A000073 -id:A000073 - cp's OEIS frontend

A008937 a(n) = Sum_{k=0..n} T(k) where T(n) are the tribonacci numbers A000073.

0, 1, 2, 4, 8, 15, 28, 52, 96, 177, 326, 600, 1104, 2031, 3736, 6872, 12640, 23249, 42762, 78652, 144664, 266079, 489396, 900140, 1655616, 3045153, 5600910, 10301680, 18947744, 34850335, 64099760, 117897840, 216847936, 398845537, 733591314, 1349284788

Offset: 0

N. J. A. Sloane, Alejandro Teruel (teruel(AT)usb.ve)

a(n+1) is the number of n-bit sequences that avoid 1100. - David Callan, Jul 19 2004 [corrected by Kent E. Morrison, Jan 08 2019]. Also the number of n-bit sequences avoiding one of the patterns 1000, 0011, 1110, ... or any binary string of length 4 without overlap at beginning and end. Strings where it is not true are: 1111, 1010, 1001, ... and their bitwise complements. - Alois P. Heinz, Jan 09 2019
Row sums of Riordan array (1/(1-x), x(1+x+x^2)). - Paul Barry, Feb 16 2005
Diagonal sums of Riordan array (1/(1-x)^2, x(1+x)/(1-x)), A104698.
A shifted version of this sequence can be found in Eqs. (4) and (3) on p. 356 of Dunkel (1925) with r = 3. (Equation (3) follows equation (4) in the paper!) The whole paper is a study of the properties of this and other similar sequences indexed by the parameter r. For r = 2, we get a shifted version of A000071. For r = 4, we get a shifted version of A107066. For r = 5, we get a shifted version of A001949. For r = 6, we get a shifted version of A172316. See also the table in A172119. - Petros Hadjicostas, Jun 14 2019
Officially, to match A000073, this should start with a(0)=a(1)=0, a(2)=1. - N. J. A. Sloane, Sep 12 2020
Numbers with tribonacci representation that is a prefix of 100100100100... . - Jeffrey Shallit, Jul 10 2024

			G.f. = x + 2*x^2 + 4*x^3 + 8*x^4 + 15*x^5 + 28*x^6 + 52*x^7 + 96*x^8 + 177*x^9 + ... [edited by _Petros Hadjicostas_, Jun 12 2019]

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 41.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Isha Agarwal, Matvey Borodin, Aidan Duncan, Kaylee Ji, Tanya Khovanova, Shane Lee, Boyan Litchev, Anshul Rastogi, Garima Rastogi, and Andrew Zhao, From Unequal Chance to a Coin Game Dance: Variants of Penney's Game, arXiv:2006.13002 [math.HO], 2020.
Kassie Archer and Noel Bourne, Pattern avoidance in compositions and powers of permutations, arXiv:2505.05218 [math.CO], 2025. See pp. 6, 9.
Erik Bates, Blan Morrison, Mason Rogers, Arianna Serafini, and Anav Sood, A new combinatorial interpretation of partial sums of m-step Fibonacci numbers, arXiv:2503.11055 [math.CO], 2025. See p. 3.
Otto Dunkel, Solutions of a probability difference equation, Amer. Math. Monthly, 32 (1925), 354-370; see pp. 356 and 369.
Jia Huang, A coin flip game and generalizations of Fibonacci numbers, arXiv:2501.07463 [math.CO], 2025. See p. 7.
Thomas Langley, Jeffrey Liese, and Jeffrey Remmel, Generating Functions for Wilf Equivalence Under Generalized Factor Order, J. Int. Seq. 14 (2011), Article # 11.4.2.
William Verreault, MacMahon Partition Analysis: a discrete approach to broken stick problems, arXiv:2107.10318 [math.CO], 2021.
Index entries for linear recurrences with constant coefficients, signature (2,0,0,-1).

Partial sums of A000073. Cf. A000213, A018921, A027084, A077908, A209972.
Row sums of A055216.
Column k = 1 of A140997 and second main diagonal of A140994.

a:=[0,1,1];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # G. C. Greubel, Sep 13 2019

a008937 n = a008937_list !! n
a008937_list = tail $ scanl1 (+) a000073_list
-- Reinhard Zumkeller, Apr 07 2012

[ n eq 1 select 0 else n eq 2 select 1 else n eq 3 select 2 else n eq 4 select 4 else 2*Self(n-1)-Self(n-4): n in [1..40] ]; // Vincenzo Librandi, Aug 21 2011

A008937 := proc(n) option remember; if n <= 3 then 2^n else 2*procname(n-1)-procname(n-4) fi; end;
a:= n-> (Matrix([[1,1,0,0], [1,0,1,0], [1,0,0,0], [1,0,0,1]])^n)[4,1]: seq(a(n), n=0..50); # Alois P. Heinz, Jul 24 2008

CoefficientList[Series[x/(1-2x+x^4), {x, 0, 40}], x]
Accumulate[LinearRecurrence[{1,1,1},{0,1,1},40]] (* Harvey P. Dale, Dec 04 2017 *)
LinearRecurrence[{2, 0, 0, -1},{0, 1, 2, 4},40] (* Ray Chandler, Mar 01 2024 *)

{a(n) = if( n<0, polcoeff( - x^3 / (1 - 2*x^3 + x^4) + x * O(x^-n), -n), polcoeff( x / (1 - 2*x + x^4) + x * O(x^n), n))}; /* Michael Somos, Aug 23 2014 */

a(n) = sum(j=0, n\2, sum(k=0, j, binomial(n-2*j,k+1)*binomial(j,k)*2^k)); \\ Michel Marcus, Sep 08 2017

def A008937_list(prec):
    P = PowerSeriesRing(ZZ, 'x', prec)
    x = P.gen().O(prec)
    return (x/(1-2*x+x^4)).list()
A008937_list(40) # G. C. Greubel, Sep 13 2019

a(n) = A018921(n-2) = A027084(n+1) + 1.
a(n) = (A000073(n+2) + A000073(n+4) - 1)/2.
From Mario Catalani (mario.catalani(AT)unito.it), Aug 09 2002: (Start)
G.f.: x/((1-x)*(1-x-x^2-x^3)).
a(n) = 2*a(n-1) - a(n-4), a(0) = 0, a(1) = 1, a(2) = 2, a(3) = 4. (End)
a(n) = 1 + a(n-1) + a(n-2) + a(n-3). E.g., a(11) = 1 + 600 + 326 + 177 = 1104. - Philippe LALLOUET (philip.lallouet(AT)orange.fr), Oct 29 2007
a(n) = term (4,1) in the 4 X 4 matrix [1,1,0,0; 1,0,1,0; 1,0,0,0; 1,0,0,1]^n. - Alois P. Heinz, Jul 24 2008
a(n) = -A077908(-n-3). - Alois P. Heinz, Jul 24 2008
a(n) = (A000213(n+2) - 1) / 2. - Reinhard Zumkeller, Apr 07 2012
a(n) = Sum_{j=0..floor(n/2)} Sum_{k=0..j} binomial(n-2j,k+1) *binomial(j,k)*2^k. - Tony Foster III, Sep 08 2017
a(n) = Sum_{k=0..floor(n/2)} (n-2*k)*hypergeom([-k,-n+2*k+1], [2], 2). - Peter Luschny, Nov 09 2017
a(n) = 2^(n-1)*hypergeom([1-n/4, 1/4-n/4, 3/4-n/4, 1/2-n/4], [1-n/3, 1/3-n/3, 2/3-n/3], 16/27) for n > 0. - Peter Luschny, Aug 20 2020
a(n-1) = T(n) + T(n-3) + T(n-6) + ... + T(2+((n-2) mod 3)), for n >= 4, where T is A000073(n+1). - Jeffrey Shallit, Dec 24 2020

A046738 Period of Fibonacci 3-step sequence A000073 mod n.

Original entry on oeis.org

1, 4, 13, 8, 31, 52, 48, 16, 39, 124, 110, 104, 168, 48, 403, 32, 96, 156, 360, 248, 624, 220, 553, 208, 155, 168, 117, 48, 140, 1612, 331, 64, 1430, 96, 1488, 312, 469, 360, 2184, 496, 560, 624, 308, 440, 1209, 2212, 46, 416, 336, 620, 1248, 168

Offset: 1

David W. Wilson

Could also be called the tribonacci Pisano periods. [Carl R. White, Oct 05 2009]
Klaska notes that n=208919=59*3541 satisfies a(n) = a(n^2). - Michel Marcus, Mar 03 2016
39, 78, 273, 546 also satisfy a(n) = a(n^2). - Michel Marcus, Mar 07 2016

T. D. Noe [1..1000] + Jean-François Alcover [1001..2000] + Zhong Ziqian [2001..20000], Table of n, a(n) for n = 1..20000
Jirí Klaška, A search for Tribonacci-Wieferich primes, Acta Mathematica Universitatis Ostraviensis, vol. 16 (2008), issue 1, pp. 15-20.
Jirí Klaška, On Tribonacci-Wieferich primes, Fibonacci Quart. 46/47 (2008/2009), no. 4, 290-297.
Jirí Klaška, Tribonacci partition formulas modulo m, Acta Mathematica Sinica, English Series, March 2010, Volume 26, Issue 3, pp 465-476.
M. E. Waddill, Some properties of a generalized Fibonacci sequence modulo m, The Fibonacci Quarterly, vol. 16, no. 4, pp. 344-353 (1978).

Cf. A106302.

Cf. A001175.

a:= proc(n) local f, k, l; l:= ifactors(n)[2];
      if nops(l)<>1 then ilcm(seq(a(i[1]^i[2]), i=l))
    else f:= [0, 0, 1];
         for k do f:=[f[2], f[3], f[1]+f[2]+f[3] mod n];
                  if f=[0, 0, 1] then break fi
         od; k
      fi
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Aug 27 2023

Table[a = {0, 1, 1}; a = a0 = Mod[a, n]; k = 0; While[k++; s = a[[3]] + a[[2]] + a[[1]]; a = RotateLeft[a]; a[[-1]] = Mod[s, n]; a != a0]; k, {n, 100}] (* T. D. Noe, Aug 28 2012 *)

from itertools import count
def A046738(n):
    a = b = (0,0,1%n)
    for m in count(1):
        b = b[1:] + (sum(b) % n,)
        if a == b:
            return m # Chai Wah Wu, Feb 27 2022

a(3^k) = 13*3^(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, 1978]. - Chai Wah Wu, Feb 25 2022

Let the prime factorization of n be p1^e1*...*pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)) [Waddill, 1978]. - Avery Diep, Aug 26 2025

A232570 Numbers k that divide tribonacci(k) (A000073(k)).

Original entry on oeis.org

1, 8, 16, 19, 32, 47, 53, 64, 103, 112, 128, 144, 155, 163, 192, 199, 208, 221, 224, 256, 257, 269, 272, 299, 311, 368, 397, 401, 419, 421, 448, 499, 512, 587, 599, 617, 640, 683, 757, 768, 773, 784, 863, 883, 896, 907, 911, 929, 936, 991, 1021, 1024

Offset: 1

Seiichi Manyama, Jun 17 2016

nonn

Inspired by A023172 (numbers k such that k divides Fibonacci(k)).
Includes all primes p such that x^3-x^2-x-1 has 3 distinct roots in the field GF(p) (A106279). - Robert Israel, Feb 07 2018
Includes 2^k for k >= 3. - Robert Israel, Jul 26 2024

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Cf. A000073, A023172, A106279.

with(LinearAlgebra[Modular]):
T:= (n, m)-> MatrixPower(m, Mod(m, <<0|1|0>,
    <0|0|1>, <1|1|1>>, float[8]), n)[1, 3]:
a:= proc(n) option remember; local k; if n=1
      then 1 else for k from 1+a(n-1)
      while T(k$2)>0 do od; k fi
    end:
seq(a(n), n=1..70);  # Alois P. Heinz, Feb 05 2018

trib = LinearRecurrence[{1, 1, 1}, {0, 0, 1}, 2000]; Reap[Do[If[Divisible[ trib[[n+1]], n], Print[n]; Sow[n]], {n, 1, Length[trib]-1}]][[2, 1]] (* Jean-François Alcover, Mar 22 2019 *)

require 'matrix'
def power(a, n, mod)
  return Matrix.I(a.row_size) if n == 0
  m = power(a, n >> 1, mod)
  m = (m * m).map{|i| i % mod}
  return m if n & 1 == 0
  (m * a).map{|i| i % mod}
end
def f(m, n)
  ary0 = Array.new(m, 0)
  ary0[0] = 1
  v = Vector.elements(ary0)
  ary1 = [Array.new(m, 1)]
  (0..m - 2).each{|i|
    ary2 = Array.new(m, 0)
    ary2[i] = 1
    ary1 << ary2
  }
  a = Matrix[*ary1]
  mod = n
  (power(a, n, mod) * v)[m - 1]
end
def a(n)
  (1..n).select{|i| f(3, i) == 0}
end

A085697 a(n) = T(n)^2, where T(n) = A000073(n) is the n-th tribonacci number.

Original entry on oeis.org

0, 0, 1, 1, 4, 16, 49, 169, 576, 1936, 6561, 22201, 75076, 254016, 859329, 2907025, 9834496, 33269824, 112550881, 380757169, 1288092100, 4357584144, 14741602225, 49870482489, 168710633536, 570743986576, 1930813074369, 6531893843049

Offset: 0

Emanuele Munarini, Jul 18 2003

In general, squaring the terms of a third-order linear recurrence with signature (x,y,z) will result in a sixth-order recurrence with signature (x^2 + y, x^2*y + z*x + y^2, x^3*z + 4*x*y*z - y^3 + 2*z^2, x^2*z^2 - x*y^2*z - z^2*y, z^2*y^2 - z^3*x, -z^4). - Gary Detlefs, Jan 10 2023

R. Schumacher, Explicit formulas for sums involving the squares of the first n Tribonacci numbers, Fib. Q., 58:3 (2020), 194-202.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,3,6,-1,0,-1).

Cf. A000073, A057597, A099463, A107239, A113300.

R:=PowerSeriesRing(Integers(), 40); [0,0] cat Coefficients(R!( x^2*(1-x-x^2-x^3)/((1-3*x-x^2-x^3)*(1+x+x^2-x^3)) )); // G. C. Greubel, Nov 20 2021

LinearRecurrence[{2,3,6,-1,0,-1},{0,0,1,1,4,16},30] (* Harvey P. Dale, Oct 26 2020 *)

t[0]:0$  t[1]:0$  t[2]:1$
t[n]:=t[n-1]+t[n-2]+t[n-3]$
makelist(t[n]^2,n,0,40); /* Emanuele Munarini, Mar 01 2011 */

@CachedFunction
def T(n): # A000073
    if (n<2): return 0
    elif (n==2): return 1
    else: return T(n-1) +T(n-2) +T(n-3)
def A085697(n): return T(n)^2
[A085697(n) for n in (0..40)] # G. C. Greubel, Nov 20 2021

G.f.: x^2*( 1-x-x^2-x^3 )/( (1-3*x-x^2-x^3)*(1+x+x^2-x^3) ).
a(n+6) = 2*a(n+5) + 3*a(n+4) + 6*a(n+3) - a(n+2) - a(n).
a(n) = (-A057597(n-2) + 3*A057597(n-1) + 6*A057597(n) + 5*A113300(n-1) - A099463(n-2))/11. - R. J. Mathar, Aug 19 2008

Offset corrected to match A000073 by N. J. A. Sloane, Sep 12 2020

Name corrected to match corrected offset by Michael A. Allen, Jun 10 2021

A107239 Sum of squares of tribonacci numbers (A000073).

Original entry on oeis.org

0, 0, 1, 2, 6, 22, 71, 240, 816, 2752, 9313, 31514, 106590, 360606, 1219935, 4126960, 13961456, 47231280, 159782161, 540539330, 1828631430, 6186215574, 20927817799, 70798300288, 239508933824, 810252920400, 2741065994769, 9272959837818, 31370198430718

Offset: 0

Jonathan Vos Post, May 17 2005

			a(7) = 71 = 0^2 + 0^2 + 1^2 + 1^2 + 2^2 + 4^2 + 7^2

R. Schumacher, Explicit formulas for sums involving the squares of the first n Tribonacci numbers, Fib. Q., 58:3 (2020), 194-202.

G. C. Greubel, Table of n, a(n) for n = 0..1000
M. Feinberg, Fibonacci-Tribonacci, Fib. Quart. 1(3) (1963), 71-74.
Z. Jakubczyk, Advanced Problems and Solutions, Fib. Quart. 51 (3) (2013) 185, H-715.
Eric Weisstein's World of Mathematics, Tribonacci Number
Index entries for linear recurrences with constant coefficients, signature (3,1,3,-7,1,-1,1).

Cf. A000073, A000213, A085697.

Cf. A107240, A107241, A107242, A107243, A107244, A107245, A107246, A107247.

R:=PowerSeriesRing(Integers(), 40); [0,0] cat Coefficients(R!( x^2*(1-x-x^2-x^3)/((1+x+x^2-x^3)*(1-3*x-x^2-x^3)*(1-x)) )); // G. C. Greubel, Nov 20 2021

b:= proc(n) option remember; `if`(n<3, [n*(n-1)/2$2],
     (t-> [t, t^2+b(n-1)[2]])(add(b(n-j)[1], j=1..3)))
    end:
a:= n-> b(n)[2]:
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 22 2021

Accumulate[LinearRecurrence[{1,1,1},{0,0,1},30]^2] (* Harvey P. Dale, Sep 11 2011 *)
LinearRecurrence[{3,1,3,-7,1,-1,1}, {0,0,1,2,6,22,71}, 30] (* Ray Chandler, Aug 02 2015 *)

@CachedFunction
def T(n): # A000073
    if (n<2): return 0
    elif (n==2): return 1
    else: return T(n-1) +T(n-2) +T(n-3)
def A107231(n): return sum(T(j)^2 for j in (0..n))
[A107239(n) for n in (0..40)] # G. C. Greubel, Nov 20 2021

a(n) = T(0)^2 + T(1)^2 + ... + T(n)^2 where T(n) = A000073(n).
From R. J. Mathar, Aug 19 2008: (Start)
a(n) = Sum_{i=0..n} A085697(i).
G.f.: x^2*(1-x-x^2-x^3)/((1+x+x^2-x^3)*(1-3*x-x^2-x^3)*(1-x)). (End)
a(n) = 1/4 - (1/11)*Sum_{R = RootOf(_Z^3-_Z^2-_Z-1)} ((3 + 7*_R + 5*_R^2)/(3*_R^2 - 2*_R - 1)*_R^(-n) - (1/44)*Sum{_R = RootOf(Z^3+_Z^2+3*_Z-1)} ((-1 - 2*_R - 9*_R^2)/(3*_R^2 + 2*_R + 3)*_R^(-n). - _Robert Israel, Mar 26 2010
a(n+1) = A000073(n)*A000073(n+1) + ( (A000073(n+1) - A000073(n-1))^2 - 1 )/4 for n>0 [Jakubczyk]. - R. J. Mathar, Dec 19 2013

A073778 a(n) = Sum_{k=0..n} T(k)*T(n-k), where T is A000073; convolution of A000073 with itself.

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 5, 12, 26, 56, 118, 244, 499, 1010, 2027, 4040, 8004, 15776, 30956, 60504, 117845, 228818, 443057, 855732, 1649022, 3171128, 6086626, 11662252, 22309543, 42614178, 81286743, 154856528, 294660040, 560052736, 1063367384, 2017030256

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 10 2002

Number of binary sequences of length n+1 that have exactly one subsequence 000. Example: a(4)=5 because we have 00010,00011,01000,10001 and 11000. Column 1 of A118390. - Emeric Deutsch, Apr 27 2006

Let (b(n)) be the p-INVERT of (1,1,1,0,0,0,0,0,0,...) using p(S) = 1 - S^2; then b(n) = a(n+3) for n >= 0. See A292324. - Clark Kimberling, Sep 15 2017

Michael De Vlieger, Table of n, a(n) for n = 0..3770, terms n = 2..1002 from Vincenzo Librandi.
Daniel Birmajer, Juan B. Gil, and Michael D. Weiner, Linear recurrence sequences and their convolutions via Bell polynomials, arXiv:1405.7727 [math.CO], 2014; see also, J. Int. Seq. 18 (2015) # 15.1.2.
Michael Dairyko, Lara Pudwell, Samantha Tyner and Casey Wynn. Non-contiguous pattern avoidance in binary trees. Electron. J. Combin. 19 (2012), no. 3, Paper 22, 21 pp. MR2967227. - From _N. J. A. Sloane_, Feb 01 2013
Omar Khadir, László Németh, and László Szalay, Tiling of dominoes with ranked colors, Results in Math. (2024) Vol. 79, Art. No. 253. See p. 2.
Emrah Kilic and Helmut Prodinger, A Note on the Conjecture of Ramirez and Sirvent, J. Int. Seq. 17 (2014) # 14.5.8.
László Németh and László Szalay, Explicit solution of system of two higher-order recurrences, arXiv:2408.12196 [math.NT], 2024. See p. 10.
José L. Ramirez and Víctor F. Sirvent, Incomplete Tribonacci Numbers and Polynomials, Journal of Integer Sequences, Vol. 17, 2014, #14.4.2.
Index entries for linear recurrences with constant coefficients, signature (2,1,0,-3,-2,-1).

Cf. A000073, A118390.

A073778:=proc(n) coeftayl(x^4/(1-x-x^2-x^3)^2, x=0, n); end proc: seq(A073778(n), n=0..40); # Wesley Ivan Hurt, Nov 17 2014

CoefficientList[Series[x^4/(1-x-x^2-x^3)^2, {x, 0, 40}], x]

a(n):= sum((k+1)*sum(binomial(j,n-3*k+2*j-4)*binomial(k,j), j,0,k), k,0,n-4);
makelist(a(n), n, 0, 30); /* Vladimir Kruchinin, Dec 14 2011 */

T(n) = ([0, 1, 0; 0, 0, 1; 1, 1, 1]^n)[1, 3]; \\ A000073
a(n) = sum(k=0, n, T(k)*T(n-k)); \\ Michel Marcus, Oct 20 2021

[( x^4/(1-x-x^2-x^3)^2 ).series(x,n+1).list()[n] for n in (0..40)] # Zerinvary Lajos, Jun 02 2009; modified by G. C. Greubel, Dec 15 2021

G.f.: x^4/(1 - x - x^2 - x^3)^2.
a(n) = Sum_{k=0..n-4} (k+1)*Sum_{j=0..k} binomial(j,n-3*k+2*j-4)*binomial(k,j). - Vladimir Kruchinin, Dec 14 2011
(n-2)*a(n) - (n-1)*a(n-1) - n*a(n-2) - (n+1)*a(n-3) = 0, n > 2. - Michael D. Weiner, Nov 18 2014

Two initial zeros inserted by Hans J. H. Tuenter, Oct 20 2021

A117546 Number of representations of n as a sum of distinct tribonacci numbers (A000073).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 3, 2, 2, 2, 2, 2, 2, 3, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 3, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2

Offset: 0

T. D. Noe and Jonathan Vos Post, Mar 28 2006

It can be shown that, like the Fibonacci numbers, the tribonacci numbers are complete; that is, a(n)>0 for all n. There is always a representation, free of three consecutive tribonacci numbers, which is analogous to the Zeckendorf representation of Fibonacci numbers. See A003726.

			a(14)=2 because 14 is both 13+1 and 7+4+2+1.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Tribonacci Number.
Eric Weisstein's World of Mathematics, Zeckendorf Representation.

Cf. A000119 (number of representations of n as a sum of distinct Fibonacci numbers).

Cf. A240844.

a117546 = p $ drop 3 a000073_list where
   p _  0     = 1
   p (k:ks) m = if m < k then 0 else p ks (m - k) + p ks m
-- Reinhard Zumkeller, Apr 13 2014

tr={1,2,4,7,13,24,44,81,149}; len=tr[[ -1]]; cnt=Table[0,{len}]; Do[v=IntegerDigits[k,2,Length[tr]]; s=Dot[tr,v]; If[s<=len, cnt[[s]]++ ], {k,2^(Length[tr])-1}]; cnt

a(0)=1 added and offset changed by Reinhard Zumkeller, Apr 13 2014

A106302 Period of the Fibonacci 3-step sequence A000073 mod prime(n).

Original entry on oeis.org

4, 13, 31, 48, 110, 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, Sep 18 2008

nonn

This sequence differs from the corresponding Lucas sequence (A106294) 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 in A106279.

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

Cf. A106273, A106279, A106294, A046738.

n=3; Table[p=Prime[i]; a=Join[{1},Table[0,{n-1}]]; 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
from sympy import prime
def A106302(n):
    a = b = (0,)*2+(1 % (p:= prime(n)),)
    for m in count(1):
        b = b[1:] + (sum(b) % p,)
        if a == b:
            return m # Chai Wah Wu, Feb 27 2022

a(n) = A046738(prime(n)).

A115390 Binomial transform of tribonacci sequence A000073.

Original entry on oeis.org

0, 0, 1, 4, 12, 34, 96, 272, 772, 2192, 6224, 17672, 50176, 142464, 404496, 1148480, 3260864, 9258528, 26287616, 74638080, 211918912, 601698560, 1708394752, 4850622592, 13772308480, 39103533056, 111026143488, 315235058688, 895042726912, 2541282959872

Offset: 0

Jonathan Vos Post, Mar 08 2006

See also A117189 Binomial transform of the tribonacci sequence A000073.

			1*0 = 0.
1*0 + 1*0 = 0.
1*0 + 2*0 + 1*1 = 1.
1*0 + 3*0 + 3*1 + 1* 1 = 4.
1*0 + 4*0 + 6*1 + 4*1 + 1*2 = 12.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
J. Pan, Multiple Binomial Transforms and Families of Integer Sequences, J. Int. Seq. 13 (2010), 10.4.2
J. Pan, Some Properties of the Multiple Binomial Transform and the Hankel Transform of Shifted Sequences, J. Int. Seq. 14 (2011) # 11.3.4, remark 14.
Eric Weisstein's World of Mathematics, Binomial Transform.
Index entries for linear recurrences with constant coefficients, signature (4,-4,2).

Cf. A000073, A117189. Trisection of A103685.

a115390 n = a115390_list !! n
a115390_list = 0 : 0 : 1 : map (* 2) (zipWith (-) a115390_list
   (tail $ map (* 2) $ zipWith (-) a115390_list (tail a115390_list)))
-- Reinhard Zumkeller, Oct 21 2011

b[0]=b[1]=0;b[2]=1;b[n_]:=b[n]=b[n-1]+b[n-2]+b[n-3]; a[n_]:=Sum[n!/(k!*(n-k)!)*b[k],{k,0,n}];Table[a[n],{n,0,27}] (* Farideh Firoozbakht, Mar 11 2006 *)

sum(sum(binomial(j-1,k-1)*2^(j-k)*binomial(n-j+k-1,2*k-1),j,k,n-k),k,1,n); /* Vladimir Kruchinin, Aug 18 2010 */

a(n) = Sum_{k=0..n} C(n,k)*A000073(k).
O.g.f.: -x^2/(-1+4*x-4*x^2+2*x^3). - R. J. Mathar, Apr 02 2008
a(n) = sum(sum(binomial(j-1,k-1)*2^(j-k)*binomial(n-j+k-1,2*k-1),j,k,n-k),k,1,n). - Vladimir Kruchinin, Aug 18 2010

A240844 Number of partitions of n into tribonacci numbers (cf. A000073).

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 6, 7, 10, 11, 14, 16, 20, 23, 28, 32, 38, 43, 50, 56, 65, 73, 83, 92, 105, 116, 131, 144, 163, 178, 199, 217, 242, 263, 291, 316, 348, 377, 413, 447, 488, 527, 573, 617, 670, 720, 779, 835, 903, 966, 1041, 1112, 1198, 1277, 1371, 1460, 1566

Offset: 0

Reinhard Zumkeller, Apr 13 2014

nonn

			a(6) = #{4+2, 4+1+1, 2+2+2, 2+2+1+1, 2+1+1+1+1, 6x1} = 6;
a(7) = #{7, 4+2+1, 4+1+1+1, 2+2+2+1, 2+2+1+1+1, 2+1+1+1+1+1, 7x1} = 7;
a(8) = #{7+1, 4+4, 4+2+2, 4+2+1+1, 4+1+1+1+1, 2+2+2+2, 2+2+2+1+1, 2+2+1+1+1+1, 2+6x1, 8x1} = 10;
a(9) = #{7+2, 7+1+1, 4+4+1, 4+2+2+1, 4+2+1+1+1, 4+5x1, 2+2+2+2+1, 2+2+2+1+1+1, 2+2+5x1, 2+7x1, 9x1} = 11;
a(10) = #{7+2+1, 7+1+1+1, 4+4+2, 4+4+1+1, 4+2+2+2, 4+2+2+1+1, 4+2+1+1+1+1, 4+6x1, 5x2, 2+2+2+2+1+1, 2+2+2+1+1+1+1, 2+2+6x1, 2+8x1, 10x1} = 14.

Cf. A117546.

a240844 = p $ drop 3 a000073_list where
   p _          0 = 1
   p ks'@(k:ks) m = if m < k then 0 else p ks' (m - k) + p ks m

cp's OEIS Frontend

A008937 a(n) = Sum_{k=0..n} T(k) where T(n) are the tribonacci numbers A000073.

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A046738 Period of Fibonacci 3-step sequence A000073 mod n.

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A232570 Numbers k that divide tribonacci(k) (A000073(k)).

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A085697 a(n) = T(n)^2, where T(n) = A000073(n) is the n-th tribonacci number.

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A107239 Sum of squares of tribonacci numbers (A000073).

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A073778 a(n) = Sum_{k=0..n} T(k)*T(n-k), where T is A000073; convolution of A000073 with itself.

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