

A331124


Number of function evaluations in a recursive calculation of Fibonacci(n).


2



1, 1, 1, 3, 6, 5, 11, 10, 15, 12, 23, 17, 27, 22, 37, 26, 38, 28, 51, 36, 53, 41, 68, 45, 67, 50, 87, 60, 86, 64, 102, 65, 93, 67, 118, 80, 116, 88, 141, 90, 131, 95, 163, 110, 155, 114, 181, 113, 163, 118, 205, 138, 198, 148, 234, 147, 211, 151, 253, 167
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OFFSET

0,4


COMMENTS

One way to calculate the Fibonacci numbers recursively is to use:
 F(0) = 0,
 F(1) = 1,
 F(2) = 1,
 F(n) = F((n + 1)/2)^2 + F((n  1)/2)^2 for odd n,
 F(n) = F(n/2) * (F(n/2  1) + F(n/2 + 1)) for even n.
Proof: it is known that F(i) * F(j) + F(i + 1) * F(j + 1) = F(i + j + 1) (see formula section of A000045):
 for even n, let i = n/2 and j = n/2  1,
 for odd n, let i = j = (n + 1)/2.
This sequence gives the number of evaluations of F for calculating F(n). It is assumed that F needs to be evaluated more than once even if it has been evaluated before (no caching).
Conjecture: for large n, this sequence is bounded by a small constant times n^(4/3).


LINKS

Rémy Sigrist, Table of n, a(n) for n = 0..10000
GeeksforGeeks, Program for Fibonacci numbers, see method 6, but erroneously states to take O(log n) operations.
Rémy Sigrist, PARI program for A331124


FORMULA

a(0) = 1,
a(1) = 1,
a(2) = 1,
a(n) = a((n + 1)/2) + a((n  1)/2) + 1, n odd and n > 2,
a(n) = a(n/2) + a(n/2  1) + a(n/2 + 1) + 1, n even.


EXAMPLE

a(5) = a(3) + a(2) + 1 = (a(2) + a(1) + 1) + a(2) + 1 = 5.


PROG

(Fortran)
program main
implicit none
integer, parameter :: pmax = 100
integer :: r, n
logical, dimension(0:pmax) :: cache
do n = 0, pmax
cache (0:2) = .true.
cache (3:pmax) = .false.
r = a(n)
write (*, fmt="(I0, ', ')", advance="no") r
end do
write (*, fmt="()")
contains
recursive function a (n) result(r)
integer, intent(in) :: n
integer :: r
if (cache(n)) then
r = 1
return
else if (mod(n, 2) == 1) then
r = a ((n + 1)/2) + a ((n  1)/2) + 1
else
r = a (n/2) + a(n/2  1) + a(n/2 + 1) + 1
end if
cache (n) = .false.
end function a
end program main
(PARI) \\ See Links section.


CROSSREFS

Cf. A000045; see A331164 for the number of function evaluations with caching.
Sequence in context: A093419 A160049 A096620 * A299209 A007479 A076535
Adjacent sequences: A331121 A331122 A331123 * A331125 A331126 A331127


KEYWORD

nonn,easy,look


AUTHOR

Thomas König, Jan 10 2020


STATUS

approved



