The Fibonacci sequence is given as

where the $m$-th term ${F}_{m}={F}_{m-1}+{F}_{m-2}$, for $m\ge 3$, is the sum of the previous two terms with ${F}_{1}=0$ and ${F}_{2}=1$. Note that ${F}_{3k+1}$ is even, for $k\in \mathbb{N}\cup \left\{0\right\}$. The other numbers are always odd. This can be easily observed by dividing each Fibonacci number ${F}_{m}$ by $2$ and writing down its remainder. Thus, we get

The remainder zero corresponds to even terms and remainder one corresponds to odd terms. Note that the pattern repeats itself as $0,1,1$, a set of three elements in $\mathbb{N}$ modulo $2$. It so happens that this property is true if we divide by any natural number $k$ instead of $2$. In that case we get a repeated pattern in $\mathbb{N}$ modulo $k$. The periodicity with which the pattern repeates is called Pisano period. The pisano period was ﬁrst introduced by Lagrange while addressing the question of in what frequency does the last digits of Fibonacci numbers repeat itself. Answering this question is equivalent to dividing each ${F}_{m}$ by $10$ and considering the remainders. It turns out that in $\mathbb{N}$ modulo $10$ the periodicity is $60$.

### 1 Modulo $n$

The set of whole numbers

are inﬁnite but they are smallest of the inﬁnity. They are listable or what mathematicians call denumerable or countable, in contrast to, all real numbers which are inﬁnite but not denumerable. Consider a $n\in \mathbb{N}$ and identify all its integral multiple, i.e, $0,n,2n,3n,\dots $, as one element denoted as $\left[0\right]$. Similarly, we identify all elements which leave the same remainder $k$ when divided by $n$, i.e., $k,n+k,2n+k,3n+k,\dots $, as one element denoted as $\left[k\right]$. Note that $k$ takes value in $\left\{0,1,\dots ,n-1\right\}$. The $\mathbb{N}$ modulo $n$ set is deﬁned as $\mathbb{N}\left[n\right]:=\left\{\left[0\right],\left[1\right],\dots ,\left[n-1\right]\right\}$. Thus, we have divided the inﬁnite set of whole numbers in to a set of exactly $n$ elements, a ﬁnite set. Mathematically, the integer modulo $n$ concept can be made precise using an equivalence relation. Given $n\in \mathbb{N}$, we deﬁne the equivalence relation $a\sim b$, called as $a$ equivalent to $b$, if $a-b$ is an integral (positive or negative) multiple of $n$. This equivalence relation will divide $\mathbb{N}\cup \left\{0\right\}$ into $n$ disjoint equivalence classes, where each equivalence class contains numbers which have the same remainder. In this sense, the Fibonacci sequence elements in $\mathbb{N}\left[2\right]$ are

Henceforth, we shall drop $\left[\cdot \right]$ symbol in the sequence, for simplicity, and write it as

Note that the sequence is periodic (or repeating) of period three (after every third term). This period of the resulting periodic sequence is called Pisano period. The Pisano period of the Fibonacci sequence in $\mathbb{N}\left[n\right]$ is given in the table below for the ﬁrst 30 numbers.

$\mathbb{N}\left[1\right]$ | $1$ |

$\mathbb{N}\left[2\right]$ | $3$ |

$\mathbb{N}\left[3\right]$ | $8$ |

$\mathbb{N}\left[4\right]$ | $6$ |

$\mathbb{N}\left[5\right]$ | $20$ |

$\mathbb{N}\left[6\right]$ | $24$ |

$\mathbb{N}\left[7\right]$ | $16$ |

$\mathbb{N}\left[8\right]$ | $12$ |

$\mathbb{N}\left[9\right]$ | $24$ |

$\mathbb{N}\left[10\right]$ | $60$ |

$\mathbb{N}\left[11\right]$ | $10$ |

$\mathbb{N}\left[12\right]$ | $24$ |

$\mathbb{N}\left[13\right]$ | $28$ |

$\mathbb{N}\left[14\right]$ | $48$ |

$\mathbb{N}\left[15\right]$ | $40$ |

$\mathbb{N}\left[16\right]$ | $24$ |

$\mathbb{N}\left[17\right]$ | $36$ |

$\mathbb{N}\left[18\right]$ | $24$ |

$\mathbb{N}\left[19\right]$ | $18$ |

$\mathbb{N}\left[20\right]$ | $60$ |

$\mathbb{N}\left[21\right]$ | $16$ |

$\mathbb{N}\left[22\right]$ | $30$ |

$\mathbb{N}\left[23\right]$ | $48$ |

$\mathbb{N}\left[24\right]$ | $24$ |

$\mathbb{N}\left[25\right]$ | $100$ |

$\mathbb{N}\left[26\right]$ | $84$ |

$\mathbb{N}\left[27\right]$ | $72$ |

$\mathbb{N}\left[28\right]$ | $48$ |

$\mathbb{N}\left[29\right]$ | $14$ |

$\mathbb{N}\left[30\right]$ | $120$ |

It is known that the resulting sequence of modulo $n$ has Pisano period at most ${n}^{2}-1$, for $n\ge 2$. However, it is still an open problem to ﬁnd a general formula for Pisano period in terms of $n$.

### 2 Fibonacci Colours

Note: This section and the next are my own creation/imagination. As on this day, to my knowledge, there is no such thing as Fibonacci Colour or Raaga. This is created only as an experiment, purely motivated by pleasure and leisure.

Let us associate Red, Green, Blue (RGB) colours corresponding to the Fibonacci sequence. Since we have chosen three colours, the possible rearrangement of these colours is $3!=6$. Thus, we can obtain six Fibonacci colours. For $3$ colours the Pisano period is $8$ and, hence, we get the following proportions for the colours:

0 | 1 | 1 | 2 | 0 | 2 | 2 | 1 |

I | II | II | III | I | III | III | II |

Thus, the three colours have to be mixed in the following proportion: I colour should be $\frac{2}{8}=25\%$, II and III colour should be each of $\frac{3}{8}=37.5\%$. Though we said there are six possible Fibonacci colours for three choice, the fact that two of them have to be mixed in same ratio reduces the possibility to $3=\frac{3!}{2!}$. This is because the number of choices for the $25\%$ proportion is three.

For instance, if we choose RGB we need to mix in the percentage $37.5\%$ each of Blue and Green, and $25\%$ of Red. This combination will give us the colour resembling dark cyan whose colour code is #406060. Note that this combination is same as RBG. If we choose BGR we need to mix in the percentage $37.5\%$ each of Red and Green, and $25\%$ of Blue. This combination will give us the colour resembling dark yellow whose colour code is #606040. Note that this combination is same as BRG. The only other possibility is GBR which is same as GRB. This combination will give us the colour whose colour code is #604840.

Suppose we choose four colours, viz., Cyan, Magenta, Yellow, Black (CMYK) then the possible rearrangement is $4!=24$, thus, giving us $24$ Fibonacci colours. For four colours the Pisano period is six and, hence, we get the following proportions for the colours:

0 | 1 | 1 | 2 | 3 | 1 |

I | II | II | III | IV | II |

Thus, the four colours have to be mixed in the following proportion: First, third and fourth colour in $\frac{1}{6}=16.7\%$ and second should be in $\frac{3}{6}=50\%$. Though we said there are $24$ possible Fibonacci colours for four choice, the fact that three of them have to be mixed in same ratio reduces the possibility to $4=\frac{4!}{3!}$. This is because the number of choices for the $50\%$ proportion is four.

For instance, if we choose the order CMYK we need to mix in the percentage $16.7\%$ each of CYK, and $50\%$ of Magenta. This combination will give us the colour resembling strong magenta whose colour code is #AA55AA. The other three colours are MYK with Cyan, CMK with Yellow and CMY with Black.

If we choose the seven rainbow colours VIBGYOR then the possible rearrangement is $7!=5040$, thus, giving us $5040$ Fibonacci colours. For seven colours the Pisano period is $16$ and, hence, we get the following proportions for the colours:

0 | 1 | 1 | 2 | 3 | 5 | 1 | 6 | 0 | 6 | 6 | 5 | 4 | 2 | 6 | 1 |

I | II | II | III | IV | VI | II | VII | I | VII | VII | VI | V | III | VII | II |

In this case, one has to mix the colours in the following proportions: First, third, sixth colour in $12.5\%$, second and seventh colour in $25\%$, fourth and ﬁfth colour in $6.25\%$. Though we said there are $5040$ possible Fibonacci colours for seven choice, the fact that some of them have to be mixed in same ratio reduces the possibility to $210=\frac{7!}{3!\cdot 2!\cdot 2!}$. For instance, if we mix VIBGYOR we get a colour resembling strong pink or strong raspberry whose colour code turns out to be #94305B. If we mix ROYGBIV we get the colour code #AE5055.

### 3 Fibonacci Raaga

The basic notes or swaram of carnatic music are seven, viz., Sa, Ri, Ga, Ma, Pa, Dha, Ni. Of these Sa and Pa are called Prakriti or nature swaras and the remaining ﬁve swaras are called vikriti or artiﬁcial swaras. It is believed that the prakriti swaras are given directly by God through Sama veda and the vikriti swaras are man-made. These seven notes are present in the nature in the sounds of peacock (Sa), bull (Ri), goat (Ga), dove/crane (Ma), cuckoo (Pa), horse (Dha) and elephant (Ni), respectively. In western music this corresponds to the following notes: C, D, E, F, G, A, B.

An octave or sthayi is the collection of swaram with diﬀerent tones (or volume) such that between any pair of similar swara with diﬀerent tone, the higher swaram is twice more than the lower swaram. The tones of Sa and Pa remains constant, Ma takes two tones while the other four swaras take three tones. This makes a total of $16$ notes:

Sa, Ri1, Ri2, Ri3, G1, G2, G3, M1, M2, Pa, D1, D2, D3, N1, N2, N3.

I do not know if it is a strange coincidence that the 7 swaras and 16 notes matches with the pisano period 16 for modulo 7. Of the sixteen notes there are only 12 scales or volumes:

Sa, Ri1, Ri2 = G1, Ri3 = G2, G3, M1, M2, Pa, D1, D2 = N1, D3 = N2, N3.

The raagas are a combination of swaras. The raagas are classiﬁed into two categories: the parent/Janaka/Melakarta raagas and the Janya raagas. There are $72$ Janaka raagas and Janya raagas are derived from the parent raagas. The Janaka raagas must have all the $8$ swaras either in an ascent (aarohanam) or in descent (avarohanam). The $72$ raagas are actually $36$ pairs where one contains $M1$ and the second contains $M2$. One of each pair has $6$ chakras (cycles) and each chakra has $6$ raagas. The $6$ raagas in each chakra diﬀer only in one of the vikriti swara.

Consider the Fibonacci sequence in $\mathbb{N}\left[7\right]$

Note that its Pisano period is $16$. Let us associate a raaga corresponding to the Fibonacci sequence, as follows:

0 | 1 | 1 | 2 | 3 | 5 | 1 | 6 | 0 | 6 | 6 | 5 | 4 | 2 | 6 | 1 |

Sa | Ri1 | Ri1 | Ga1 | Ma1 | Da1 | Ri1 | Ni1 | Sa | Ni1 | Ni1 | Da1 | Pa | Ga1 | Ni1 | Ri1 |

C | D | D | E | F | A | D | B | C | B | B | A | G | E | B | D |