A219642 Number of steps to reach 0 starting with n and using the iterated process: x -> x - (number of 1's in Zeckendorf expansion of x).

0, 1, 2, 3, 3, 4, 4, 5, 6, 6, 7, 7, 7, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 16, 16, 17, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 21, 21, 21, 22, 22, 22, 22, 23, 24, 24, 25, 25, 25, 26, 26, 26, 27, 27, 27, 28, 28, 29, 29

Offset: 0

Author

Antti Karttunen, Nov 24 2012

Keywords

• nonn

Comments

See A014417 for the Fibonacci number system representation, also known as Zeckendorf expansion.

Links

• A. Karttunen, Table of n, a(n) for n = 0..10946

Crossrefs

Cf. A007895, A014417, A219640, A219641, A219643-A219645, A219648. Analogous sequence for binary system: A071542, for factorial number system: A219652.

Programs

PARI

A007895(n)=if(n<4, n>0, my(k=2,s,t); while(fibonacci(k++)<=n,); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s)
a(n)=my(s); while(n, n-=A007895(n); s++); s \\ Charles R Greathouse IV, Sep 02 2015

Python

from sympy import fibonacci
def a007895(n):
    k=0
    x=0
    while n>0:
        k=0
        while fibonacci(k)<=n: k+=1
        x+=10**(k - 3)
        n-=fibonacci(k - 1)
    return str(x).count("1")
def a219641(n): return n - a007895(n)
l=[0]
for n in range(1, 101):
    l.append(1 + l[a219641(n)])
print(l) # Indranil Ghosh, Jun 09 2017

Formula

a(0)=0; for n>0, a(n) = 1+a(A219641(n)).