A000078 -id:A000078 - cp's OEIS frontend

A106295 Period of the Lucas 4-step sequence A073817 mod n.

1, 5, 26, 10, 312, 130, 342, 20, 78, 1560, 120, 130, 84, 1710, 312, 40, 4912, 390, 6858, 1560, 4446, 120, 12166, 260, 1560, 420, 234, 1710, 280, 1560, 61568, 80, 1560, 24560, 17784, 390, 1368, 34290, 1092, 1560, 240, 22230, 162800, 120, 312, 60830, 103822

Offset: 1

T. D. Noe, May 02 2005

nonn

This sequence is the same as the period of Fibonacci 4-step sequence (A000078) mod n for n<563 because the discriminant of the characteristic polynomial x^4-x^3-x^2-x-1 is -563. The two sequences differ only at n that are multiples of 563.

Chai Wah Wu, Table of n, a(n) for n = 1..1486
Marcellus E. Waddill, Some Properties of the Tetranacci Sequence Modulo m, The Fibonacci Quarterly, vol. 30, no. 3, 232-238 (1992).
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number

Cf. A000078, A073817, A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1).

n=4; Table[p=i; a=Join[Table[ -1, {n-1}], {n}]; a=Mod[a, p]; a0=a; k=0; While[k++; s=Mod[Plus@@a, p]; a=RotateLeft[a]; a[[n]]=s; a!=a0]; k, {i, 60}]

from itertools import count
def A106295(n):
    a = b = (4%n,1%n,3%n,7%n)
    s = sum(b) % n
    for m in count(1):
        b, s = b[1:] + (s,), (s+s-b[0]) % n
        if a == b:
            return m # Chai Wah Wu, Feb 22-27 2022

Let the prime factorization of n be p1^e1...pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)).

a(2^k) = 5*2^(k-1) for k > 0. If a(p) != a(p^2) for p prime, then a(p^k) = p^(k-1)*a(p) for k > 0 [Waddill, 1992]. - Chai Wah Wu, Feb 25 2022

A100532 The first four numbers of this sequence are the primes 2,3,5,7. The other terms are calculated by adding the previous four terms.

Original entry on oeis.org

2, 3, 5, 7, 17, 32, 61, 117, 227, 437, 842, 1623, 3129, 6031, 11625, 22408, 43193, 83257, 160483, 309341, 596274, 1149355, 2215453, 4270423, 8231505, 15866736, 30584117, 58952781, 113635139, 219038773, 422210810, 813837503, 1568722225, 3023809311, 5828579849

Offset: 1

Parthasarathy Nambi, Nov 24 2004

			The fifth term is 2 + 3 + 5 + 7 = 17.

Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1).

Cf. A000078, A001631.

[n le 4 select NthPrime(n) else Self(n-1)+Self(n-2)+Self(n-3)+Self(n-4): n in [1..41]]; // G. C. Greubel, Jun 30 2022

LinearRecurrence[{1,1,1,1}, {2,3,5,7}, 40] (* G. C. Greubel, Jun 30 2022 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; 1,1,1,1]^(n-1)*[2;3;5;7])[1,1] \\ Charles R Greathouse IV, Nov 01 2018

@CachedFunction
def a(n): # a = A100532
    if (n<5): return nth_prime(n)
    else: return sum( a(n-j) for j in (1..4))
[a(n) for n in (1..40)] # G. C. Greubel, Jun 30 2022

a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) where n >= 5 and a(1) = 2, a(2) = 3, a(3) = 5 and a(4) = 7.

G.f.: x*(1-x)*(2+3*x+3*x^2) / ( 1-x-x^2-x^3-x^4 ). - R. J. Mathar, Feb 03 2011

A254990 4-bonacci word. Fixed point of morphism 0->01, 1->02, 2->03, 3->0.

Original entry on oeis.org

0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 3

Offset: 0

Ondrej Turek, Feb 11 2015

Special case of k-bonacci word for k = 4 (see crossrefs).
The lengths of iterations S(i) are Tetranacci numbers (A000078).
Set S(0) = 0; S(1) = 0,1; S(2) = 0,1,0,2; S(3) = 0,1,0,2,0,1,0,3; for n >= 4: S(n) = S(n-1) S(n-2) S(n-3) S(n-4). The sequence is the limit S(infinity).

			The iterates are:
0
01
0102
01020103
010201030102010
01020103010201001020103010201
01020103010201001020103010201010201030102010010201030102
...

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Elena Barcucci, Luc Belanger and Srecko Brlek, On tribonacci sequences, Fib. Q., 42 (2004), 314-320. See Section 4.
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, J. Int. Seq. 18 (2015) # 15.3.4

Cf. A000078 (lengths of iterations).
Cf. A003849 (k=2, Fibonacci word), A080843 (k=3, Tribonacci word).
Cf. A316837, A316838, A316839, A316840.

Nest[Flatten[#/.{0->{0,1},1->{0,2},2->{0,3},3->0}]&,0,7] (* Harvey P. Dale, Mar 26 2015 *)

A251656 4-step Fibonacci sequence starting with 1,0,1,0.

Original entry on oeis.org

1, 0, 1, 0, 2, 3, 6, 11, 22, 42, 81, 156, 301, 580, 1118, 2155, 4154, 8007, 15434, 29750, 57345, 110536, 213065, 410696, 791642, 1525939, 2941342, 5669619, 10928542, 21065442, 40604945, 78268548, 150867477, 290806412, 560547382, 1080489819

Offset: 0

Arie Bos, Dec 06 2014

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

Other 4-step Fibonacci sequences are A000078, A000288, A001630, A001631, A001648, A073817, A100532, A251654, A251655, A251703, A251704, A251705.

Cf. A000336.

NB. see A251655 for the program and apply it to 1,0,1,0.

LinearRecurrence[Table[1, {4}], {1, 0, 1, 0}, 36] (* Michael De Vlieger, Dec 09 2014 *)

a(n+4) = a(n)+a(n+1)+a(n+2)+a(n+3).
G.f.: (-1+x+2*x^3)/(-1+x+x^2+x^3+x^4) . - R. J. Mathar, Mar 28 2025
a(n) = A000078(n+3)-A000078(n+2)-2*A000078(n). - R. J. Mathar, Mar 28 2025

A251703 4-step Fibonacci sequence starting with 1,1,0,0.

Original entry on oeis.org

1, 1, 0, 0, 2, 3, 5, 10, 20, 38, 73, 141, 272, 524, 1010, 1947, 3753, 7234, 13944, 26878, 51809, 99865, 192496, 371048, 715218, 1378627, 2657389, 5122282, 9873516, 19031814, 36685001, 70712613, 136302944, 262732372, 506432930, 976180859

Offset: 0

Arie Bos, Dec 07 2014

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

Other 4-step Fibonacci sequences are A000078, A000288, A001630, A001631, A001648, A073817, A100532, A251654, A251655, A251656, A251704, A251705.

NB. see A251655 for the program and apply it to 1,1,0,0.

LinearRecurrence[Table[1, {4}], {1, 1, 0, 0}, 36] (* Michael De Vlieger, Dec 09 2014 *)

a(n+4) = a(n) + a(n+1) + a(n+2) + a(n+3).
G.f.: (-1+2*x^2+2*x^3)/(-1+x+x^2+x^3+x^4) . - R. J. Mathar, Mar 28 2025
a(n) = A000078(n+3)-2*A000078(n+1)-2*A000078(n). - R. J. Mathar, Mar 28 2025

A000102 a(n) = number of compositions of n in which the maximum part size is 4.

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 5, 12, 27, 59, 127, 269, 563, 1167, 2400, 4903, 9960, 20135, 40534, 81300, 162538, 324020, 644282, 1278152, 2530407, 5000178, 9863763, 19427976, 38211861, 75059535, 147263905, 288609341, 565047233, 1105229439, 2159947998, 4217784107, 8230006378

Offset: 0

N. J. A. Sloane

a(n) is also the number of binary sequences of length n-1 in which the longest run of consecutive 0's is exactly three. - Geoffrey Critzer, Nov 06 2008

			For example, a(6)=5 counts 1+1+4, 2+4, 4+2, 4+1+1, 1+4+1. - _David Callan_, Dec 09 2004
a(6)=5 because there are 5 binary sequences of length 5 in which the longest run of consecutive 0's is exactly 3; 00010, 00011, 01000, 10001, 11000. - _Geoffrey Critzer_, Nov 06 2008

J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 155.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
J. L. Yucas, Counting special sets of binary Lyndon words, Ars Combin., 31 (1991), 21-29.

T. D. Noe, Table of n, a(n) for n=0..200
Nick Hobson, Python program for this sequence
J. L. Yucas, Counting special sets of binary Lyndon words, Ars Combin., 31 (1991), 21-29. (Annotated scanned copy)
Index entries for linear recurrences with constant coefficients, signature (2, 1, 0, -2, -3, -2, -1).

a:= n-> (Matrix(7, (i,j)-> if i+1=j then 1 elif j=1 then [2, 1, 0, -2, -3, -2, -1][i] else 0 fi)^n)[1,5]: seq(a(n), n=0..40); # Alois P. Heinz, Oct 07 2008

a[n_] := MatrixPower[ Table[ Which[i+1 == j, 1, j == 1, {2, 1, 0, -2, -3, -2, -1}[[i]], True, 0], {i, 1, 7}, {j, 1, 7}], n][[1, 5]]; Table[a[n], {n, 0, 34}] (* Jean-François Alcover, May 28 2013, after Alois P. Heinz *)
LinearRecurrence[{2,1,0,-2,-3,-2,-1},{0,0,0,0,1,2,5},40] (* Harvey P. Dale, Jul 01 2013 *)

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

a(n) = 2*a(n-1) + a(n-2) - 2*a(n-4) - 3*a(n-5) - 2*a(n-6) - a(n-7). Convolution of tribonacci and tetranacci numbers (A000073 and A000078). - Franklin T. Adams-Watters, Jan 13 2006

More terms from Sascha Kurz, Aug 15 2002

Definition improved by David Callan and Franklin T. Adams-Watters

A104577 Indices of prime generalized tetranacci numbers, A073817.

Original entry on oeis.org

2, 3, 8, 9, 16, 19, 24, 27, 46, 68, 71, 78, 107, 198, 309, 377, 477, 1057, 1631, 2419, 3974, 4293, 8247, 10513, 10709, 12011, 15042, 30543, 31607, 39664, 47552, 145858

Offset: 1

T. D. Noe, Mar 16 2005

nonn

The sequence of generalized tetranacci numbers is defined as beginning with 1, 3, 7, 15. Subsequent terms are the sum of the previous four terms. Note that the sequence of these generalized tetranacci numbers has many more primes than the tetranacci sequence A000078 (whose prime indices are in A104534).

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

Cf. A104576 (indices of prime generalized tribonacci numbers).

a={-1, -1, -1, 4}; Do[s=Plus@@a; a=RotateLeft[a]; a[[4]]=s; If[PrimeQ[s], Print[n]], {n, 30000}]

A124168 Union of all n-Fibonacci sequences, that is, all sequences s(0) = s(1) = ... = s(n-2) = 0, s(n-1) = 1 and for k >= n, s(k) = s(k-1) + ... + s(k-n).

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 13, 15, 16, 21, 24, 29, 31, 32, 34, 44, 55, 56, 61, 63, 64, 81, 89, 108, 120, 125, 127, 128, 144, 149, 208, 233, 236, 248, 253, 255, 256, 274, 377, 401, 464, 492, 504, 509, 511, 512, 610, 773, 912, 927, 976, 987, 1004, 1016, 1021, 1023, 1024

Offset: 1

Carlos Alves, Dec 03 2006

nonn

Note that an n-Fibonacci sequence contains the numbers 2^k numbers for kA001792 (for n large)...

Noe and Post conjectured that the only positive terms that are common to any two distinct n-step Fibonacci sequences are the powers of 2 that begin each sequence and 13 (in 2- and 3-step) and 504 (in 3- and 7-step). Perhaps we should also include 8 (in 2- and 4-step). - T. D. Noe, Dec 05 2006

T. D. Noe, Table of n, a(n) for n=1..1000
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.

Cf. A000045, A000073, A000078, A001591, A001592, A124257.

Cf. A227880 (primes here).

NFib25[nfb_] := Transpose[NestList[Join[Drop[ #, {1}], {Plus @@ #}] &, Map[If[ # == nfb, 1, 0] &, Range[nfb]], 25]][[ -1]]; Union[Flatten[Map[NFib25, Range[2, 20]]]][[Range[100]]]
NFib[nfb_, lim_] := Module[{f = 2^Range[0, nfb - 1]}, While[f[[-1]] <= lim, AppendTo[f, Total[Take[f, -nfb]]]]; Most[f]]; lim = 12; Union[Flatten[Table[NFib[i, 2^lim], {i, 2, lim + 1}]]] (* T. D. Noe, Oct 25 2013 *)

Union(A000045, A000073, A000078, A001591, A001592, ...)

Edited by N. J. A. Sloane, Dec 15 2006

A168084 Fibonacci 13-step numbers.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8191, 16381, 32760, 65516, 131024, 262032, 524032, 1048000, 2095872, 4191488, 8382464, 16763904, 33525760, 67047424, 134086657, 268156933, 536281106, 1072496696

Offset: 1

Vladimir Joseph Stephan Orlovsky, Nov 18 2009

G. C. Greubel, Table of n, a(n) for n = 1..1000
Martin Burtscher, Igor Szczyrba, Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, J. Int. Seq. 18 (2015) # 15.4.7.
Index entries for linear recurrences with constant coefficients, signature (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1).

Cf. A000078, A001591, A001592, A122189, A079262, A168083.

k:=13:a:=taylor((z^(k-1)-z^(k))/(1-2*z+z^(k+1)),z=0,51);for p from 0 to 50 do j(p):=coeff(a,z,p):od :seq(j(p),p=0..50); k:=13:for n from 0 to 50 do l(n):=sum((-1)^i*binomial(n-k+1-k*i,i)*2^(n-k+1-(k+1)*i),i=0..floor((n-k+1)/(k+1)))-sum((-1)^i*binomial(n-k-k*i,i)*2^(n-k-(k+1)*i),i=0..floor((n-k)/(k+1))):od:seq(l(n),n=0..50); # Richard Choulet, Feb 22 2010

a={1,0,0,0,0,0,0,0,0,0,0,0,0};Flatten[Prepend[Table[s=Plus@@a;a=RotateLeft[a];a[[ -1]]=s,{n,60}],Table[0,{m,Length[a]-1}]]]
LinearRecurrence[{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, 50]
With[{nn=13},LinearRecurrence[Table[1,{nn}],Join[Table[0,{nn-1}],{1}],50]] (* Harvey P. Dale, Aug 17 2013 *)

Another form of the g.f. f: f(z)=(z^(k-1)-z^(k))/(1-2*z+z^(k+1)) with k=13. then a(n)=sum((-1)^i*binomial(n-k+1-k*i,i)*2^(n-k+1-(k+1)*i),i=0..floor((n-k+1)/(k+1)))-sum((-1)^i*binomial(n-k-k*i,i)*2^(n-k-(k+1)*i),i=0..floor((n-k)/(k+1))) with k=13 and convention sum(alpha(i),i=m..n)=0 for m>n. - Richard Choulet, Feb 22 2010

A220547 Number of ways to reciprocally link elements of an n X 2 array either to themselves or to exactly one horizontal, vertical or antidiagonal neighbor.

Original entry on oeis.org

2, 8, 29, 108, 401, 1490, 5536, 20569, 76424, 283953, 1055026, 3919944, 14564533, 54114452, 201061985, 747044834, 2775641472, 10312882481, 38317465040, 142368356257, 528968939938, 1965381541064, 7302365621709, 27131904192124

Offset: 1

R. H. Hardin, Dec 15 2012

nonn

Column 2 of A220553.

			Some solutions for n=3 0=self 2=n 3=ne 4=w 6=e 7=sw 8=s (reciprocal directions total 10):
..8..8....0..7....6..4....6..4....8..0....0..0....0..0....0..0....0..8....6..4
..2..2....3..8....0..0....6..4....2..8....8..8....0..8....0..0....8..2....8..8
..6..4....0..2....0..0....0..0....0..2....2..2....0..2....0..0....2..0....2..2

R. H. Hardin, Table of n, a(n) for n = 1..210

Cf. A220553, A000078.

Empirical: a(n) = 3*a(n-1) + 3*a(n-2) - a(n-3) - a(n-4).
Empirical g.f.: x*(1 + x)*(2 - x^2) / (1 - 3*x - 3*x^2 + x^3 + x^4). - Colin Barker, Mar 13 2018
Empirical: a(n) = A000078(2*n + 3). - Greg Dresden, Jan 10 2021

cp's OEIS Frontend

A106295 Period of the Lucas 4-step sequence A073817 mod n.

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A100532 The first four numbers of this sequence are the primes 2,3,5,7. The other terms are calculated by adding the previous four terms.

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A254990 4-bonacci word. Fixed point of morphism 0->01, 1->02, 2->03, 3->0.

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A251656 4-step Fibonacci sequence starting with 1,0,1,0.

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A251703 4-step Fibonacci sequence starting with 1,1,0,0.

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A000102 a(n) = number of compositions of n in which the maximum part size is 4.

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A104577 Indices of prime generalized tetranacci numbers, A073817.

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A124168 Union of all n-Fibonacci sequences, that is, all sequences s(0) = s(1) = ... = s(n-2) = 0, s(n-1) = 1 and for k >= n, s(k) = s(k-1) + ... + s(k-n).

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