A001644 -id:A001644 - cp's OEIS frontend

A127214 a(n) = 2^ntribonacci(n) or (2^n)A001644(n+1).

2, 12, 56, 176, 672, 2496, 9088, 33536, 123392, 453632, 1669120, 6139904, 22585344, 83083264, 305627136, 1124270080, 4135714816, 15213527040, 55964073984, 205867974656, 757300461568, 2785785413632, 10247716470784, 37696978288640, 138671105769472

Offset: 1

Artur Jasinski, Jan 09 2007

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,4,8).

Cf. A087131, A127210, A127211, A127212, A127213, A127215, A127216.

I:=[2,12,56]; [n le 3 select I[n] else 2*Self(n-1) + 4*Self(n-2) + 8*Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 18 2017

Table[Tr[MatrixPower[2*{{1, 1, 1}, {1, 0, 0}, {0, 1, 0}}, x]], {x, 1, 20}]
LinearRecurrence[{2, 4, 8}, {2, 12, 56}, 50] (* G. C. Greubel, Dec 18 2017 *)

x='x+O('x^30); Vec(-2*x*(12*x^2+4*x+1)/(8*x^3+4*x^2+2*x-1)) \\ G. C. Greubel, Dec 18 2017

a(n) = Trace of matrix [({2,2,2},{2,0,0},{0,2,0})^n].
a(n) = 2^n * Trace of matrix [({1,1,1},{1,0,0},{0,1,0})^n].
From Colin Barker, Sep 02 2013: (Start)
a(n) = 2*a(n-1) + 4*a(n-2) + 8*a(n-3).
G.f.: -2*x*(12*x^2+4*x+1)/(8*x^3+4*x^2+2*x-1). (End)

More terms from Colin Barker, Sep 02 2013

A104576 Indices of primes in A001644 (the Lucas 3-step numbers).

Original entry on oeis.org

2, 3, 4, 7, 8, 9, 10, 12, 20, 30, 33, 66, 76, 77, 82, 87, 98, 180, 205, 360, 553, 719, 766, 1390, 1879, 1999, 4033, 5620, 16506, 17436, 23676, 24428, 27758, 31932, 58199, 67661, 85040, 102023, 185595

Offset: 1

T. D. Noe, Mar 16 2005

The sequence of generalized tribonacci numbers is defined as beginning with 1, 3, 7. Subsequent terms are the sum of the previous three terms. Note that the sequence of these generalized tribonacci numbers has many more primes than the tribonacci sequence A000073 (whose prime indices are in A092835).

a(40) > 2*10^5. - Robert Price, Dec 24 2013

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. A104577 (indices of prime generalized tetranacci numbers).

Cf. A000073, A001644, A092835, A105762.

A001644(a(n)) = A105762(n). - Arthur O'Dwyer, Jul 26 2024

a(37)-a(39) from Robert Price, Dec 24 2013

Name clarified by Arthur O'Dwyer, Jul 25 2024

A127215 a(n) = 3^ntribonacci(n) or (3^n)A001644(n+1).

Original entry on oeis.org

3, 27, 189, 891, 5103, 28431, 155277, 859491, 4743603, 26158707, 144374805, 796630059, 4395548511, 24254435799, 133832255589, 738466498755, 4074759563139, 22483948079115, 124063275771981, 684563868232731, 3777327684782127, 20842766314284447

Offset: 1

Artur Jasinski, Jan 09 2007

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,9,27).

Cf. A087131, A127210, A127211, A127212, A127213, A127214, A127216.

I:=[3,27,189]; [n le 3 select I[n] else 3*Self(n-1) + 9*Self(n-2) + 27*Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 18 2017

Table[Tr[MatrixPower[3*{{1, 1, 1}, {1, 0, 0}, {0, 1, 0}}, x]], {x, 1, 20}]
LinearRecurrence[{3, 9, 27}, {3, 27, 189}, 50] (* G. C. Greubel, Dec 18 2017 *)

x='x+O('x^30); Vec(-3*x*(27*x^2+6*x+1)/(27*x^3+9*x^2+3*x-1)) \\ G. C. Greubel, Dec 18 2017

a(n) = Trace of matrix [({3,3,3},{3,0,0},{0,3,0})^n].
a(n) = 3^n * Trace of matrix [({1,1,1},{1,0,0},{0,1,0})^n].
From Colin Barker, Sep 02 2013: (Start)
a(n) = 3*a(n-1) + 9*a(n-2) + 27*a(n-3).
G.f.: -3*x*(27*x^2+6*x+1)/(27*x^3+9*x^2+3*x-1). (End)

More terms from Colin Barker, Sep 02 2013

A106293 Period of the Lucas 3-step sequence A001644 mod n.

Original entry on oeis.org

1, 1, 13, 4, 31, 13, 48, 8, 39, 31, 10, 52, 168, 48, 403, 16, 96, 39, 360, 124, 624, 10, 553, 104, 155, 168, 117, 48, 140, 403, 331, 32, 130, 96, 1488, 156, 469, 360, 2184, 248, 560, 624, 308, 20, 1209, 553, 46, 208, 336, 155, 1248, 168, 52, 117, 310, 48, 4680, 140

Offset: 1

T. D. Noe, May 02 2005

This sequence can differ from the corresponding Fibonacci sequence (A046738) only when n is a multiple of 2 or 11 because the discriminant of the characteristic polynomial x^3-x^2-x-1 is -44. [Clarified by Avery Diep, Aug 22 2025]

a(n) divides A046738(n). - Avery Diep, Aug 22 2025

T. D. Noe, Table of n, a(n) for n=1..1000
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.

Cf. A046738 (period of Fibonacci 3-step sequence mod n), A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1).

n=3; 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}]

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

A106299 Primes that do not divide any term of the Lucas 3-step sequence A001644.

Original entry on oeis.org

2, 103, 199, 211, 421, 757, 883, 907, 991, 1021, 1123, 1237, 1543, 1567, 1621, 1699, 1753, 1873, 2113, 2539, 2731, 2797, 2803, 3391, 3433, 3463, 3499, 3613, 3631, 3793, 3853, 3919, 4093, 4591, 4723, 4933, 4951, 4987, 5107, 5179, 5527, 5791, 5839, 6073

Offset: 1

T. D. Noe, May 02 2005

nonn

If a prime p divides a term a(k) of this sequence, then k must be less than the period of the sequence mod p. Hence these primes are found by computing A001644(k) mod p for increasing k and stopping when either A001644(k) mod p = 0 or the end of the period is reached. Interestingly, for all of these primes except 211, the period of the sequence A001644(k) mod p is (p-1)/d, where d is a small integer. The only other exceptional primes less than 1000000 are 23977 and 47093.

Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.

Cf. A053028 (primes not dividing any Lucas number), A106300 (primes not dividing any Lucas 4-step number), A106301 (primes not dividing any Lucas 5-step number).

n=3; lst={}; Table[p=Prime[i]; a=Join[Table[ -1, {n-1}], {n}]; a=Mod[a, p]; a0=a; While[s=Mod[Plus@@a, p]; a=RotateLeft[a]; a[[n]]=s; !(a==a0 || s==0)]; If[s>0, AppendTo[lst, p]], {i, 1000}]; lst

A074678 a(n) = Sum_{j=0..floor(n/2)} (-1)^(j+floor(n/2))*S(2j+q), where S(n) are generalized tribonacci numbers (A001644) and q = (1-(-1)^n)/2.

Original entry on oeis.org

3, 1, 0, 6, 11, 15, 28, 56, 103, 185, 340, 630, 1159, 2127, 3912, 7200, 13243, 24353, 44792, 82390, 151539, 278719, 512644, 942904, 1734271, 3189817, 5866988, 10791078, 19847887, 36505951, 67144912, 123498752, 227149619, 417793281

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 30 2002

a(n) is the convolution of S(n) with the sequence (1,0,-1,0,1,0,-1,0,....) A056594.

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

Cf. A001644, A056594.

a:=[3,1,0,6,11];; for n in [6..40] do a[n]:=a[n-1]+2*a[n-3]+a[n-4] +a[n-5]; od; a; # G. C. Greubel, Apr 02 2019

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (3-2*x-x^2)/(1-x-2*x^3-x^4-x^5) )); // G. C. Greubel, Apr 02 2019

CoefficientList[Series[(3-2*x-x^2)/(1-x-2*x^3-x^4-x^5), {x, 0, 40}], x]
LinearRecurrence[{1,0,2,1,1}, {3,1,0,6,11}, 40] (* G. C. Greubel, Apr 02 2019 *)

my(x='x+O('x^40)); Vec((3-2*x-x^2)/(1-x-2*x^3-x^4-x^5)) \\ G. C. Greubel, Apr 02 2019

((3-2*x-x^2)/(1-x-2*x^3-x^4-x^5)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 02 2019

a(n) = Sum_{j=0..floor(n/2)} (-1)^(j+floor(n/2))*S(2j+q), where S(n) are generalized tribonacci numbers (A001644) and q = (1-(-1)^n)/2.
a(n) = a(n-1) + 2*a(n-3) + a(n-4) + a(n-5), a(0)=3, a(1)=1, a(2)=0, a(3)=6, a(4)=11.
G.f.: (3 - 2*x - x^2)/(1 - x - 2*x^3 - x^4 - x^5).

A075298 Inverted (definition in A075193) generalized tribonacci numbers A001644.

Original entry on oeis.org

1, 1, -5, 5, 1, -11, 15, -3, -23, 41, -21, -43, 105, -83, -65, 253, -271, -47, 571, -795, 177, 1189, -2161, 1149, 2201, -5511, 4459, 3253, -13223, 14429, 2047, -29699, 42081, -10335, -61445, 113861, -62751, -112555, 289167, -239363, -162359, 690889, -767893, -85355, 1544137, -2226675, 597183, 3173629

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Sep 13 2002

a(n) = -C(n+1), C(n)=reflected generalized tribonacci numbers A073145.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Curtis Cooper, S. Miller, Peter J. C. Moses, M. Sahin, and T. Thanatipanonda, On Identities of Ruggles, Horadam, Howard, and Young, Preprint 2016.
Index entries for linear recurrences with constant coefficients, signature (-1,-1,1).

Cf. A001644, A073145, A075193.

a:=[1,1,-5];; for n in [4..50] do a[n]:=-a[n-1]-a[n-2]+a[n-3]; od; a; # G. C. Greubel, Apr 09 2019

R:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (1+2*x-3*x^2)/(1+x+x^2-x^3) )); // G. C. Greubel, Apr 09 2019

CoefficientList[Series[(1+2x-3x^2)/(1+x+x^2-x^3), {x, 0, 50}], x]

my(x='x+O('x^50)); Vec((1+2*x-3*x^2)/(1+x+x^2-x^3)) \\ G. C. Greubel, Apr 09 2019

((1+2*x-3*x^2)/(1+x+x^2-x^3)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Apr 09 2019

a(n) = -a(n-1) - a(n-2) + a(n-3), a(0)=1, a(1)=1, a(2)=-5.
G.f.: (1+2*x-3*x^2)/(1+x+x^2-x^3).
a(n) = A078046(n) + 3*A078046(n-1). - R. J. Mathar, Sep 20 2020

A106294 Period of the Lucas 3-step sequence A001644 mod prime(n).

Original entry on oeis.org

1, 13, 31, 48, 10, 168, 96, 360, 553, 140, 331, 469, 560, 308, 46, 52, 3541, 1860, 1519, 5113, 5328, 3120, 287, 8011, 3169, 680, 51, 1272, 990, 12883, 5376, 5720, 18907, 3864, 7400, 2850, 8269, 162, 9296, 2494, 32221, 10981, 36673, 4656, 3234, 198, 5565

Offset: 1

T. D. Noe, May 02 2005

nonn

This sequence differs from the corresponding Fibonacci sequence (A106302) at n=1 and 5 because these correspond to the primes 2 and 11, which are the prime factors of -44, the discriminant of the characteristic polynomial x^3-x^2-x-1. We have a(n) < prime(n) for the primes 2, 11 and A106279.

For a prime p, the period depends on the zeros of x^3-x^2-x-1 mod p. If there are 3 zeros, then the period is < p. If there are no zeros, then the period is p^2+p+1 or a simple fraction of p^2+p+1. Also note that the period can be prime, as for p=3, 5, 31, 59, 71, 89, 97, 157, 223. When the period is prime, the orbits have a simple structure. [From T. D. Noe, Sep 18 2008]

Eric Weisstein's World of Mathematics, Fibonacci n-Step Number

Cf. A106273, A106279, A106302.

n=3; Table[p=Prime[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}]

a(n) = A106293(prime(n)).

A073446 Product L(n)S(n), where L(n) are Lucas numbers and S(n) are Lucas 3-step numbers = A000032(n) A001644(n).

Original entry on oeis.org

6, 1, 9, 28, 77, 231, 702, 2059, 6157, 18316, 54489, 162185, 482678, 1436397, 4274853, 12722028, 37861085, 112675763, 335326230, 997940307, 2969899037, 8838503884, 26303639349, 78280380217, 232964641030, 693309407681

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 01 2002

a(n) is also the trace of the matrix R^n, where R is the Kronecker product of the Fibonacci matrix (Fibomatrix): first row (1,1), second row (1,0), times the Tribomatrix: first row (1,1,0), second row (1,0,1), third row (1,0,0).

a(n) is semiprime iff n is an element of A001606 (an index of a prime Lucas number) and an element of A104576 (an index of a prime Lucas 3-step number). The only known such are n = 2, 4, 7, 8 (through 67661). - Jonathan Vos Post, May 10 2005

Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Elia, Derived Sequences, The Tribonacci Recurrence and Cubic Forms, The Fibonacci Quarterly 39.2 (2001): 107-109.
F. T. Howard, A Tribonacci Identity, The Fibonacci Quarterly 39.4 (2001): 352-357.
Index entries for linear recurrences with constant coefficients, signature (1,4,5,2,-1,1).

Cf. A000032, A000040, A001358, A001606, A001644, A104576.

a:=[6,1,9,28,77,231];; for n in [7..40] do a[n]:=a[n-1]+4*a[n-2] +5*a[n-3]+2*a[n-4]-a[n-5]+a[n-6]; od; a; # G. C. Greubel, Feb 19 2019

m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (6-5*x-16*x^2-15*x^3-4*x^4+x^5)/(1-x-4*x^2-5*x^3-2*x^4+x^5-x^6) )); // G. C. Greubel, Feb 19 2019

CoefficientList[Series[(6-5x-16x^2-15x^3-4x^4+x^5)/(1-x-4x^2-5x^3-2x^4 +x^5-x^6), {x, 0, 50}], x]

my(x='x+O('x^40)); Vec((6-5*x-16*x^2-15*x^3-4*x^4+x^5)/(1-x-4*x^2-5*x^3-2*x^4+x^5-x^6)) \\ G. C. Greubel, Feb 19 2019

((6-5*x-16*x^2-15*x^3-4*x^4+x^5)/(1-x-4*x^2-5*x^3-2*x^4+x^5-x^6)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Feb 19 2019

a(n) = a(n-1)+4*a(n-2)+5*a(n-3)+2*a(n-4)-a(n-5)+a(n-6), a(0)=6, a(1)=1, a(2)=9, a(3)=28, a(4)=77, a(5)=231.

G.f.: (6-5*x-16*x^2-15*x^3-4*x^4+x^5)/(1-x-4*x^2-5*x^3-2*x^4+x^5-x^6).

A073728 a(n) = Sum_{k=0..n} S(k), where S(n) are the tribonacci generalized numbers A001644.

Original entry on oeis.org

3, 4, 7, 14, 25, 46, 85, 156, 287, 528, 971, 1786, 3285, 6042, 11113, 20440, 37595, 69148, 127183, 233926, 430257, 791366, 1455549, 2677172, 4924087, 9056808, 16658067, 30638962, 56353837, 103650866, 190643665, 350648368, 644942899

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 06 2002

Robert Israel, Table of n, a(n) for n = 0..3399
Daniel Birmajer, Juan B. Gil, and Michael D. Weiner, Linear recurrence sequences with indices in arithmetic progression and their sums, arXiv:1505.06339 [math.NT], 2015.
Index entries for linear recurrences with constant coefficients, signature (1,1,1).

Partial sums of A001644.

Cf. A000073.

I:=[3,4,7]; [n le 3 select I[n] else Self(n-1)+Self(n-2) +Self(n-3): n in [1..40]]; // Vincenzo Librandi, Mar 27 2015

A:= gfun[rectoproc]({a(n)=a(n-1)+a(n-2)+a(n-3), a(0)=3, a(1)=4, a(2)=7},a(n),remember):
seq(A(n),n=0..100); # Robert Israel, Mar 26 2015

CoefficientList[Series[(3+x)/(1-x-x^2-x^3), {x, 0, 40}], x]

my(x='x+O('x^40)); Vec((3+x)/(1-x-x^2-x^3)) \\ G. C. Greubel, Apr 09 2019

((3+x)/(1-x-x^2-x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 09 2019

a(n) = a(n-1) + a(n-2) + a(n-3), a(0)=3, a(1)=4, a(2)=7.
G.f.: (3+x)/(1-x-x^2-x^3).
a(n) = 3*T(n+1) + T(n), where T(n) are the tribonacci numbers A000073.
a(n) = (S(n+3) - S(n+1))/2, where S(n) = A001644(n). - Michael D. Weiner, Mar 27 2015
a(n) = Sum_{r root of x^3-x^2-x-1} r^n/(-r^2+r+2). - Fabian Pereyra, Nov 21 2024

cp's OEIS Frontend

A127214 a(n) = 2^n*tribonacci(n) or (2^n)*A001644(n+1).

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A104576 Indices of primes in A001644 (the Lucas 3-step numbers).

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A127215 a(n) = 3^n*tribonacci(n) or (3^n)*A001644(n+1).

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A106293 Period of the Lucas 3-step sequence A001644 mod n.

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A106299 Primes that do not divide any term of the Lucas 3-step sequence A001644.

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A074678 a(n) = Sum_{j=0..floor(n/2)} (-1)^(j+floor(n/2))*S(2j+q), where S(n) are generalized tribonacci numbers (A001644) and q = (1-(-1)^n)/2.

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A075298 Inverted (definition in A075193) generalized tribonacci numbers A001644.

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A127214 a(n) = 2^ntribonacci(n) or (2^n)A001644(n+1).

A127215 a(n) = 3^ntribonacci(n) or (3^n)A001644(n+1).

A073446 Product L(n)S(n), where L(n) are Lucas numbers and S(n) are Lucas 3-step numbers = A000032(n) A001644(n).