A008998 -id:A008998 - cp's OEIS frontend

A098588 a(n) = 2^n for n = 0..4; for n > 4, a(n) = 2*a(n-1) + a(n-5).

1, 2, 4, 8, 16, 33, 68, 140, 288, 592, 1217, 2502, 5144, 10576, 21744, 44705, 91912, 188968, 388512, 798768, 1642241, 3376394, 6941756, 14272024, 29342816, 60327873, 124032140, 255006036, 524284096, 1077911008, 2216149889

Offset: 0

Paul Barry, Sep 16 2004

a(n) equals the number of n-length words on {0,1,2} such that 0 appears only in a run whose length is a multiple of 5. - Milan Janjic, Feb 17 2015

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,0,0,1).

Cf. A008998, A008999.

I:=[1,2,4,8,16]; [n le 5 select I[n] else 2*Self(n-1) +Self(n-5): n in [1..30]]; // G. C. Greubel, Feb 03 2018

CoefficientList[Series[1/(1-2*x-x^5), {x,0,50}], x] (* or *) LinearRecurrence[{2,0,0,0,1}, {1,2,4,8,16}, 50] (* G. C. Greubel, Feb 03 2018 *)

x='x+O('x^30); Vec(1/(1-2*x-x^5)) \\ G. C. Greubel, Feb 03 2018

G.f.: 1/(1-2*x-x^5).
a(n) = Sum_{k=0..floor(n/4)} Sum_{i=0..n} binomial(n-4k, i)binomial(i, k).
G.f.: G(0), where G(k)= 1 + x*(2+x^4)/(1 - x*(2+x^4)/(x*(2+x^4) + 1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 03 2013
Lim_{n->infinity} a(n)/a(n+1) = 0.486389... is a real root of -1 + 2Z + Z^5 = 0. - Sergei N. Gladkovskii, Jul 03 2013

A180675 The Ca3 sums of the Pell-Jacobsthal triangle A013609.

Original entry on oeis.org

1, 1, 1, 9, 97, 1041, 11169, 119833, 1285697, 13794337, 148000449, 1587907625, 17036776865, 182788823089, 1961154631009, 21041371248697, 225754408665729, 2422135536207937, 25987269043538817, 278819307278968905

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n+3) represent the Ca3 sums of the Pell-Jacobsthal triangle A013609. See A180662 for information about these camel and other chess sums.

Index entries for linear recurrences with constant coefficients, signature (11, -3, 1).

Cf. A077949 (Ca1), A008998 (Ca2), A180675 (Ca3), A092467 (Ca4).

nmax:=20: a(0):=1: a(1):=1: a(2):=1: for n from 3 to nmax do a(n) := 11*a(n-1)-3*a(n-2)+a(n-3) od: seq(a(n),n=0..nmax);

LinearRecurrence[{11,-3,1},{1,1,1},20] (* Harvey P. Dale, Jun 23 2013 *)

a(n) = 11*a(n-1)-3*a(n-2)+a(n-3) with a(0)=1, a(1)=1 and a(2)=1.
a(n+2) = add(A013609(n+2*k,3*k),k=0..floor(n)).
GF(x) = (1-10*x-7*x^2)/(1-11*x+3*x^2-x^3).

A214260 First differences of A052980.

Original entry on oeis.org

0, 1, 3, 6, 13, 29, 64, 141, 311, 686, 1513, 3337, 7360, 16233, 35803, 78966, 174165, 384133, 847232, 1868629, 4121391, 9090014, 20048657, 44218705, 97527424, 215103505, 474425715, 1046378854, 2307861213

Offset: 0

Philippe Deléham, Jul 22 2012

nonn

1 -> 123, 2 -> 12, 3 -> 2, starting with 1 gives the sequence: 1, 123, 123122, 1231221231212, ... the n-th term has a(n) digits.

Ternary words of length n-1 with subwords (0,1), (1,1) and (1,2) not allowed. - Olivier Gérard, Aug 28 2012

Index entries for linear recurrences with constant coefficients, signature (2,0,1).

Cf. A008998, A052980, A064353, A078061.

LinearRecurrence[{2,0,1},{0,1,3},30] (* Harvey P. Dale, Sep 04 2017 *)

Recurrence: a(0) = 0, a(1) = 1, a(2) = 3, a(n+1) = 2*a(n) + a(n-2).
G.f.: x*(1+x)/(1-2*x-x^3).
a(n) = A052980(n) + A052980(n-2) = A052980(n+1) - A052980(n).
a(n+1) = A078061(n)*(-1)^n.
a(0) = 0, a(n) = A008998(n-1) + A008998(n-2) for n>0.
a(n+1) = Sum_{k=0..n} C(n-k, floor(k/2))*2^(n-k-floor(k/2)).

A317495 Triangle read by rows: T(0,0) = 1; T(n,k) =2 * T(n-1,k) + T(n-3,k-1) for k = 0..floor(n/3); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, 2, 4, 8, 1, 16, 4, 32, 12, 64, 32, 1, 128, 80, 6, 256, 192, 24, 512, 448, 80, 1, 1024, 1024, 240, 8, 2048, 2304, 672, 40, 4096, 5120, 1792, 160, 1, 8192, 11264, 4608, 560, 10, 16384, 24576, 11520, 1792, 60, 32768, 53248, 28160, 5376, 280, 1, 65536, 114688, 67584, 15360, 1120, 12

Offset: 0

Zagros Lalo, Jul 30 2018

The numbers in rows of the triangle are along a "second layer" of skew diagonals pointing top-left in center-justified triangle given in A013609 ((1+2*x)^n) and along a "second layer" of skew diagonals pointing top-right in center-justified triangle given in A038207 ((2+x)^n), see links. (Note: First layer skew diagonals in center-justified triangles of coefficients in expansions of (1+2*x)^n and (2+x)^n are given in A128099 and A207538 respectively.)
The coefficients in the expansion of 1/(1-2x-x^3) are given by the sequence generated by the row sums.
The row sums give A008998 and Pisot sequences E(4,9), P(4,9) when n > 1, see A020708.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 2.205569430400..., when n approaches infinity.

			Triangle begins:
       1;
       2;
       4;
       8,      1;
      16,      4;
      32,     12;
      64,     32,      1;
     128,     80,      6;
     256,    192,     24;
     512,    448,     80,      1;
    1024,   1024,    240,      8;
    2048,   2304,    672,     40;
    4096,   5120,   1792,    160,     1;
    8192,  11264,   4608,    560,    10;
   16384,  24576,  11520,   1792,    60;
   32768,  53248,  28160,   5376,   280,   1;
   65536, 114688,  67584,  15360,  1120,  12;
  131072, 245760, 159744,  42240,  4032,  84;
  262144, 524288, 372736, 112640, 13440, 448, 1;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 358, 359.

Row sums give A008998, A020708.
Cf. A013609
Cf. A038207
Cf. A317494
Cf. A128099, A207538.
Cf. A000079 (column 0), A001787 (column 1), A001788 (column 2), A001789 (column 3), A003472 (column 4).

Flat(List([0..20],n->List([0..Int(n/3)],k->2^(n-3*k)/(Factorial(n-3*k)*Factorial(k))*Factorial(n-2*k)))); # Muniru A Asiru, Jul 31 2018

/* As triangle */ [[2^(n-3*k)/(Factorial(n-3*k)*Factorial(k))* Factorial(n-2*k): k in [0..Floor(n/3)]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 05 2018

t[n_, k_] := t[n, k] = 2^(n - 3k)/((n - 3 k)! k!) (n - 2 k)!; Table[t[n, k], {n, 0, 18}, {k, 0, Floor[n/3]} ]  // Flatten
t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, 2 t[n - 1, k] + t[n - 3, k - 1]]; Table[t[n, k], {n, 0, 18}, {k, 0, Floor[n/3]}] // Flatten

T(n,k) = 2^(n - 3k) / ((n - 3k)! k!) * (n - 2k)! where n >= 0 and k = 0..floor(n/3).

A332647 a(n) = 2*a(n-1) + a(n-3) with a(0) = 3, a(1) = 2, a(2) = 4.

Original entry on oeis.org

3, 2, 4, 11, 24, 52, 115, 254, 560, 1235, 2724, 6008, 13251, 29226, 64460, 142171, 313568, 691596, 1525363, 3364294, 7420184, 16365731, 36095756, 79611696, 175589123, 387274002, 854159700, 1883908523, 4155091048, 9164341796, 20212592115, 44580275278, 98324892352

Offset: 0

Greg Dresden, Feb 18 2020

a(n) is the number of ways to tile a bracelet of length n with black trominos, and black or white squares.

Colin Barker, Table of n, a(n) for n = 0..1000
Greg Dresden and Michael Tulskikh, Tilings of 2 X n boards with dominos and L-shaped trominos, Washington & Lee University (2021).
Helmut Prodinger, On third-order Pell polynomials, arXiv:2011.04388 [math.NT], 2020.
Index entries for linear recurrences with constant coefficients, signature (2,0,1).

Cf. A008998, A052980. Equals one more than A080204.

a:=[3,2,4]; [n le 3 select a[n] else 2*Self(n-1)+Self(n-3):n in [1..33]]; // Marius A. Burtea, Feb 18 2020

LinearRecurrence[{2, 0, 1}, {3, 2, 4}, 50]

Vec((3 - 4*x) / (1 - 2*x - x^3) + O(x^30)) \\ Colin Barker, Feb 18 2020

polsym(x^3-2*x^2-1, 44) \\ Joerg Arndt, May 28 2020

a(n) = 2*a(n-1) + a(n-3).
a(n) = w1^n + w2^n + w3^n where w1,w2,w3 are the three roots of x^3-2x^2-1=0.
For n>2, a(n) = round(w1^n) for w1 the single real root of x^3-2x^2-1=0.
G.f.: (3 - 4*x) / (1 - 2*x - x^3). - Colin Barker, Feb 18 2020
a(n) = A008998(n) + 2*A008998(n-3) = 3*A008998(n) - 4*A008998(n-1).
a(n) = (5*b(n) - b(n-1) - b(n-2))/2 where b(n) = A052980(n). - Greg Dresden, Mar 10 2020
a(n) = A080204(n) + 1. - Greg Dresden, May 27 2020

A078061 Expansion of (1-x)/(1+2*x+x^3).

Original entry on oeis.org

1, -3, 6, -13, 29, -64, 141, -311, 686, -1513, 3337, -7360, 16233, -35803, 78966, -174165, 384133, -847232, 1868629, -4121391, 9090014, -20048657, 44218705, -97527424, 215103505, -474425715, 1046378854, -2307861213, 5090148141, -11226675136, 24761211485, -54612571111, 120451817358

Offset: 0

N. J. A. Sloane, Nov 17 2002

Shanzhen Gao, Keh-Hsun Chen, Tackling Sequences From Prudent Self-Avoiding Walks, FCS'14, The 2014 International Conference on Foundations of Computer Science.
Index entries for linear recurrences with constant coefficients, signature (-2,0,-1).

CoefficientList[Series[(1-x)/(1+2x+x^3),{x,0,40}],x] (* or *) LinearRecurrence[ {-2,0,-1},{1,-3,6},40] (* Harvey P. Dale, Dec 15 2017 *)

Vec((1-x)/(1+2*x+x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

a(n) = (-1)^n*sum{k=0..n, C(n-k,floor(k/2))*2^(n-k-floor(k/2))}. [Paul Barry, Oct 20 2009]

(-1)^n*a(n) = A008998(n)+A008998(n-1). - R. J. Mathar, Jul 08 2022

A144401 Padovan ( A000931) version of A038137: expansion of polynomials as antidiagonal: p(x,n)=1/(1-x-x^3)^n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 1, 4, 6, 6, 3, 1, 5, 10, 13, 11, 4, 1, 6, 15, 24, 27, 18, 6, 1, 7, 21, 40, 55, 51, 30, 9, 1, 8, 28, 62, 100, 116, 94, 50, 13, 1, 9, 36, 91, 168, 231, 234, 171, 81, 19, 1, 10, 45, 128, 266, 420, 505, 460, 303, 130, 28, 1, 11, 55, 174, 402, 714, 987, 1065

Offset: 1

Roger L. Bagula and Gary W. Adamson, Oct 03 2008

Row sums are: 1, 2, 4, 9, 20, 44, 97, 214, 472, 1041, 2296, 5064, 11169, 24634, 54332 (cf. A008998).
These polynomials are sort of pseudo-combinations with the last element Padovan instead of one.
If you subtract the binomial triangle sequence you get:
{0},
{0, 0},
{0, 0, 0},
{0, 0, 0, 1},
{0, 0, 0, 2, 2},
{0, 0, 0, 3, 6, 3},
{0, 0, 0, 4, 12, 12, 5},
{0, 0, 0, 5, 20, 30, 23, 8},
{0, 0, 0, 6, 30, 60, 66, 42, 12}

			{1},
{1, 1},
{1, 2, 1},
{1, 3, 3, 2},
{1, 4, 6, 6, 3},
{1, 5, 10, 13, 11, 4},
{1, 6, 15, 24, 27, 18, 6},
{1, 7, 21, 40, 55, 51, 30, 9},
{1, 8, 28, 62, 100, 116, 94, 50, 13},
{1, 9, 36, 91, 168, 231, 234, 171, 81, 19},
{1, 10, 45, 128, 266, 420, 505, 460, 303, 130, 28},
{1, 11, 55, 174, 402, 714, 987, 1065, 879, 527, 208, 41},
{1, 12, 66, 230, 585, 1152, 1792, 2220, 2175, 1640, 906, 330, 60},
{1, 13, 78, 297, 825, 1782, 3072, 4278, 4815, 4320, 3006, 1539, 520, 88},
{1, 14, 91, 376, 1133, 2662, 5028, 7752, 9807, 10122, 8391, 5424, 2586, 816, 129}

Cf. A000931, A038137.

Clear[f, b, a, g, h, n, t]; f[t_, n_] = 1/(1 - t - t^3)^n; a = Table[Table[SeriesCoefficient[Series[f[t, m], {t, 0, 30}], n], {n, 0, 30}], {m, 1, 31}]; b = Table[Table[a[[n - m + 1]][[m]], {m, 1, n }], {n, 1, 15}]; Flatten[b]

p(x,n)=1/(1-x-x^3)^n; t(n,m)=anti_diagonal_expansion(p(x,n)).

A228815 Symmetric triangle, read by rows, related to Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 2, 5, 5, 2, 3, 10, 14, 10, 3, 5, 20, 36, 36, 20, 5, 8, 38, 83, 106, 83, 38, 8, 13, 71, 182, 281, 281, 182, 71, 13, 21, 130, 382, 690, 834, 690, 382, 130, 21, 34, 235, 778, 1606, 2268, 2268, 1606, 778, 235, 34, 55, 420, 1546, 3586, 5780, 6750

Offset: 0

Philippe Deléham, Oct 30 2013

Triangles satisfying the same recurrence: A091533, A091562, A185081, A205575, A209137, A209138.

			Triangle begins :
0
1, 1
1, 2, 1
2, 5, 5, 2
3, 10, 14, 10, 3
5, 20, 36, 36, 20, 5
8, 38, 83, 106, 83, 38, 8
13, 71, 182, 281, 281, 182, 71, 13
21, 130, 382, 690, 834, 690, 382, 130, 21
34, 235, 778, 1606, 2268, 2268, 1606, 778, 235, 34
55, 420, 1546, 3586, 5780, 6750, 5780, 3586, 1546, 420, 55

Cf. A000045 (1st column), A001629 (2nd column), A008998, A152011, A261055 (3rd column).

G.f.: x*(1+y)/(1-x-x*y-x^2-x^2*y-x^2*y^2).
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k) + T(n-2,k-1) + T(n-2,k-2), T(0,0) = 0, T(1,0) = T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n.
Sum_{k = 0..n} T(n,k)*x^k = A000045(n), 2*A015518(n), 3*A015524(n), 4*A200069(n) for x = 0, 1, 2, 3 respectively.
Sum_{k = 0..floor(n/2)} T(n-k,k) = A008998(n+1).

A257365 Triangle, read by rows, T(n,k) = Sum_{m=0..(n-k)/2} C(k,m)C(n-2m,k).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 6, 8, 4, 1, 1, 8, 16, 13, 5, 1, 1, 10, 28, 32, 19, 6, 1, 1, 12, 44, 68, 55, 26, 7, 1, 1, 14, 64, 128, 136, 86, 34, 8, 1, 1, 16, 88, 220, 296, 241, 126, 43, 9, 1, 1, 18, 116, 352, 584, 592, 393, 176, 53, 10, 1

Offset: 0

Vladimir Kruchinin, Apr 21 2015

From Emanuele Munarini, Feb 21 2017: (Start)
T(n,k) is the number of lattice paths from (0,0) to (n,k) using steps X=(1,0), D=(1,1) and E=(3,1).
Row sums = A008998.
Central coefficients = A006139. (End)

			1;
1, 1;
1, 2, 1;
1, 4, 3, 1;
1, 6, 8, 4, 1;
1, 8, 16, 13, 5, 1;

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened)
James East, Nicholas Ham, Lattice paths and submonoids of Z^2, arXiv:1811.05735 [math.CO], 2018.

Cf. A006139.

Table[Sum[Binomial[k, m] Binomial[n - 2 m, k], {m, 0, (n - k)/2}], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Apr 21 2015 *)

T(n,k):=sum(binomial(k,m)*binomial(n-2*m,k),m,0,(n-k)/2);

G.f.: 1/(1-y-x*(1+y^2)).
From Emanuele Munarini, Feb 21 2017: (Start)
G.f. for the triangle: 1/(1-x-x*y-x^3*y).
Recurrence: T(n+3,k+1) = T(n+2,k+1) + T(n+2,k) + T(n,k). (End)

A335242 a(n) = 2*a(n-1) + a(n-3) for n >= 4, with initial values a(0) = 1, a(1) = 0, a(2) = 2, and a(3) = 3.

Original entry on oeis.org

1, 0, 2, 3, 6, 14, 31, 68, 150, 331, 730, 1610, 3551, 7832, 17274, 38099, 84030, 185334, 408767, 901564, 1988462, 4385691, 9672946, 21334354, 47054399, 103781744, 228897842, 504850083, 1113481910, 2455861662, 5416573407, 11946628724, 26349119110, 58114811627

Offset: 0

Greg Dresden, May 28 2020

a(n) is the number of ways to tile this 2 X n strip (with one extra square added at the top left) with dominoes and L-shaped trominoes (also called polyominoes):
._
|| _  
|||_||| . . .
|||_||| . . .

			a(2) = 2 thanks to the following two tilings (where the L-shaped trominoes are tiled with X's and the dominoes are left blank):
._            _
|X|_         | |_
|X|X|  and   |_|X|
|_ _|        |X X|

Index entries for linear recurrences with constant coefficients, signature (2,0,1).

Cf. A052980, A008998.

LinearRecurrence[{2, 0, 1}, {1, 0, 2, 3}, 40]

a(n) = 2*a(n-1) + a(n-3) for n >= 4.
a(n) = A008998(n-2) + A052980(n-2) for n >= 2.
G.f.: (2*x^3-2*x^2+2*x-1)/(x^3+2*x-1).

cp's OEIS Frontend

A098588 a(n) = 2^n for n = 0..4; for n > 4, a(n) = 2*a(n-1) + a(n-5).

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A180675 The Ca3 sums of the Pell-Jacobsthal triangle A013609.

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A214260 First differences of A052980.

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A317495 Triangle read by rows: T(0,0) = 1; T(n,k) =2 * T(n-1,k) + T(n-3,k-1) for k = 0..floor(n/3); T(n,k)=0 for n or k < 0.

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A332647 a(n) = 2*a(n-1) + a(n-3) with a(0) = 3, a(1) = 2, a(2) = 4.

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A078061 Expansion of (1-x)/(1+2*x+x^3).

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A144401 Padovan ( A000931) version of A038137: expansion of polynomials as antidiagonal: p(x,n)=1/(1-x-x^3)^n.

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A257365 Triangle, read by rows, T(n,k) = Sum_{m=0..(n-k)/2} C(k,m)C(n-2m,k).