0,3

The craft of coding, Weird recursive algorithms, zib(n).

Thomas Scheuerle, Plot of a(n) - (5/4)*n for n = 1..200.

Thomas Scheuerle, Plot of a(n) - (5/4)*n for n = 1..100000.

a(0)=1 and a(1)=1 and a(2)=2.

a(2*n+1) = a(n) + a(n-1) + 1, for n>=1.

a(2*n) = a(n) + a(n+1) + 1, for n>=2.

From Thomas Scheuerle, Dec 19 2023: (Start)

a(4^n+1) = (5*4^n - 8)/4, for n > 1.

a(2*4^n-1) = (5*4^n - 8)/2, for n > 0. (End)

a[0] = 1; a[1] = 1; a[2] = 2; a[n_Integer] := a[n] = Which[n == 0, 1, n == 1, 1, n == 2, 2, Mod[n, 2] == 1 && n >= 3, a[Quotient[(n-1), 2]] + a[Quotient[(n-1), 2]-1]+1, Mod[n, 2] == 0 && n >= 4, a[Quotient[n, 2]] + a[Quotient[n, 2]+ 1]+1]; result = Table[a[n], {n, 0, 66}] (* James C. McMahon, Dec 19 2023 *)

(Python)

from functools import cache

@cache

def a(n):

if n < 3: return [1, 1, 2][n]

x, y = (n-1)>>1, n>>1

if n & 1 == 1: return a(x) + a(x-1) + 1

else: return a(y) + a(y+1) + 1

print([a(n) for n in range(0, 67)])

(MATLAB)

function a = A368275( max_n )

a = [1, 1, 2];

for m = 4:max_n

n = m-1;

if mod(n, 2) == 0

a(m) = a(1 + (n/2)) + a(2 + (n/2)) + 1;

else

a(m) = a(1 + (n-1)/2) + a((n-1)/2) + 1;

end

end

end % Thomas Scheuerle, Dec 19 2023

Cf. A000045, A083420, A016813.

Sequence in context: A194357 A165783 A354046 * A289272 A073318 A254047

Adjacent sequences: A368272 A368273 A368274 * A368276 A368277 A368278

Darío Clavijo, Dec 19 2023

approved