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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).
12

%I #22 Mar 09 2021 19:13:58

%S 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,

%T 14,14,14,15,15,16,16,17,17,17,18,18,18,19,19,19,20,20,21,21,21,22,22,

%U 22,22,23,24,24,25,25,25,26,26,26,27,27,27,28,28,29,29

%N 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).

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

%H A. Karttunen, <a href="/A219642/b219642.txt">Table of n, a(n) for n = 0..10946</a>

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

%o (Scheme with memoization macro definec from _Antti Karttunen_'s Intseq-library):

%o (definec (A219642 n) (if (zero? n) n (+ 1 (A219642 (A219641 n)))))

%o (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)

%o a(n)=my(s); while(n, n-=A007895(n); s++); s \\ _Charles R Greathouse IV_, Sep 02 2015

%o (Python)

%o from sympy import fibonacci

%o def a007895(n):

%o k=0

%o x=0

%o while n>0:

%o k=0

%o while fibonacci(k)<=n: k+=1

%o x+=10**(k - 3)

%o n-=fibonacci(k - 1)

%o return str(x).count("1")

%o def a219641(n): return n - a007895(n)

%o l=[0]

%o for n in range(1, 101):

%o l.append(1 + l[a219641(n)])

%o print(l) # _Indranil Ghosh_, Jun 09 2017

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

%K nonn

%O 0,3

%A _Antti Karttunen_, Nov 24 2012