A000078 -id:A000078 - cp's OEIS frontend

A220469 Fibonacci 14-step numbers, a(n) = a(n-1) + a(n-2) + ... + a(n-14).

1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16383, 32765, 65528, 131052, 262096, 524176, 1048320, 2096576, 4193024, 8385792, 16771072, 33541120, 67080192, 134156288, 268304384, 536592385, 1073152005, 2146238482, 4292345912, 8584429728

Offset: 1

Ruskin Harding, Feb 20 2013

nonn

Also called tetradecanacci numbers. In previous similar sequences, a(1),...,a(n-1) have been set equal to zero and a(n)=1. For example, A168084 (Fibonacci 13-step numbers) has 12 0's as the first 12 terms and a(13)=1.

M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, J. Int. Seq. 18 (2015) # 15.4.7
Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4
Index entries for linear recurrences with constant coefficients, signature (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1).

Cf. A000045 (Fibonacci), A000073 (tribonacci), A000078 (tetranacci), A001591 (pentanacci).

FibonacciSequence[n_, kMax_] := Module[{a, s}, a = Join[{1}, Table[0, {n - 1}]]; lst = {}; Table[s = Plus @@ a; a = RotateLeft[a]; a[[n]] = s, {k, 1, kMax}]]; FibonacciSequence[14, 50] (* T. D. Noe, Feb 20 2013 *)
Drop[LinearRecurrence[PadRight[{},14,1],Join[PadRight[{},13,0],{1}],50],13] (* Harvey P. Dale, Feb 25 2013 *)
LinearRecurrence[{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},{1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096},35] (* Ray Chandler, Aug 03 2015 *)

A298371 a(n) = Sum_{m=0..n} Sum_{i=0..m} iC(m-i,i)C(m-i,n-m-i).

Original entry on oeis.org

0, 0, 1, 3, 7, 17, 40, 90, 198, 430, 922, 1956, 4115, 8597, 17853, 36883, 75856, 155396, 317228, 645580, 1310132, 2652072, 5356277, 10795351, 21716195, 43608549, 87429944, 175025918, 349901074, 698604058, 1393149486, 2775103948, 5522129511, 10977608425

Offset: 0

Vladimir Kruchinin, Jan 17 2018

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,0,-1,-4,-3,-2,-1).

Cf. A000078.

a(n):=sum(sum(i*binomial(m-i,i)*binomial(m-i,n-m-i),i,0,m),m,0,n);

concat(vector(2), Vec(x^2*(1 + x) / (1 - x - x^2 - x^3 - x^4)^2 + O(x^40))) \\ Colin Barker, Jan 18 2018

G.f.: x^3*(1 + x) / (1 - x - x^2 - x^3 - x^4)^2.

a(n) = 2*a(n-1) + a(n-2) - a(n-4) - 4*a(n-5) - 3*a(n-6) - 2*a(n-7) - a(n-8) for n>7. - Colin Barker, Jan 18 2018

A309703 G.f. A(x) satisfies: A(x) = A(x^2) / (1 - x - x^2 - x^3 - x^4).

Original entry on oeis.org

1, 1, 3, 5, 13, 22, 48, 88, 184, 342, 684, 1298, 2556, 4880, 9506, 18240, 35366, 67992, 131446, 253044, 488532, 941014, 1815334, 3497924, 6745360, 12999632, 25063130, 48306046, 93123674, 179492482, 346003572, 666925774, 1285580868, 2478002696, 4776580902, 9207090240

Offset: 0

Ilya Gutkovskiy, Aug 13 2019

nonn

Cf. A000078, A018819, A173285, A309702.

nmax = 35; A[] = 1; Do[A[x] = A[x^2]/(1 - x - x^2 - x^3 - x^4) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
nmax = 35; CoefficientList[Series[Product[1/(1 - x^(2^k) - x^(2^(k + 1)) - x^(3 2^k) - x^(2^(k + 2))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x]

G.f.: Product_{k>=0} 1/(1 - x^(2^k) - x^(2^(k+1)) - x^(3*2^k) - x^(2^(k+2))).

A349903 Array read by ascending antidiagonals. Inverse Euler transform of the right-shifted k-bonacci numbers.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 1, -1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 2, -1, 0, 0, 0, 0, 1, 2, 2, 0, 0, 0, 0, 0, 0, 1, 3, 4, 0, 0, 0, 0, 0, 0, 1, 2, 6, 5, 0, 0, 0, 0, 0, 0, 0, 1, 4, 10, 8, 0, 0, 0, 0, 0, 0, 0, 1, 2, 7, 18, 11, 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 14, 31, 18, 0, 0

Offset: 0

Peter Luschny, Dec 05 2021

			Array starts:
[0] 0, 1, 0, -1, 0,  0, 0,  0,  0,  0,  0,  0,   0, ...
[1] 0, 1, 1,  0, 0, -1, 0,  0,  0,  0,  0,  0,   0, ...
[2] 0, 1, 1,  1, 2,  2, 4,  5,  8, 11, 18, 25,  40, ...
[3] 0, 0, 1,  1, 2,  3, 6, 10, 18, 31, 56, 96, 172, ...
[4] 0, 0, 0,  1, 1,  2, 4,  7, 14, 26, 50, 93, 178, ...
[5] 0, 0, 0,  0, 1,  1, 2,  4,  8, 15, 30, 58, 114, ...
[6] 0, 0, 0,  0, 0,  1, 1,  2,  4,  8, 16, 31,  62, ...
[7] 0, 0, 0,  0, 0,  0, 1,  1,  2,  4,  8, 16,  32, ...
[8] 0, 0, 0,  0, 0,  0, 0,  1,  1,  2,  4,  8,  16, ...
[9] 0, 0, 0,  0, 0,  0, 0,  0,  1,  1,  2,  4,   8, ...
.
Compare the rows with the columns of A349802.

Rows are the inverse Euler transforms of A063524, A057427, A000045, A000073, A000078, A001591, A001592.

Cf. A092921, A349802

read transforms;
F := proc(n, k) option remember;
     ifelse(k < 2, k, add(F(n, k-j), j = 1..min(n, k))) end:
Frow := (n, len) -> [seq(0, j = 0..n-3), seq(F(n, k), k = 0..len)]:
Arow := (n, len) -> EULERi(Frow(n, len)):
for n from 0 to 9 do Arow(n, 14 - n) od;

A356807 Tetranacci sequence beginning with 3, 7, 12, 24.

Original entry on oeis.org

3, 7, 12, 24, 46, 89, 171, 330, 636, 1226, 2363, 4555, 8780, 16924, 32622, 62881, 121207, 233634, 450344, 868066, 1673251, 3225295, 6216956, 11983568, 23099070, 44524889, 85824483, 165432010, 318880452, 614661834, 1184798779, 2283773075, 4402114140, 8485347828

Offset: 1

Greg Dresden and Hangyu Liang, Aug 29 2022

By "Tetranacci sequence" we mean a sequence in which each term is the sum of the four previous terms.
For n>1, a(n) is the number of ways to tile this figure of length n with squares, dominoes, trominoes, and tetraminoes:
   _
  |||_________       _
  |||_|||_|| ... ||

			Here is one of the a(6) = 89 ways to tile this figure of length 6 with tiles of length <= 4, this one using three squares, one domino, and one tromino:
   ___
  | |_|_______
  |_|_____|_|_|

Paolo Xausa, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1).

Cf. A022120, A100683.

LinearRecurrence[{1, 1, 1, 1}, {3, 7, 12, 24}, 50] (* Paolo Xausa, Aug 30 2024 *)

a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4).
a(n) = 5*b(n+2) + 2*b(n+1) - 2*b(n-2) for b(n) = A000078(n) the tetranacci numbers.
a(n) = L(n+2) - F(n-2) + Sum_{k=0..n-3} a(k)*F(n-k-1), for L(n) and F(n) the Lucas and Fibonacci numbers.
G.f.: x*(-2*x^3 - 2*x^2 - 4*x - 3)/(x^4 + x^3 + x^2 + x - 1). - Chai Wah Wu, Aug 30 2022

A093175 Tetranacci numbers starting with first four squares.

Original entry on oeis.org

1, 4, 9, 16, 30, 59, 114, 219, 422, 814, 1569, 3024, 5829, 11236, 21658, 41747, 80470, 155111, 298986, 576314, 1110881, 2141292, 4127473, 7955960, 15335606, 29560331, 56979370, 109831267, 211706574, 408077542, 786594753, 1516210136, 2922589005

Offset: 1

Jun Mizuki (suzuki32(AT)sanken.osaka-u.ac.jp), May 11 2004

nonn

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

Cf. A086192.

a[1] = 1; a[2] = 4; a[3] = 9; a[4] = 16; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3] + a[n - 4]; Table[ a[n], {n, 33}] (* Robert G. Wilson v *)
LinearRecurrence[{1,1,1,1},Range[4]^2,40] (* Harvey P. Dale, Oct 16 2012 *)

a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4).

G.f.: -x*(1+x)*(2*x^2+2*x+1)/(-1+x+x^2+x^3+x^4). a(n)=2*A000078(n+3)+2*A000078(n)-A001631(n+1). [R. J. Mathar, Apr 20 2009]

More terms from Robert G. Wilson v and Labos Elemer, May 12 2004, May 11 2004

A111432 Fibonacci(tetranacci(n)).

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 3, 21, 610, 514229, 225851433717, 16641027750620563662096, 13180872826374098837632191485015125807374171, 284812298108489611757988937681460995615380088782304890986477195645969271404032323901

Offset: 0

Jonathan Vos Post, Nov 13 2005

			a(0) = Fibonacci(tetranacci(0)) = A000045(A000078(0)) = A000045(0) = 0.
a(1) = Fibonacci(tetranacci(1)) = A000045(A000078(1)) = A000045(0) = 0.
a(2) = Fibonacci(tetranacci(2)) = A000045(A000078(2)) = A000045(0) = 0.
a(3) = Fibonacci(tetranacci(3)) = A000045(A000078(3)) = A000045(1) = 1.
a(4) = Fibonacci(tetranacci(4)) = A000045(A000078(4)) = A000045(1) = 1.
a(5) = A000045(A000078(5)) = A000045(2) = 1.
a(6) = A000045(A000078(6)) = A000045(4) = 3.
a(7) = A000045(A000078(7)) = A000045(8) = 21.
a(8) = A000045(A000078(8)) = A000045(15) = 610.
a(9) = A000045(A000078(9)) = A000045(29) = 514229.
a(10) = A000045(A000078(10)) = A000045(56) = 225851433717.

Cf. A000045, A000078, A066178, A007570.

Fibonacci[LinearRecurrence[{1,1,1,1},{0,0,0,1},14]] (* Harvey P. Dale, Aug 03 2017 *)

a(n) = A000045(A000078(n)).

A255014 Abelian complexity function of the 4-bonacci word (A254990).

Original entry on oeis.org

4, 4, 6, 4, 7, 6, 7, 4, 7, 7, 8, 6, 8, 7, 7, 4, 7, 7, 8, 7, 8, 8, 7, 7, 8, 8, 7, 8, 7, 7, 4, 7, 7, 8, 7, 8, 8, 8, 7, 8, 8, 8, 8, 7, 7, 7, 7, 8, 8, 8, 8, 7, 8, 8, 8, 7, 8, 7, 7, 4, 7, 8, 9, 7, 8, 9, 9, 7, 8, 10, 10, 8, 8, 8, 8, 7, 9, 10, 9, 8, 9, 9, 8, 8, 9, 10, 7, 8, 7, 8, 7, 8, 9, 9, 8, 8, 8, 8, 8, 7

Offset: 1

Ondrej Turek, Feb 12 2015

nonn

For all n, a(n) either equals 4 or belongs to {6,7,...,16}; value 5 is never attained.
a(n)=4 if and only if n = T(k)+T(k-4)+T(k-8)+T(k-12)+...+T(4+(k mod 4)) for a certain k>=4, where T(i) are tetranacci numbers A000078.
a(n)=6 only for n = 3,6,12.
Each value from the set {7,8,...,16} is attained infinitely often.

			From _Wolfdieter Lang_, Mar 26 2015: (Start)
a(1) = 4 because the one letter factor words of A254990 are 0, 1, 2, 3 with the set of occurrence tuples (Parikh vectors) {(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)} of cardinality 4. See the Turek links.
a(2) = 4 because the set of occurrence tuples for the two letter factors 00, 01, 10, 02, 20, 03, 30 of A254990 is {(2, 0, 0, 0), (1, 1, 0, 0), (1, 0, 1, 0), (1, 0, 0, 1)} of cardinality 4. (End)

K. Brinda, Abelian complexity of infinite words, bachelor thesis, Czech Technical University in Prague, 2011.
K. Brinda, Abelian complexity of infinite words and Abelian return words, Research project, Czech Technical University in Prague, 2012.
F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
O. Turek, Abelian complexity function of the Tribonacci word, arXiv:1309.4810 [math.CO], 2013.
O. Turek, Abelian complexity function of the Tribonacci word, Journal of Integer Sequences, Vol. 18 (2015), Article 15.3.4.

Cf. A000078 (tetranacci numbers).

Cf. A216190 (abelian complexity of tribonacci word), A254990 (4-bonacci word).

A364145 a(n) is the sum of the first 2*n nonzero n-bonacci numbers.

Original entry on oeis.org

0, 2, 7, 28, 116, 480, 1968, 8000, 32320, 130048, 521984, 2092032, 8377344, 33529856, 134164480, 536756224, 2147237888, 8589410304, 34358624256, 137436594176, 549750833152, 2199012769792, 8796071002112, 35184325951488, 140737391886336, 562949752094720

Offset: 0

Muhammad Adam Dombrowski and Greg Dresden, Jul 10 2023

For our purposes, for n > 0 fixed we define the k-th n-bonacci number T(n,k) as equal to 0 for k <= 0, equal to 1 for k=1, and then equal to the sum of the previous n numbers for k > 1. For n=2, then, we get T(2,k) equal to F(n) = A000045(n), the Fibonacci numbers. For n=3, then, T(3,k) is the tribonacci numbers, and so on.

a(n) is thus defined as Sum_{k=1..2*n} T(n,k).

			For n=3, a(3) is the sum of the first 6 nonzero tribonacci numbers, found at A000073. This gives a(3) = 1 + 1 + 2 + 4 + 7 + 13 = 28.

Index entries for linear recurrences with constant coefficients, signature (8,-20,16).

Cf. A000045, A000073, A000078, A092921, A144406.

T[n_, k_] := SeriesCoefficient[Series[x/(1 - Sum[x^i, {i, 1, n}]), {x, 0, k + 1}], k]; Table[Sum[T[n, k], {k, 1, 2n}], {n, 1, 30}]

a(n) = (2*4^n - (n-1)*2^n)/4 for n>=1.
a(n) = Sum_{i=1..2*n} A092921(n,i).
G.f.: -x*(12*x^2-9*x+2)/((4*x-1)*(2*x-1)^2). - Alois P. Heinz, Jul 11 2023
E.g.f.: exp(2*x)*(1 - 2*x - cosh(2*x) + 5*sinh(2*x))/4. - Stefano Spezia, Jul 12 2023

A365293 a(n) = n!*tetranacci(n+3).

Original entry on oeis.org

1, 1, 4, 24, 192, 1800, 20880, 282240, 4354560, 75479040, 1455148800, 30855686400, 713712384000, 17884003737600, 482619020083200, 13954193180928000, 430360865206272000, 14102295149150208000, 489295008086556672000, 17919783031425859584000

Offset: 0

Enrique Navarrete, Aug 31 2023

nonn

a(n) is the number of ways to partition [n] into blocks of size at most 4, order the blocks, and order the elements within each block.

			a(5) = 1800 since the number of ways to partition [5] into blocks of size at most 4, order the blocks, and order the elements within each block are the following:
1) 1234,5: 10 such ordered blocks; 240 ways;
2) 123,4,5: 60 such ordered blocks; 360 ways;
3) 123,45: 20 such ordered blocks; 240 ways;
4) 12,34,5: 90 such ordered blocks; 360 ways;
5) 12,3,4,5: 240 such ordered blocks; 480 ways;
6) 1,2,3,4,5: 120 such ordered blocks; 120 ways.

Cf. A000078, A000142, A002866, A005442, A276924, A364324.

Table[n! SeriesCoefficient[1/(1-x-x^2-x^3-x^4),{x,0,n}],{n,0,19}] (* Stefano Spezia, Aug 31 2023 *)

E.g.f.: 1/(1-x-x^2-x^3-x^4).

a(n) = A000142(n) * A000078(n+3).

cp's OEIS Frontend

A220469 Fibonacci 14-step numbers, a(n) = a(n-1) + a(n-2) + ... + a(n-14).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

A298371 a(n) = Sum_{m=0..n} Sum_{i=0..m} i*C(m-i,i)*C(m-i,n-m-i).

Views

Author

Keywords

Links

Crossrefs

Programs

Maxima

PARI

Formula

A309703 G.f. A(x) satisfies: A(x) = A(x^2) / (1 - x - x^2 - x^3 - x^4).

Views

Author

Keywords

Crossrefs

Programs

Mathematica

Formula

A349903 Array read by ascending antidiagonals. Inverse Euler transform of the right-shifted k-bonacci numbers.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

A356807 Tetranacci sequence beginning with 3, 7, 12, 24.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A093175 Tetranacci numbers starting with first four squares.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A111432 Fibonacci(tetranacci(n)).

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

A255014 Abelian complexity function of the 4-bonacci word (A254990).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A364145 a(n) is the sum of the first 2*n nonzero n-bonacci numbers.

Views

Author

Keywords

Comments

Examples

A298371 a(n) = Sum_{m=0..n} Sum_{i=0..m} iC(m-i,i)C(m-i,n-m-i).