Math. Soc. (2014), Vol. 142, No. 7, 2337-2352. MR3195758.

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

Formula

G.f.: (4-x^3)/(1-x^3-x^4). - Christian G. Bower, Dec 23 1999

a(n) = (x_1)^n + (x_2)^n + (x_3)^n + (x_4)^n where (x_i) 1 <= i <= 4 are the roots of x^4 = x + 1. - Benoit Cloitre, Oct 27 2003

Let M = the 4 X 4 matrix [0,1,0,0; 0,0,1,0; 0,0,0,1; 1,1,0,0]; then a(n) = the leftmost term of M^n * [4,0,0,3]. Example: a(13) = 13 since M^13 * [4,0,0,3] = [13,21,18,20]. - Gary W. Adamson, Nov 14 2006

a(0) = 4 and a(n) = n*Sum_{k=1..floor(n/3)} binomial(k,n-3*k)/k for n > 0. - Seiichi Manyama, Mar 04 2019

From Aleksander Bosek, Mar 10 2019: (Start)

a(n+10) = a(n+5) + 2*a(n+3) + a(n).

a(n+11) = a(n+6) + 3*a(n+1) + 2*a(n).

a(n+12) = a(n+10) + 5*a(n+5) + a(n).

a(n+12) = 3*a(n+5) + a(n+3) + a(n).

a(n+13) = 3*a(n+6) + 2*a(n+1) + a(n).

a(n+14) = 2*a(n+8) + 3*a(n+3) + a(n).

a(n+15) = 2*a(n+7) + 4*a(n+5) + a(n).

a(n+15) = 2*a(n+9) + 4*a(n+1) + 3*a(n).

a(n+19) = a(n+17) + 5*a(n+5) + a(n).

a(n+20) = 5*a(n+10) + 6*a(n+5) + a(n).

a(n+22) = a*(n+21) + 5*a(n+5) + a(n).

a(n+25) = 2*a(n+21) + 5*a(n+5) + a(n).

a((s+4)*n+m) = Sum_{l=0..n} binomial(n-l,l)*a(s*n+l+m) for every m,s > 0.

a(m) = Sum_{l=0..n}(-1)^{n-l}*binomial(n-l,l)*a(m+n+3*l) for every m > 0. (End)

a(n) = 4*A017817(n) - A017817(n-3). - R. J. Mathar, Aug 10 2021

Example

a(11) = 11 because a(7) = 7 and a(8) = 4.

Mathematica

LinearRecurrence[{0, 0, 1, 1}, {4, 0, 0, 3}, 60] (* G. C. Greubel, Mar 04 2019 *)

Prog

(PARI) polsym(x^4-x-1, 55) \\ Joerg Arndt, Mar 04 2019
(Magma) I:=[4, 0, 0, 3]; [n le 4 select I[n] else Self(n-3) +Self(n-4): n in [1..60]]; // G. C. Greubel, Mar 04 2019
(SageMath) ((4-x^3)/(1-x^3-x^4)).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Mar 04 2019
(GAP) a:=[4, 0, 0, 3];; for n in [5..60] do a[n]:=a[n-3]+a[n-4]; od; Print(a); # Muniru A Asiru, Mar 09 2019

Crossrefs

Column 3 of A306646.

Cf. A001608, A052338.

Cf. A087935, A087936.

Sequence in context: A266270 A091467 A224861 * A306770 A389779 A308278

Adjacent sequences: A050440 A050441 A050442 * A050444 A050445 A050446

Keyword

easy,nonn

Author

Tony Davie (ad(AT)dcs.st-and.ac.uk), Dec 23 1999

Extensions

More terms from Christian G. Bower, Dec 23 1999

More terms from Benoit Cloitre, Oct 27 2003

Status

approved