In honour of Fibonacci

---

NOTE: This transcription was contributed by Martin P.M. van der Burgt, who has devised a process for producing transcripts automatically. Although its markup is incomplete, we believe it serves a useful purpose by virtue of its searchability and its accessibility to text-reading software. It will be replaced by a fully marked-up version when time permits. —HR

---

In honour of Fibonacci.

Studying an artificial intelligence approach to programming the other day —I read the most weird documents!— I was reminded of the Fibonacci sequence. given by

	Fi = 0 , F2 = 1 ,
	Fn = Fn-1 + Fn-2 (-inf < n < +inf)

For N ≥ 2 the relation

R:

x = FN

is trivially established by the program

	y, x, i := 0, 1, 2 {Y = Fi-1 and x = Fi and 2 ≤ i ≤ N};	(1)
	do i ≠ N → y, x, i := x, x + y, i + l .od {R}

a program with a time-complexity proportional to N; I remembered —although I did not know the formulae— that R can also be established in a number of operations proportional to log(N) and wondered —as a matter of fact: I still wonder— how proponents of “program transformations” propose to transform the linear algorithm (1) into the logarithmic one.

Yesterday evening I was wondering whether I could reconstruct the logarithmic scheme for the Fibonacci sequence, and whether similar schemes existed for higher order recurrence relations (for a k ≥ 2 ):

	F1 = F2 = ... = Fk-1 = 0 , Fk = 1
	Fn = Fn-1 + ... + Fn-k (-inf < n < +inf)	(2)

Eventually I found a way of deriving these schemes. For k = 2 , the normal Fibonacci numbers, the method leads to the well-known formulae

	F2j = fj2 + Fj+12
	F2j+1 = (2Fj + Fj+1) * Fj+1 or F2j-1 = (2fj+1 - Fj) * Fj .

This note is written, because I liked my general derivation. I shall describe it for k = 3 .
Because for k = 3 we have F₁ = F₂ = 0 and F₃ = l we may write

Fn = F3 * Fn + (F2 + F1)* Fn-1 + F2 * Fn-2

(3)

From (3) we deduce the truth of

Fn = Fi+3* Fn-i + (Fi+2 + Fi+1)* Fn-i-1+ Fi+2* Fn-i-2

(4)

for i = 0 . The truth of (4) for all positive values of i is derived by mathematical induction; the induction step consists of
1)     substituting F_n-i-1 + F_n-i-2 + F_n-i-3 for F_n-i
2)     combining after rearrangement F_i+3 + F_i+2 + F_i+1 for F_i+4 .
(The proof for negative values of i is done by performing the induction step the other way round.) Substituting in (4) n = 2j and i = j-l we get

F2j = Fj2 * Fj+1 + (Fj+1 + Fj)* Fj + Fj+1* Fj-1

and, by substituting F_j+2 - F_j+1 - F_j for F_j-1 , and subsequent rearranging

F2j = Fj2 + (2Fj+2 - Fj+1)* Fj+1

(5)

Substituting in (4) n = 2j+l and i = j-l we get

	F2j+l = Fj+2 + (Fj+1 + Fj)* Fj+1 + Fj+1 * Fj
	= (2Fj + Fj+1) * Fj+1 + Fj+22	(6)

Formulae (5) and (6) were the ones I was after.

Note. For k = 4 the analogue to (4) is F_n = F_i+4 * F_n-i + (F_i+3 + F_i+2 + F_i+1)* F_n-i-1 + (F_i+3 + F_i+2)* F_n-i-2 + F_i+3 * F_n-i-3 (End of note.)

.

Plataanstraat 5	prof.dr.Edsger W.Dijkstra
5671 AL Nuenen	Burroughs Research Fellow

The Netherlands

---

Transcribed by Martin P.M. van der Burgt
Last revision 2015-01-29 .