## Tribonacci Numbers

### September 14, 2012

We use iterate from the Standard Prelude to calculate the tribonacci sequence:

(define (tribs n)   (iterate n + 1 1 2))

Thus, the first 25 elements of the tribonacci sequence are:

> (tribs 25) (1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 5768 10609  19513 35890 66012 121415 223317 410744 755476 1389537)

We use the matrix datatype from the Standard Prelude, matrix multiplication from the exercise on matrices and matrix powering from the exercise on fibonacci numbers to define this calculation of the nth tribonacci number:

(define (trib n)   (matrix-ref     (matrix-power       #( #(1 1 0)          #(1 0 1)          #(1 0 0))       n) 2 0))

Here’s an example (that’s all one big number):

> (trib 1000) 149995252232719672994127119633436824577569749158277812578756625414 806969052829656874238599632454281061578352939019541212503423640707 076075654939096072721522668597272334783989205780788704954034154039 4345570010550821354375819311674972209464069786275283520364029575324

Finally, we calculate the tribonacci constant as the ratio between two successive tribonacci numbers:

> (inexact->exact (/ (trib 1000) (trib 999))) 1.8392867552141612

You can run the program at http://programmingpraxis.codepad.org/mxq30iLw.

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### 8 Responses to “Tribonacci Numbers”

1. Paul said

A python version that uses the numpy library for matrices

import itertools as IT
import numpy

M = numpy.matrix([[1,1,0], [1, 0 ,1], [1, 0, 0]], dtype=float)

def tribonacci():
a, b, c = 0, 0, 1
while 1:
yield c
a, b, c = b, c, a + b + c

def tripow(N):
return int((M ** N)[0,0])

print list(IT.islice(tribonacci(), 10)) #-> [1, 1, 2, 4, 7, 13, 24, 44, 81, 149]
print tripow(9)                         # -> 149
x = list(IT.islice(tribonacci(), 1000))
print x[-1] / float(x[-2])              # -> 1.83928675521

2. Paul said

Another Python version with the power of a matrix explicitly in the code.

import itertools as IT

M = [[1,1,0], [1, 0 ,1], [1, 0, 0]]

def vecvec(a, b):
return sum(ai*bi for ai, bi in zip(a, b))

def matmul(a, b):
""" number of columns in a must be equal to number of rows in b"""
bt = zip(*b)
return [[vecvec(ai, btj) for btj in bt] for ai in a]

def mpow(a, nn):
""""matrix a in the power nn
Build the 1, 2, 4, 8 ... powers of a and find binary of nn
Finally multiply the required powers of a
"""
powers = [a]
binary = []
while nn:
binary.append(nn & 1)
nn /= 2
if nn:
powers.append(matmul(powers[-1], powers[-1]))
mats = (powi for powi, bi in zip(powers, binary) if bi)
return reduce(lambda a, b: matmul(a, b), mats)

def tribonacci():
a, b, c = 0, 0, 1
while 1:
yield c
a, b, c = b, c, a + b + c

def tripow(N):
return mpow(M, N)[0][0]

print list(IT.islice(tribonacci(), 20)) #-> [1, 1, 2, 4, 7, 13, 24, 44, 81, 149]
print tripow(9)                         # -> 149
x = list(IT.islice(tribonacci(), 1000))
print x[-1] / float(x[-2])              # -> 1.83928675521
print tripow(35) / float(tripow(34))    # -> 1.83928675521

3. Jan Van lent said

Calculation of the limit of the ratio as one of the roots the characteristic polynomial.
This is also the real eigenvalue of the powering matrix.

/* Maxima transcript */

(%i40) r : rhs(solve(z^3-z^2-z-1, z)[3]);
sqrt(11)   19 1/3            4             1
(%o40)           (-------- + --)    + -------------------- + -
3/2     27          sqrt(11)   19 1/3   3
3                 9 (-------- + --)
3/2     27
3
(%i41) string(optimize(r));
(%o41)  block([%1],%1:(sqrt(11)/sqrt(3)^3+19/27)^(1/3),%1+4/(9*%1)+1/3)
(%i42) float(r);
(%o42)                         1.839286755214161


$r = left({{\sqrt{11}}\over{3^{{{3}\over{2}}}}}+{{19}\over{27}}\right)^{{{1}\over{3}}}+{{4}\over{9\,\left({{\sqrt{11}}\over{3^{{{3}\over{2}}}}}+{{19}\over{27}}\right)^{{{1}\over{3}}}}}+{{1}\over{3}}$

4. Graham said

It’s been a while since I’ve used GNU Octave (Matlab clone), so I thought I’d give it a try. Not particularly clever, but it was good practice to brush up on working with matrices.

format long;

function t = tribo(n)
a = 0; b = 0; t = 1;
for i = 2:n
tmpa = a; tmpb = b; tmpt = t;
a = b; b = t; t = a + b + t;
endfor
return;
endfunction

function t = tripow(n)
T = [1, 1, 0; 1, 0, 1; 1, 0, 0];
t = (T^n)(1, 1);
return;
endfunction

function l = lim(n)
l = tripow(n) / tripow(n - 1);
return;
endfunction

5. treeowl said

As I mentioned in a recent comment on the fibonacci post, it’s actually impossible to achieve O(log n) performance for arbitrarily large n. The sequence is (to a close approximation) exponential, so the lengths of the binary or decimal values increase linearly. It is thus impossible for any algorithm to run in sublinear time.

6. Another Python solution:

def gen_tribonacci():
a, b, c = 0, 0, 1
while True:
yield c
a, b, c = b, c, a + b + c

def nth_tribonacci(n):
m = matpow([[1, 1, 0],
[1, 0, 1],
[1, 0, 0]], n)
return m[2][0]

def matpow(m, n):
if n == 1:
return m
if n % 2:
return matmult(m, matpow(matmult(m, m), n / 2))
return matpow(matmult(m, m), n / 2)

def matmult(m1, m2):
return [[dot(row, col) for col in zip(*m2)] for row in m1]

def dot(a, b):
return sum([x * y for x, y in zip(a, b)])

def tribonacci_const(n):
ts = take(n, gen_tribonacci())
return float(ts[-1]) / ts[-2]

def take(n, g):
result = []
for _ in xrange(n):
result.append(g.next())
return result

if __name__ == '__main__':
print(take(10, gen_tribonacci()))
print(nth_tribonacci(8))
print(tribonacci_const(1000))

7. #include
using namespace std;

int main()
{
unsigned long int Counter;
unsigned long int result = 0;
unsigned long int numa, numb, numc;
numa = 0;
numb = 0;
numc = 1;
unsigned long int number;
cout << "Qual o numero que desejas conhecer da sequencia tribonacci?" << endl <> number;
cout << endl << endl;
for(Counter = 0; Counter < number – 1; Counter++)
{
result = numa + numb + numc;
numa = numb;
numb = numc;
numc = result;
}
cout << "O Valor Tribonacci do numero " << number << " e igual a: " << result;
return 0;
}