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
Examples
a(11) = 11 because a(7) = 7 and a(8) = 4.
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).
Programs
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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
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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
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Mathematica
LinearRecurrence[{0,0,1,1}, {4,0,0,3}, 60] (* G. C. Greubel, Mar 04 2019 *) -
PARI
polsym(x^4-x-1,55) \\ Joerg Arndt, Mar 04 2019
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Sage
((4-x^3)/(1-x^3-x^4)).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Mar 04 2019
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)
Extensions
More terms from Christian G. Bower, Dec 23 1999
More terms from Benoit Cloitre, Oct 27 2003
Comments