A179243 Numbers that have three terms in their Zeckendorf representation.

12, 17, 19, 20, 25, 27, 28, 30, 31, 32, 38, 40, 41, 43, 44, 45, 48, 49, 50, 52, 59, 61, 62, 64, 65, 66, 69, 70, 71, 73, 77, 78, 79, 81, 84, 93, 95, 96, 98, 99, 100, 103, 104, 105, 107, 111, 112, 113, 115, 118, 124, 125, 126, 128, 131, 136, 148, 150, 151, 153, 154, 155

Offset: 1

Author

Emeric Deutsch, Jul 05 2010

Keywords

• nonn
• look

Examples

			12 = 1+3+8;
17 = 1+3+13;
19 = 1+5+13;
20 = 2+5+13;
25 = 21+3+1;
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A035517, A007895, A179242, A179244, A179245, A179246, A179247, A179248, A179249, A179250, A179251, A179252, A179253.

Programs

Haskell

a179243 n = a179243_list !! (n-1)
a179243_list = filter ((== 3) . a007895) [1..]
-- Reinhard Zumkeller, Mar 10 2013

Maple

with(combinat): B := proc (n) local A, ct, m, j: A := proc (n) local i: for i while fibonacci(i) <= n do n-fibonacci(i) end do end proc: ct := 0: m := n: for j while 0 < A(m) do ct := ct+1: m := A(m) end do: ct+1 end proc: Q := {}: for i from fibonacci(7)-1 to 160 do if B(i) = 3 then Q := `union`(Q, {i}) else end if end do: Q

Mathematica

zeck = DigitCount[Select[Range[2000], BitAnd[#, 2*#] == 0 &], 2, 1];
Position[zeck, 3] // Flatten (* Jean-François Alcover, Jan 30 2018 *)

Python

from math import comb, isqrt
from sympy import fibonacci, integer_nthroot
def A179243(n): return fibonacci(2+(r:=n-1-comb((m:=integer_nthroot(6*n, 3)[0])+(t:=(n>comb(m+2, 3)))+1, 3))-comb((k:=isqrt(b:=r+1<<1))+(b>k*(k+1)), 2))+fibonacci((a:=isqrt(s:=n-comb(m-(t^1)+2, 3)<<1))+((s<<2)>(a<<2)*(a+1)+1)+3)+fibonacci(m+t+5) # Chai Wah Wu, Apr 09 2025

Formula

A007895(a(n)) = 3. - Reinhard Zumkeller, Mar 10 2013