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

%I A050443 #78 Apr 24 2025 10:46:30
%S A050443 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,
%T A050443 169,219,277,319,388,496,596,707,884,1092,1303,1591,1976,2395,2894,
%U A050443 3567,4371,5289,6461,7938,9660,11750,14399,17598,21410,26149,31997
%N A050443 a(0)=4, a(1)=0, a(2)=0, a(3)=3; thereafter a(n) = a(n-3) + a(n-4).
%C A050443 Related to Perrin sequence. a(p) is divisible by p for primes p.
%C A050443 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
%C A050443 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
%D A050443 David Wells, "Prime Numbers, the Most Mysterious Figures in Math", John Wiley & Sons, Inc.; 2005, p. 103.
%H A050443 Seiichi Manyama, <a href="/A050443/b050443.txt">Table of n, a(n) for n = 0..10000</a> (terms 0..1000 from T. D. Noe)
%H A050443 Sadjia Abbad and Hacène Belbachir, <a href="https://math.colgate.edu/~integers/z38/z38.pdf">The r-Fibonacci polynomial and its companion sequences linked with some classical sequences</a>, Integers (2025), Vol. 25, Art. No. A38. See p. 17.
%H A050443 Mihaly Bencze, Dan Saracino, and Allen Stenger, <a href="https://www.jstor.org/stable/2589334">Solution of Problem 10655: A Recurrence Generating Multiples of Primes</a>, American Mathematical Monthly 107 (2000) 281-282.
%H A050443 Johann Cigler, <a href="https://arxiv.org/abs/2212.02118">Recurrences for certain sequences of binomial sums in terms of (generalized) Fibonacci and Lucas polynomials</a>, arXiv:2212.02118 [math.NT], 2022.
%H A050443 Gregory T. Minton, <a href="http://dx.doi.org/10.1090/S0002-9939-2014-12168-X">Linear recurrence sequences satisfying congruence conditions</a>, Proc. Amer. Math. Soc. (2014), Vol. 142, No. 7, 2337-2352. MR3195758.
%H A050443 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,1,1).
%F A050443 G.f.: (4-x^3)/(1-x^3-x^4). - _Christian G. Bower_, Dec 23 1999
%F A050443 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
%F A050443 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
%F A050443 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
%F A050443 From _Aleksander Bosek_, Mar 10 2019: (Start)
%F A050443 a(n+10) = a(n+5) + 2*a(n+3) + a(n).
%F A050443 a(n+11) = a(n+6) + 3*a(n+1) + 2*a(n).
%F A050443 a(n+12) = a(n+10) + 5*a(n+5) + a(n).
%F A050443 a(n+12) = 3*a(n+5) + a(n+3) + a(n).
%F A050443 a(n+13) = 3*a(n+6) + 2*a(n+1) + a(n).
%F A050443 a(n+14) = 2*a(n+8) + 3*a(n+3) + a(n).
%F A050443 a(n+15) = 2*a(n+7) + 4*a(n+5) + a(n).
%F A050443 a(n+15) = 2*a(n+9) + 4*a(n+1) + 3*a(n).
%F A050443 a(n+19) = a(n+17) + 5*a(n+5) + a(n).
%F A050443 a(n+20) = 5*a(n+10) + 6*a(n+5) + a(n).
%F A050443 a(n+22) = a*(n+21) + 5*a(n+5) + a(n).
%F A050443 a(n+25) = 2*a(n+21) + 5*a(n+5) + a(n).
%F A050443 a((s+4)*n+m) = Sum_{l=0..n} binomial(n-l,l)*a(s*n+l+m) for every m,s > 0.
%F A050443 a(m) = Sum_{l=0..n}(-1)^{n-l}*binomial(n-l,l)*a(m+n+3*l) for every m > 0. (End)
%F A050443 a(n) = 4*A017817(n) - A017817(n-3). - _R. J. Mathar_, Aug 10 2021
%e A050443 a(11) = 11 because a(7) = 7 and a(8) = 4.
%t A050443 LinearRecurrence[{0,0,1,1}, {4,0,0,3}, 60] (* _G. C. Greubel_, Mar 04 2019 *)
%o A050443 (PARI) polsym(x^4-x-1,55) \\ _Joerg Arndt_, Mar 04 2019
%o A050443 (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
%o A050443 (Sage) ((4-x^3)/(1-x^3-x^4)).series(x, 60).coefficients(x, sparse=False) # _G. C. Greubel_, Mar 04 2019
%o A050443 (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
%Y A050443 Column 3 of A306646.
%Y A050443 Cf. A001608, A052338.
%Y A050443 Cf. A087935, A087936.
%K A050443 easy,nonn
%O A050443 0,1
%A A050443 Tony Davie (ad(AT)dcs.st-and.ac.uk), Dec 23 1999
%E A050443 More terms from _Christian G. Bower_, Dec 23 1999
%E A050443 More terms from _Benoit Cloitre_, Oct 27 2003