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 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 (list; graph; refs; listen; history; text; internal format)
 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: A096620 A093419 A160049 * A299209 A007479 A076535 Adjacent sequences:  A331121 A331122 A331123 * A331125 A331126 A331127 KEYWORD nonn,easy,look AUTHOR Thomas König, Jan 10 2020 STATUS approved

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Last modified June 3 07:14 EDT 2020. Contains 334799 sequences. (Running on oeis4.)