A078010 -id:A078010 - cp's OEIS frontend

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

N. J. A. Sloane, Nov 17 2002

Number of sequences of codewords of total length n from the code C={0,10,110,111}. E.g., a(3)=5 corresponds to the sequences 000, 010, 100, 110 and 111. - Paul Barry, Jan 23 2004
In other words: number of compositions of n into 1 kind of 1's and 2's and two kinds of 3's. - Joerg Arndt, Jun 25 2011
Diagonal sums of number Pascal-(1,2,1) triangle A081577. - Paul Barry, Jan 24 2005
For n>0: a(n) = A173593(2*n+1) - A173593(2*n); a(n+1) = A173593(2*n) - A173593(2*n-1). - Reinhard Zumkeller, Feb 22 2010
Sums of 3 successive terms are powers of 2. - Mark Dols, Aug 20 2010
For n > 2, a(n) is the number of quaternary sequences of length n (i) starting with q(0)=0; (ii) ending with q(n-1)=0 or 3 and (iii) in which all triples (q(i), q(i+1), q(i+2)) contain digits 0 and 3; cf. A294627. - Wojciech Florek, Jul 30 2018

			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

S. Roman, Introduction to Coding and Information Theory, Springer-Verlag, 1996, p. 42

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).

Apart from signs, same as A077972.
Cf. A139217 and A139218.
Cf. A078010.
Cf. A294627.

[Round(2^(n+2)/7): n in [0..40]]; // Vincenzo Librandi, Jun 25 2011

seq(round(2^(n+2)/7),n=0..25); # Mircea Merca, Dec 28 2010

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 *)

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 */

Vec(1/(1-x-x^2-2*x^3) + O(x^100)) \\ Altug Alkan, Oct 31 2015

def A077947(n): return (k:=(m:=1<=7) # Chai Wah Wu, Jan 21 2023

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

Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A033138 a(n) = floor(2^(n+2)/7).

Original entry on oeis.org

1, 2, 4, 9, 18, 36, 73, 146, 292, 585, 1170, 2340, 4681, 9362, 18724, 37449, 74898, 149796, 299593, 599186, 1198372, 2396745, 4793490, 9586980, 19173961, 38347922, 76695844, 153391689, 306783378, 613566756, 1227133513, 2454267026

Offset: 1

Clark Kimberling

Previous name was: "Base 2 digits are, in order, the first n terms of the periodic sequence with initial period 1,0,0".
Here we let p = 3 to produce the above sequence, but p can be an arbitrary natural number. By letting p = 2, 4, 6, 7 we produce A000975, A083593, A195904 and A117302. We denote by U[p,n,m] the number of cases in which the first player in a game of Russian roulette gets killed when p players use a gun with n chambers and m bullets. They never rotate the cylinder after the game starts.
The chambers can be represented by the list {1,2,...,n}. We are going to calculate the following (0), (1),...(t) separately. (0) The first player gets killed when one bullet is in the first chamber and the remaining (m-1)- bullets are in {2,3,...,n}. We have binomial(n-1,m-1) cases for this. (1) The first gets killed when one bullet is in the (p+1)st chamber and the rest of the bullets are in {p+2,..,n}. We have binomial(n-p-1,m-1) cases for this. We continue to calculate and the last is (t), where t = floor((n-m)/p). (t) The first gets killed when one bullet is in the (pt+1)st chamber and the remaining bullets are in {pt+2,...,n}. We have binomial(n-pt-1,m-1) cases for this.
Therefore U[p,n,m] = Sum_{z=0..t} binomial(n-pz-1,m-1), where t = floor((n-m)/p). Let A[p,n] be the number of the cases in which the first player gets killed when p players use a gun with n chambers and the number of bullets can be from 1 to n. Then A[p,n] = Sum_{m=1..n} U[p,n,m]. - Ryohei Miyadera, Tomohide Hashiba, Yuta Nakagawa, Hiroshi Matsui, Jun 04 2006
Partial sums of A077947. - Mircea Merca, Dec 28 2010
a(n+1) is the number of partitions of n into two kinds of part 1 and one kind of part 2. - Joerg Arndt, Mar 10 2015
A078010(n) = b(n+1) - 2*b(n) + b(n-1) where b=A078010. - Michael Somos, Nov 18 2020

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 926
S. Klavzar, Structure of Fibonacci cubes: a survey, Institute of Mathematics, Physics and Mechanics Jadranska 19, 1000 Ljubljana, Slovenia; Preprint series Vol. 49 (2011), 1150 ISSN 2232-2094.
Index entries for linear recurrences with constant coefficients, signature (2,0,1,-2).

Cf. A000975, A078010, A083593, A195904, A117302, A023001, A111662, A077947.

[Round((4*2^n-2)/7): n in [1..40]]; // Vincenzo Librandi, Jun 25 2011

seq(iquo(2^n,7),n=3..34); # Zerinvary Lajos, Apr 20 2008

U[p_, n_, m_, v_]:=Block[{t}, t=Floor[(1+p-m+n-v)/p];Sum[Binomial[n-v-p*z,m-1],{z,0,t-1}]]; A[p_,n_,v_]:=Sum[U[p,n,k,v],{k,1,n}]; (* Here we let p = 3 to produce the above sequence, but this code can produce A000975, A083593, A195904, A117302 for p = 2,4,6,7. *) Table[A[3,n,1], {n,1,20}] (* Ryohei Miyadera, Tomohide Hashiba, Yuta Nakagawa, Hiroshi Matsui, Jun 04 2006 *)

a(n)=2^(n+2)\7 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 2*a(n-1) + a(n-3) - 2*a(n-4). -John W. Layman
G.f.: 1/((1-x^3)*(1-2*x)); a(n) = sum{k=0..floor(n/3), 2^(n-3*k)}; a(n) = Sum_{k=0..n} 2^k*( cos(2*Pi*(n-k)/3 + Pi/3)/3 + sqrt(3)*sin(2*Pi*(n-k)/3 + Pi/3)/3 + 1/3 ). - Paul Barry, Apr 16 2005
a(n) = floor(2^(n+2)/7). - Gary Detlefs, Sep 06 2010
a(n) = floor((4*2^n - 1)/7) = ceiling((4*2^n - 4)/7) = round((4*2^n - 2)/7) = round((8*2^n - 5)/14); a(n) = a(n-3) + 2^(n-1), n>3. - Mircea Merca, Dec 28 2010
a(n) = 4/7*2^n - 5/21*cos(2/3*Pi*n) + 1/21*3^(1/2)*sin(2/3*Pi*n)-1/3. - Leonid Bedratyuk, May 13 2012

Edited by Jeremy Gardiner, Oct 08 2011

New name (using formula form Gary Detlefs) from Joerg Arndt, Mar 10 2015

A122016 Riordan array(1, x(1+2x)/(1-x)).

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 0, 3, 6, 1, 0, 3, 15, 9, 1, 0, 3, 24, 36, 12, 1, 0, 3, 33, 90, 66, 15, 1, 0, 3, 42, 171, 228, 105, 18, 1, 0, 3, 51, 279, 579, 465, 153, 21, 1, 0, 3, 60, 414, 1200, 1500, 828, 210, 24, 1, 0, 3, 69, 576, 2172, 3858, 3258, 1344, 276, 27, 1

Offset: 0

Philippe Deléham, Sep 24 2006

Triangle T(n,k), 0 <= k <= n, read by rows given by [0,3,-2,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. Rising and falling diagonals are A078010 and A122552.

			Triangle begins:
  1;
  0, 1;
  0, 3,  1;
  0, 3,  6,   1;
  0, 3, 15,   9,    1;
  0, 3, 24,  36,   12,    1;
  0, 3, 33,  90,   66,   15,   1;
  0, 3, 42, 171,  228,  105,  18,   1;
  0, 3, 51, 279,  579,  465, 153,  21,  1;
  0, 3, 60, 414, 1200, 1500, 828, 210, 24, 1;

Huyile Liang, Jinyang Zhang, and Yu Wang, Some properties of the matrix related to q-coloured coordination number, Filomat (2024) Vol. 38, No. 4, 1465-1477. See p. 1466.

Cf. Diagonals A000012, A008585, A062741, columns A000007, A122553, A122709, row sums A026150.

Cf. A078010, A084938, A102900, A117575, A122117, A122552, A147518.

T[n_,k_]:=SeriesCoefficient[(1-x)/(1-(y+1)*x-2*y*x^2),{x,0,n},{y,0,k}]; Table[T[n,k],{n,0,10},{k,0,n}]//Flatten (* Stefano Spezia, Dec 27 2023 *)

Sum_{k=0..n} T(n,k)*x^(n-k) = A026150(n), A102900(n) for x = 1, 2.
T(n,k) = T(n-1,k) + T(n-1,k-1) + 2*T(n-2,k-1). - Philippe Deléham, Sep 25 2006
G.f.: (1-x)/(1-(y+1)*x-2*y*x^2). - Philippe Deléham, Jan 31 2012
Sum_{k=0..n} T(n,k)*x^k = A117575(n+1), A000007(n), A026150(n), A122117(n), A147518(n) for x = -1, 0, 1, 2, 3 respectively. - Philippe Deléham, Jan 31 2012

More terms from Stefano Spezia, Dec 27 2023

A122552 a(0)=a(1)=a(2)=1, a(n) = a(n-1) + a(n-2) + 2*a(n-3) for n > 2.

Original entry on oeis.org

1, 1, 1, 4, 7, 13, 28, 55, 109, 220, 439, 877, 1756, 3511, 7021, 14044, 28087, 56173, 112348, 224695, 449389, 898780, 1797559, 3595117, 7190236, 14380471, 28760941, 57521884, 115043767, 230087533, 460175068, 920350135, 1840700269, 3681400540

Offset: 0

Philippe Deléham, Sep 20 2006

Equals INVERT transform of (1, 0, 3, 0, 3, 0, 3, ...). - Gary W. Adamson, Apr 27 2009
No term is divisible by 3. - Vladimir Joseph Stephan Orlovsky, Mar 24 2011
For n > 3, a(n) is the number of quaternary sequences of length n-1 starting with q(0) = 0, in which all triples (q(i), q(i+1), q(i+2)) contain digits 0 and 3; cf. A294627. - Wojciech Florek, Jul 30 2018
For n > 0, a(n) is the number of ways to tile a strip of length n with squares, dominoes, and two colors of trominoes, with the restriction that the first tile cannot be a domino. - Greg Dresden and Bora Bursalı, Aug 31 2023

			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(5=4+1) = 13 of them start with 0: 003x, 030x, 03y0, 0y30, 0330, where x = 0, 1, 2, 3 and y = 1, 2.

Wojciech Florek, Table of n, a(n) for n = 0..2000
Wojciech Florek, A class of generalized Tribonacci sequences applied to counting problems, Appl. Math. Comput., 338 (2018), 809-821.
Index entries for linear recurrences with constant coefficients, signature (1,1,2).

Cf. A294627.

a:=[1,1,1];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+2*a[n-3]; od; a; # Muniru A Asiru, Jul 30 2018

seq(coeff(series((1-x^2)/(1-x-x^2-2*x^3), x,n+1),x,n),n=0..40); # Muniru A Asiru, Aug 02 2018

LinearRecurrence[{1, 1, 2}, {1, 1, 1}, 40]
CoefficientList[ Series[(x^2 - 1)/(2x^3 + x^2 + x - 1), {x, 0, 35}], x] (* Robert G. Wilson v, Jul 30 2018 *)

Vec((1-x^2)/(1-x-x^2-2*x^3)+O(x^99)) \\ Charles R Greathouse IV, Jan 17 2012

from sage.combinat.sloane_functions import recur_gen3; it = recur_gen3(1,1,1,1,1,2); [next(it) for i in range(30)] # Zerinvary Lajos, Jun 25 2008

a(3*n) = 2*a(3*n-1)+2, a(3*n+1) = 2*a(3*n)-1, a(3*n+2) = 2*a(3*n+1)-1, a(0)=1.
G.f.: (1-x^2)/(1-x-x^2-2*x^3).
a(n) = ((-1)^n*A130815(n+2) + 3*2^n)/7. - R. J. Mathar, Nov 30 2008
From Paul Curtz, Oct 02 2009: (Start)
a(n) = A140295(n+2)/4.
a(n+1) - 2a(n) = period 3: repeat -1,-1,2 = -A061347.
a(n) - a(n-1) = 0,0,3,3,6,15,27,54,111,... = 3*A077947.
a(n) - a(n-2) = 0,3,6,9,21,42,81,....
a(n) - a(n-3) = 3,6,12,24,... = A007283 = 3*A000079.
a(3n) + a(3n+1) + a(3n+2) = 3,24,192,... = A103333(n+1) = A140295(3n) + A140295(3n+1) + A140295(3n+2).
See A078010, A139217, A139218. (End)

Corrected by T. D. Noe, Nov 01 2006, Nov 07 2006

Typo in definition corrected by Paul Curtz, Oct 02 2009

cp's OEIS Frontend

A077947 Expansion of 1/(1 - x - x^2 - 2*x^3).

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A033138 a(n) = floor(2^(n+2)/7).

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A122016 Riordan array(1, x*(1+2*x)/(1-x)).

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A122552 a(0)=a(1)=a(2)=1, a(n) = a(n-1) + a(n-2) + 2*a(n-3) for n > 2.

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A122016 Riordan array(1, x(1+2x)/(1-x)).