A077947 Expansion of 1/(1 - x - x^2 - 2*x^3).
1, 1, 2, 5, 9, 18, 37, 73, 146, 293, 585, 1170, 2341, 4681, 9362, 18725, 37449, 74898, 149797, 299593, 599186, 1198373, 2396745, 4793490, 9586981, 19173961, 38347922, 76695845, 153391689, 306783378, 613566757, 1227133513, 2454267026, 4908534053, 9817068105
Offset: 0
Examples
It is shown in A294627 that there are 42 quaternary sequences (i.e. build from four digits 0, 1, 2, 3) and having both 0 and 3 in every (consecutive) triple. Only a(4) = 9 of them start with 0 and end with 0 or 3: 0030, 0033, 0130, 0230, 0300, 0303, 0310, 0320, 0330. - _Wojciech Florek_, Jul 30 2018
References
- S. Roman, Introduction to Coding and Information Theory, Springer-Verlag, 1996, p. 42
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- G. Cerda-Morales, A note on modified third-order Jacobsthal Numbers, arxiv:1905.00725 [math.CO], 2019.
- M. H. Cilasun, An Analytical Approach to Exponent-Restricted Multiple Counting Sequences, arXiv preprint arXiv:1412.3265 [math.NT], 2014.
- M. H. Cilasun, Generalized Multiple Counting Jacobsthal Sequences of Fermat Pseudoprimes, Journal of Integer Sequences, Vol. 19, 2016, #16.2.3.
- Charles K. Cook and Michael R. Bacon, Some identities for Jacobsthal and Jacobsthal-Lucas numbers satisfying higher order recurrence relations, Annales Mathematicae et Informaticae, 41 (2013) pp. 27-39.
- W. Florek, A class of generalized Tribonacci sequences applied to counting problems, Appl. Math. Comput., 338 (2018), 809-821.
- S. Kitaev, J. Remmel and M. Tiefenbruck, Marked mesh patterns in 132-avoiding permutations I, arXiv preprint arXiv:1201.6243 [math.CO], 2012. - From _N. J. A. Sloane_, May 09 2012
- Sergey Kitaev, Jeffrey Remmel and Mark Tiefenbruck, Quadrant Marked Mesh Patterns in 132-Avoiding Permutations II, Electronic Journal of Combinatorial Number Theory, Volume 15 #A16; see also arXiv preprint, arXiv:1302.2274 [math.CO], 2013.
- Vladimir Victorovich Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565, arXiv:1009.2565 [math.CO], 2010.
- Evren Eyican Polatlı and Yüksel Soykan, On generalized third-order Jacobsthal numbers, Asian Res. J. of Math. (2021) Vol. 17, No. 2, 1-19, Article No. ARJOM.66022.
- Yüksel Soykan, Summing Formulas For Generalized Tribonacci Numbers, arXiv:1910.03490 [math.GM], 2019.
- Anthony Zaleski and Doron Zeilberger, On the Intriguing Problem of Counting (n+1,n+2)-Core Partitions into Odd Parts, arXiv:1712.10072 [math.CO], 2017.
- Index entries for linear recurrences with constant coefficients, signature (1,1,2).
Programs
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Magma
[Round(2^(n+2)/7): n in [0..40]]; // Vincenzo Librandi, Jun 25 2011
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Maple
seq(round(2^(n+2)/7),n=0..25); # Mircea Merca, Dec 28 2010
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Mathematica
CoefficientList[Series[1/(1 - x - x^2 - 2*x^3), {x, 0, 100}], x] (* or *) LinearRecurrence[{1, 1, 2}, {1, 1, 2}, 70] (* Vladimir Joseph Stephan Orlovsky, Jun 28 2011 *) -
Maxima
a(n):=sum(sum(binomial(k,j)*binomial(j,n-3*k+2*j)*2^(k-j),j,0,k),k,1,n); /* Vladimir Kruchinin, Sep 07 2010 */
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PARI
Vec(1/(1-x-x^2-2*x^3) + O(x^100)) \\ Altug Alkan, Oct 31 2015
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Python
def A077947(n): return (k:=(m:=1<
Formula
G.f.: 1/((1-2*x)*(1+x+x^2)).
a(n) = a(n-1)+a(n-2)+2*a(n-3). - Paul Curtz, May 23 2008
a(n) = round(2^(n+2)/7). - Mircea Merca, Dec 28 2010
a(n) = 4*2^n/7 + 3*cos(2*Pi*n/3)/7 + sqrt(3)*sin(2*Pi*n/3)/21. - Paul Barry, Jan 23 2004
Convolution of A000079 and A049347. a(n) = Sum_{k=0..n} 2^k*2*sqrt(3)*cos(2*Pi(n-k)/3+Pi/6)/3. - Paul Barry, May 19 2004
a(n) = sum(sum(binomial(k,j)*binomial(j,n-3*k+2*j)*2^(k-j),j,0,k),k,1,n), n>0. - Vladimir Kruchinin, Sep 07 2010
Partial sums of A078010 starting (1, 0, 1, 3, 4, 9, ...). - Gary W. Adamson, May 13 2013
a(n) = (1/14)*(2^(n + 3) + (-1)^n*((-1)^floor(n/3) + 4*(-1)^floor((n + 1)/3) + 2*(-1)^floor((n + 2)/3) + (-1)^floor((n + 4)/3))). - John M. Campbell, Dec 23 2016
a(n) = (1/63)*(9*2^(2 + n) + (-1)^n*(2 + 9*floor(n/6) - 32*floor((n + 5)/6) + 24*floor((n + 7)/6) + 20*floor((n + 8)/6) - 10*floor((n + 9)/6) - 27*floor((n + 10)/6) + 14*floor((n + 11)/6) + 3*floor((n + 13)/6) - 2*floor((n + 14)/6) + floor((n + 15)/6))). - John M. Campbell, Dec 23 2016
7*a(n) = 2^(n+2) + A167373(n+1). - R. J. Mathar, Feb 06 2020
a(n) = T(n+1) + 2*(a(1)*T(n-1) + a(2)*T(n-2) + ... + a(n-2)*T(2) + a(n-1)*T(1)) for T(n) = A000073(n), the tribonacci numbers. - Greg Dresden and Bora Bursalı, Sep 14 2023
Extensions
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021
Comments