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a(n) = n minus (number of 1's in Zeckendorf expansion of n).
14

%I #28 Mar 19 2021 10:49:45

%S 0,0,1,2,2,4,4,5,7,7,8,9,9,12,12,13,14,14,16,16,17,20,20,21,22,22,24,

%T 24,25,27,27,28,29,29,33,33,34,35,35,37,37,38,40,40,41,42,42,45,45,46,

%U 47,47,49,49,50,54,54,55,56,56,58,58,59,61,61,62,63,63,66

%N a(n) = n minus (number of 1's in Zeckendorf expansion of n).

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

%H Antti Karttunen, <a href="/A219641/b219641.txt">Table of n, a(n) for n = 0..10000</a>

%H Paul Baird-Smith, Alyssa Epstein, Kristen Flint, and Steven J. Miller, <a href="https://arxiv.org/abs/1809.04881">The Zeckendorf Game</a>, arXiv:1809.04881 [math.NT], 2018.

%F a(n) = n - A007895(n).

%t zeck = DigitCount[Select[Range[0, 500], BitAnd[#, 2*#] == 0&], 2, 1];

%t Range[0, Length[zeck]-1] - zeck (* _Jean-François Alcover_, Jan 25 2018 *)

%o (Scheme): (define (A219641 n) (- n (A007895 n)))

%o (Python)

%o from sympy import fibonacci

%o def a(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 print([n - a(n) for n in range(101)]) # _Indranil Ghosh_, Jun 09 2017

%Y Cf. A007895, A014417. A022342 gives the positions of records, resulting the same sequence with duplicates removed: A219640. A035336 gives the positions of values that occur only once: A219639. Cf. also A219637, A219642. Analogous sequence for binary system: A011371, for factorial number system: A219651.

%K nonn

%O 0,4

%A _Antti Karttunen_, Nov 24 2012