A076264 -id:A076264 - cp's OEIS frontend

A099503 Expansion of 1/(1-4*x+x^3).

1, 4, 16, 63, 248, 976, 3841, 15116, 59488, 234111, 921328, 3625824, 14269185, 56155412, 220995824, 869714111, 3422701032, 13469808304, 53009519105, 208615375388, 820991693248, 3230957253887, 12715213640160, 50039862867392

Offset: 0

Paul Barry, Oct 20 2004

A transform of A000302 under the mapping g(x) ->(1/(1+x^3)) * g(x/(1+x^3)).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,0,-1).

Cf. A000071, A000302, A076264, A099504.

[n le 3 select 4^(n-1) else 4*Self(n-1) -Self(n-3): n in [1..30]]; // G. C. Greubel, Aug 03 2023

CoefficientList[Series[1/(1-4x+x^3),{x,0,30}],x]  (* Harvey P. Dale, Apr 01 2011 *)
LinearRecurrence[{4,0,-1}, {1,4,16}, 30] (* G. C. Greubel, Aug 03 2023 *)

@CachedFunction
def a(n): # a = A099503
    if (n<3): return 4^n
    else: return 4*a(n-1) - a(n-3)
[a(n) for n in range(31)] # G. C. Greubel, Aug 03 2023

a(n) = 4*a(n-1) - a(n-3).

a(n) = Sum_{k=0..floor(n/3)} binomial(n-2*k, k)*(-1)^k*4^(n-3*k).

A099504 Expansion of 1/(1-5*x+x^3).

Original entry on oeis.org

1, 5, 25, 124, 615, 3050, 15126, 75015, 372025, 1844999, 9149980, 45377875, 225044376, 1116071900, 5534981625, 27449863749, 136133246845, 675131252600, 3348206399251, 16604898749410, 82349362494450, 408398606072999

Offset: 0

Paul Barry, Oct 20 2004

A transform of A000351 under the mapping g(x)->(1/(1+x^3))g(x/(1+x^3)).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,0,-1).

Cf. A000071, A000351, A076264, A099503.

[n le 3 select 5^(n-1) else 5*Self(n-1) -Self(n-3): n in [1..30]]; // G. C. Greubel, Aug 03 2023

A099504:=n->sum(binomial(n-2*i, i)*(-1)^i*5^(n-3*i), i=0..floor(n/3)); seq(A099504(n), n=0..30); # Wesley Ivan Hurt, Dec 03 2013

Table[Sum[Binomial[n-2*i,i]*(-1)^i*5^(n-3*i), {i,0,Floor[n/3]}], {n,0, 30}] (* Wesley Ivan Hurt, Dec 03 2013 *)
LinearRecurrence[{5,0,-1}, {1,5,25}, 30] (* G. C. Greubel, Aug 03 2023 *)

@CachedFunction
def a(n): # a = A099504
    if (n<3): return 5^n
    else: return 5*a(n-1) - a(n-3)
[a(n) for n in range(31)] # G. C. Greubel, Aug 03 2023

a(n) = 5*a(n-1) - a(n-3).

a(n) = Sum_{k=0..floor(n/3)} binomial(n-2*k, k)*(-1)^k*5^(n-3*k).

A225682 Triangle read by rows: T(n,k) (0 <= k <= n) = chi(k)*binomial(n,k), where chi(k) = 1,-1,0 according as k == 0,1,2 mod 3.

Original entry on oeis.org

1, 1, -1, 1, -2, 0, 1, -3, 0, 1, 1, -4, 0, 4, -1, 1, -5, 0, 10, -5, 0, 1, -6, 0, 20, -15, 0, 1, 1, -7, 0, 35, -35, 0, 7, -1, 1, -8, 0, 56, -70, 0, 28, -8, 0, 1, -9, 0, 84, -126, 0, 84, -36, 0, 1, 1, -10, 0, 120, -210, 0, 210, -120, 0, 10, -1, 1, -11, 0, 165, -330, 0, 462, -330, 0, 55, -11, 0, 1, -12, 0, 220, -495, 0, 924, -792, 0, 220, -66, 0, 1

Offset: 0

Murray R. Bremner and N. J. A. Sloane, May 17 2013

Corresponding to row n of this triangle, define a generating function G_n(x) = 1/(Sum_{k=0..n} T(n,k)*x^k).
Then G_n(x) is the g.f. for the number of words of length n over an alphabet of size n which do not contain any strictly decreasing factor (consecutive subword) of length 3.
For example, G_2, G_3, G_4, G_5, G_6 are g.f.'s for A000079, A076264, A072335, A200781, A200782.

			Triangle begins:
[1],
[1, -1],
[1, -2, 0],
[1, -3, 0, 1],
[1, -4, 0, 4, -1],
[1, -5, 0, 10, -5, 0],
[1, -6, 0, 20, -15, 0, 1],
[1, -7, 0, 35, -35, 0, 7, -1],
[1, -8, 0, 56, -70, 0, 28, -8, 0],
...

Cf. A000079, A076264, A072335, A200781, A200782.

f:=proc(n) local k,s;
s:=k->if k mod 3 = 0 then 1 elif k mod 3 = 1 then -1 else 0; fi;
[seq(s(k)*binomial(n,k),k=0..n)];
end;
[seq(f(n),n=0..12)];

chi[k_] := Switch[Mod[k, 3], 0, 1, 1, -1, 2, 0]; t[n_, k_] := chi[k]*Binomial[n, k]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 14 2014 *)

A018919 Define the generalized Pisot sequence T(a(0),a(1)) by: a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n). This is T(3,9).

Original entry on oeis.org

3, 9, 26, 75, 216, 622, 1791, 5157, 14849, 42756, 123111, 354484, 1020696, 2938977, 8462447, 24366645, 70160958, 202020427, 581694636, 1674922950, 4822748423, 13886550633, 39984728949, 115131438424, 331507764639, 954538564968

Offset: 0

R. K. Guy

Let M denotes the 4 X 4 matrix = row by row (1,1,1,1)(1,1,1,0)(1,1,0,0)(1,0,0,0) and A(n) the vector (x(n),y(n),z(n),t(n))=M^n*A where A is the vector (1,1,1,1) then a(n)=y(n+1). - Benoit Cloitre, Apr 02 2002

Not to be confused with the Pisot T(3,9) sequence, which is A000244. - R. J. Mathar, Feb 13 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
Index entries for linear recurrences with constant coefficients, signature (3,0,-1).
Index entries for Pisot sequences

Cf. A076264.

Tiv:=[3,9]; [n le 2 select Tiv[n] else Ceiling(Self(n-1)^2/Self(n-2))-1: n in [1..40]]; // Bruno Berselli, Feb 17 2016

CoefficientList[Series[- (x^2 - 3)/(x^3 - 3 x + 1), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 16 2013 *)
RecurrenceTable[{a[1] == 3, a[2] == 9, a[n] == Ceiling[a[n-1]^2/a[n-2]] - 1}, a, {n, 30}] (* Bruno Berselli, Feb 17 2016 *)
LinearRecurrence[{3,0,-1},{3,9,26},30] (* Harvey P. Dale, Feb 06 2019 *)

T(a0, a1, maxn) = a=vector(maxn); a[1]=a0; a[2]=a1; for(n=3, maxn, a[n]=ceil(a[n-1]^2/a[n-2])-1); a
T(3, 9, 30) \\ Colin Barker, Feb 14 2016

For n>1, a(n) = ceiling(a(n-1)^2/a(n-2)) - 1.
For n>2, a(n) = 3*a(n-1) - a(n-3).
G.f.: -(x^2-3) / (x^3-3*x+1). - Colin Barker, Dec 13 2012

A052545 Expansion of (1-x)^2/(1-3*x+x^3).

Original entry on oeis.org

1, 1, 4, 11, 32, 92, 265, 763, 2197, 6326, 18215, 52448, 151018, 434839, 1252069, 3605189, 10380728, 29890115, 86065156, 247814740, 713554105, 2054597159, 5915976737, 17034376106, 49048531159, 141229616740, 406654474114

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

(1, 4, 11, 32, ...) = INVERT transform of (1, 3, 4, 5, 6, 7, ...).

G. C. Greubel, Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 481
Index entries for linear recurrences with constant coefficients, signature (3,0,-1).

Cf. A215448. First differences of A052536.

a:=[1,1,4];; for n in [4..40] do a[n]:=3*a[n-1]-a[n-3]; od; a; # G. C. Greubel, May 08 2019

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1-x)^2/(1-3*x+x^3) )); // G. C. Greubel, May 08 2019

spec := [S,{S=Sequence(Prod(Z,Union(Z,Sequence(Z)),Sequence(Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);

LinearRecurrence[{3,0,-1}, {1,1,4}, 40] (* G. C. Greubel, May 08 2019 *)

my(x='x+O('x^40)); Vec((1-x)^2/(1-3*x+x^3)) \\ G. C. Greubel, May 08 2019

TOP = 33
a = [1]*TOP
a[2]=4
for n in range(3,TOP):
    print(a[n-3], end=',')
    a[n] = 3*a[n-1] - a[n-3]
# Alex Ratushnyak, Aug 10 2012

((1-x)^2/(1-3*x+x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, May 08 2019

G.f.: (1-x)^2/(1-3*x+x^3).
a(n) = 3*a(n-1) - a(n-3), with a(0)=a(1)=1, a(2)=4.
a(n) = Sum_{alpha = RootOf(1-3*x+x^3)} (-1/9 * (-1+2*alpha^2-2*alpha) * alpha^(-1-n)).
a(n) = A076264(n) - 2*A076264(n-1) + A076264(n-2). - R. J. Mathar, Nov 28 2011

A120747 Sequence relating to the 11-gon (or hendecagon).

Original entry on oeis.org

0, 1, 4, 14, 50, 175, 616, 2163, 7601, 26703, 93819, 329615, 1158052, 4068623, 14294449, 50221212, 176444054, 619907431, 2177943781, 7651850657, 26883530748, 94450905714, 331837870408, 1165858298498, 4096053203771, 14390815650209, 50559786403254

Offset: 1

Gary W. Adamson, Jul 01 2006

The hendecagon is an 11-sided polygon. The preferred word in the OEIS is 11-gon.
The lengths of the diagonals of the regular 11-gon are r[k] = sin(k*Pi/11)/sin(Pi/11), 1 <= k <= 5, where r[1] = 1 is the length of the edge.
The value of limit(a(n)/a(n-1),n=infinity) equals the longest diagonal r[5].
The a(n) equal the matrix elements M^n[1,2], where M = Matrix([[1,1,1,1,1], [1,1,1,1,0], [1,1,1,0,0], [1,1,0,0,0], [1,0,0,0,0]]). The characteristic polynomial of M is (x^5 - 3x^4 - 3x^3 + 4x^2 + x - 1) with roots x1 = -r[4]/r[3], x2 = -r[2]/r[4], x3 = r[1]/r[2], x4 = r[3]/r[5] and x5 = r[5]/r[1].
Note that M^4*[1,0,0,0,0] = [55, 50, 41, 29, 15] which are all terms of the 5-wave sequence A038201. This is also the case for the terms of M^n*[1,0,0,0,0], n>=1.

			From _Johannes W. Meijer_, Aug 03 2011: (Start)
The lengths of the regular hendecagon edge and diagonals are:
  r[1] = 1.000000000, r[2] = 1.918985948, r[3] = 2.682507066,
  r[4] = 3.228707416, r[5] = 3.513337092.
The first few rows of the T(n,k) array are, n>=1, 1 <= k <=5:
    0,   0,   0,   0,   1, ...
    1,   1,   1,   1,   1, ...
    1,   2,   3,   4,   5, ...
    5,   9,  12,  14,  15, ...
   15,  29,  41,  50,  55, ...
   55, 105, 146, 175, 190, ...
  190, 365, 511, 616, 671, ... (End)

G. C. Greubel, Table of n, a(n) for n = 1..1000
Jay Kappraff, Slavik Jablan, Gary W. Adamson and Radmila Sazdanovich, Golden Fields, Generalized Fibonacci Sequences and Chaotic Matrices, Forma, Vol. 19 No. 4, pp. 367-387, 2004.
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31, MR 1439165
Eric Weisstein's World of Mathematics, Hendecagon.
Index entries for linear recurrences with constant coefficients, signature (3,3,-4,-1,1).

Cf. A006358, A033304, A038201, A052963, A065941, A066170.
From Johannes W. Meijer, Aug 03 2011: (Start)
Cf. A006358 (T(n+2,1) and T(n+1,5)), A069006 (T(n+1,2)), A038342 (T(n+1,3)), this sequence (T(n,4)) (m=5: hendecagon or 11-gon).
Cf. A000045 (m=2; pentagon or 5-gon); A006356, A006054 and A038196 (m=3: heptagon or 7-gon); A006357, A076264, A091024 and A038197 (m=4: enneagon or 9-gon); A006359, A069007, A069008, A069009, A070778 (m=6; tridecagon or 13-gon); A025030 (m=7: pentadecagon or 15-gon); A030112 (m=8: heptadecagon or 17-gon). (End)

R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x^2*(1+x-x^2)/(1-3*x-3*x^2+4*x^3+x^4-x^5) )); // G. C. Greubel, Nov 13 2022

nmax:=27: m:=5: for k from 1 to m-1 do T(1,k):=0 od: T(1,m):=1: for n from 2 to nmax do for k from 1 to m do T(n,k):= add(T(n-1,k1), k1=m-k+1..m) od: od: for n from 1 to nmax/3 do seq(T(n,k), k=1..m) od; for n from 1 to nmax do a(n):=T(n,4) od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Aug 03 2011

LinearRecurrence[{3, 3, -4, -1, 1}, {0, 1, 4, 14, 50}, 41] (* G. C. Greubel, Nov 13 2022 *)

def A120747_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1+x-x^2)/(1-3*x-3*x^2+4*x^3+x^4-x^5) ).list()
A120747_list(40) # G. C. Greubel, Nov 13 2022

a(n) = 3*a(n-1) + 3*a(n-2) - 4*a(n-3) - a(n-4) + a(n-5).
G.f.: x^2*(1+x-x^2)/(1-3*x-3*x^2+4*x^3+x^4-x^5). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 12 2009
From Johannes W. Meijer, Aug 03 2011: (Start)
a(n) = T(n,4) with T(n,k) = Sum_{k1 = 6-k..6} T(n-1, k1), T(1,1) = T(1,2) = T(1,3) = T(1,4) = 0 and T(1,5) = 1, n>=1 and 1 <= k <= 5. [Steinbach]
Sum_{k=1..5} T(n,k)*r[k] = r[5]^n, n>=1. [Steinbach]
r[k] = sin(k*Pi/11)/sin(Pi/11), 1 <= k <= 5. [Kappraff]
Sum_{k=1..5} T(n,k) = A006358(n-1).
Limit_{n -> 00} T(n,k)/T(n-1,k) = r[5], 1 <= k <= 5.
sequence(sequence( T(n,k), k=2..5), n>=1) = A038201(n-4).
G.f.: (x^2*(x - x1)*(x - x2))/((x - x3)*(x - x4)*(x - x5)*(x - x6)*(x - x7)) with x1 = phi, x2 = (1-phi), x3 = r[1] - r[3], x4 = r[3] - r[5], x5 = r[5] - r[4], x6 = r[4] - r[2], x7 = r[2], where phi = (1 + sqrt(5))/2 is the golden ratio A001622. (End)

Edited and information added by Johannes W. Meijer, Aug 03 2011

A052963 a(0)=1, a(1)=2, a(2)=5, a(n) = 3*a(n+2) - a(n+3).

Original entry on oeis.org

1, 2, 5, 14, 40, 115, 331, 953, 2744, 7901, 22750, 65506, 188617, 543101, 1563797, 4502774, 12965221, 37331866, 107492824, 309513251, 891207887, 2566130837, 7388879260, 21275429893, 61260158842, 176391597266, 507899361905

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Equals the INVERT transform of the Pell sequence prefaced with a "1": (1, 1, 2, 5, 12, 29, ...). - Gary W. Adamson, May 01 2009

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1034
Elena Barcucci, Antonio Bernini, and Renzo Pinzani, Sequences from Fibonacci to Catalan: A combinatorial interpretation via Dyck paths, RAIRO-Theor. Inf. Appl. (2024) Vol. 58, Art. No. 8. See p. 14.
Jean-Luc Baril, Pamela E. Harris, and José L. Ramírez, Flattened Catalan Words, arXiv:2405.05357 [math.CO], 2024. See p. 14.
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; arXiv preprint, arXiv:1302.2274 [math.CO], 2013.
Index entries for linear recurrences with constant coefficients, signature (3,0,-1).

Cf. A066170, A076264.

spec := [S,{S=Sequence(Union(Prod(Sequence(Union(Prod(Z,Z),Z)),Z),Z))},unlabeled ]: seq(combstruct[count ](spec,size=n), n=0..20);

LinearRecurrence[{3,0,-1},{1,2,5},30] (* Harvey P. Dale, Dec 26 2015 *)

G.f.: -(-1+x+x^2)/(1-3*x+x^3).
a(n) = Sum((1/9)*(1+2*_alpha+_alpha^2)*_alpha^(-1-n), _alpha=RootOf(1-3*_Z+_Z^3)). [in Maple notation]
a(n)/a(n-1) tends to 2.8793852... = 1/(2*cos(4*Pi/9)), a root of x^3 -3x^2 + 1 (the characteristic polynomial of the 3 X 3 matrix). The latter polynomial is a factor (with (x + 1)) of the 4th degree polynomial of A066170: x^4 - 2x^3 - 3x^2 + x + 1. Given the 3 X 3 matrix [0 1 0 / 0 0 1 / -1 0 3], (M^n)*[1 1 1] = [a(n-2), a(n-1), a(n)]. - Gary W. Adamson, Feb 29 2004
a(n) = A076264(n)-A076264(n-1)-A076264(n-2). - R. J. Mathar, Feb 27 2019

More terms from James Sellers, Jun 05 2000

A200781 G.f.: 1/(1-5x+10x^3-5*x^4).

Original entry on oeis.org

1, 5, 25, 115, 530, 2425, 11100, 50775, 232275, 1062500, 4860250, 22232375, 101698250, 465201250, 2127983750, 9734098125, 44526969375, 203681015625, 931704015625, 4261920875000, 19495429065625, 89178510250000, 407931862578125, 1866014626609375, 8535765175875000, 39045399804843750, 178606512071015625, 817004981729375000

Offset: 0

R. H. Hardin, Nov 22 2011

nonn

Number of words of length n over an alphabet of size 5 which do not contain any strictly decreasing factor (consecutive subword) of length 3. For alphabets of size 2, 3, 4, 6 see A000079, A076264, A072335, A200782.

Equivalently, number of 0..4 arrays x(0..n-1) of n elements without any two consecutive increases.

			Some solutions for n=5:
..1....3....4....0....1....0....4....0....2....1....4....1....2....2....4....4
..3....4....4....2....1....0....3....3....1....4....1....1....4....4....3....3
..3....1....0....2....0....2....0....3....3....0....4....3....0....1....4....4
..2....0....2....4....4....0....3....2....0....0....3....2....0....2....1....3
..4....4....2....2....0....3....3....2....1....0....4....1....3....1....0....2

R. H. Hardin and N. J. A. Sloane, Table of n, a(n) for n = 0..249 [The first 210 terms were computed by R. H. Hardin]
A. Burstein and T. Mansour, Words restricted by 3-letter generalized multipermutation patterns, Annals. Combin., 7 (2003), 1-14. See Th. 3.13.
Index entries for linear recurrences with constant coefficients, signature (5,0,-10,5).

The g.f. corresponds to row 5 of triangle A225682.
Column 4 of A200785.
Cf. A076264, A072335, A200782.

Vec(1/(1-5*x+10*x^3-5*x^4) + O(x^30)) \\ Jinyuan Wang, Mar 10 2020

a(n) = 5*a(n-1) - 10*a(n-3) + 5*a(n-4).

Edited by N. J. A. Sloane, May 21 2013

A215448 a(0)=1, a(1)=0, a(n) = a(n-1) + a(n-2) + Sum_{i=0...n-1} a(i).

Original entry on oeis.org

1, 0, 2, 5, 15, 43, 124, 357, 1028, 2960, 8523, 24541, 70663, 203466, 585857, 1686908, 4857258, 13985917, 40270843, 115955271, 333879896, 961368845, 2768151264, 7970573896, 22950352843, 66082907265, 190278147899, 547884090854, 1577569365297, 4542429947992

Offset: 0

Alex Ratushnyak, Aug 10 2012

For the general recurrence X(n) = 3*X(n-1) - X(n-3) we get Sum_{k=3..n} X(k) = 3*Sum_{k=2..n-1} X(k) - Sum_{k=0..n-3} X(k), which implies the following summation formula: X(n) - X(n-1) - X(n-2) - X(2) + X(1) + X(0) = Sum_{k=2..n-1} X(k). Similarly from the formula X(n) + X(n-3) = 3*X(n-1) we deduce the following relations: Sum_{k=0..2*n-1} X(3*k) = 3*Sum_{k=0..n-1} X(6*k+2), Sum_{k=0..2*n-1} X(3*k+1) = 3*Sum_{k=1..n} X(6*k), and Sum_{k=0..2*n-1} X(3*k+2) = 3*Sum_{k=1..n} X(6*k-2). Lastly from the formula X(n)-X(n-1)=(X(n-1)-X(n-3))+X(n-1) we obtain the relations: Sum_{k=2..2*n+1} (-1)^(k-1)*X(k) = X(2*n) - X(0) + Sum_{k=1..n} X(2*k) and Sum_{k=3..2n} (-1)^k*X(k) = X(2*n-1) - X(1) + Sum_{k=2..n} X(2*k-1). - Roman Witula, Aug 27 2012

Index entries for linear recurrences with constant coefficients, signature (3,0,-1).

Cf. A052536: same formula, seed {0, 1}, first term removed.
Cf. A122100: same formula, seed {0,-1}, first two terms removed.
Cf. A052545: same formula, seed {1, 1}.

LinearRecurrence[{3,0,-1},{1,0,2},30] (* Harvey P. Dale, Jan 26 2017 *)

a = [1]*33
a[1]=0
sum = a[0]+a[1]
for n in range(2,33):
    print(a[n-2], end=', ')
    a[n] = a[n-1] + a[n-2] + sum
    sum += a[n]

a(0)=1, a(1)=0, for n>=2, a(n) = a(n-1) + a(n-2) + (a(0)+...+a(n-1)).
Conjecture: a(n) = +3*a(n-1) -a(n-3) = A076264(n) -3 *A076264(n-1) +2*A076264(n-2). G.f. (2*x-1)*(x-1) / ( 1-3*x+x^3 ). - R. J. Mathar, Aug 11 2012
Proof of the above conjecture: we have a(n) - a(n-1) = a(n-1) + a(n-2) + (a(0) + ... + a(n-1)) - a(n-2) - a(n-3) - (a(0) + ... + a(n-2)), which after simple algebra implies a(n) - a(n-1) = 2*a(n-1) - a(n-3), so the Mathar's formula holds true (see also Witula's comment above). - Roman Witula, Aug 27 2012

A119851 Triangle read by rows: T(n,k) is the number of ternary words of length n containing k 012's (n >= 0, 0 <= k <= floor(n/3)).

Original entry on oeis.org

1, 3, 9, 26, 1, 75, 6, 216, 27, 622, 106, 1, 1791, 387, 9, 5157, 1350, 54, 14849, 4566, 267, 1, 42756, 15102, 1179, 12, 123111, 49113, 4833, 90, 354484, 157622, 18798, 536, 1, 1020696, 500520, 70317, 2775, 15, 2938977, 1575558, 255231, 13068, 135

Offset: 0

Emeric Deutsch, May 26 2006

Row n has 1+floor(n/3) terms.
Sum of entries in row n is 3^n (A000244).
Sum_{k>=0} k*T(n,k) = (n-2)*3^(n-3) = A027741(n-1).

			T(4,1)=6 because we have 0012, 0120, 0121, 0122, 1012 and 2012.
Triangle starts:
    1;
    3;
    9;
   26,   1;
   75,   6;
  216,  27;
  622, 106, 1;

Cf. A000244, A076264 (=T(n,0)), A119852 (=T(n,1)), A027741.

G:=1/(1-3*z+z^3-t*z^3): Gser:=simplify(series(G,z=0,20)): P[0]:=1: for n from 1 to 15 do P[n]:=sort(coeff(Gser,z^n)) od: for n from 0 to 15 do seq(coeff(P[n],t,j),j=0..floor(n/3)) od; # yields sequence in triangular form

{ T(n,k) = sum(j=0,n-3*k, if((n-3*k-j)%2,0, binomial(-(k-1),j) *
binomial(j,(n-3*k-j)/2) * (-3)^((3*j+3*k-n)/2) )) } \\ Max Alekseyev

T(n,k) = Sum_{j=0..n-3k} binomial(-(k-1),j) * binomial(j,(n-3k-j)/2) * (-3)^((3j+3k-n)/2). - Max Alekseyev
G.f.: G(t,z) = 1/(1-3z+z^3-tz^3).
Recurrence relation: T(n,k) = 3*T(n-1,k) - T(n-3,k) + T(n-3,k-1) for n >= 3.

cp's OEIS Frontend

A099503 Expansion of 1/(1-4*x+x^3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

A099504 Expansion of 1/(1-5*x+x^3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

SageMath

Formula

A225682 Triangle read by rows: T(n,k) (0 <= k <= n) = chi(k)*binomial(n,k), where chi(k) = 1,-1,0 according as k == 0,1,2 mod 3.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

A018919 Define the generalized Pisot sequence T(a(0),a(1)) by: a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n). This is T(3,9).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A052545 Expansion of (1-x)^2/(1-3*x+x^3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Python

Sage

Formula

A120747 Sequence relating to the 11-gon (or hendecagon).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

SageMath

Formula

Extensions

A052963 a(0)=1, a(1)=2, a(2)=5, a(n) = 3*a(n+2) - a(n+3).

A200781 G.f.: 1/(1-5x+10x^3-5*x^4).