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

1,1

Links

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

Formula

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

Example

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

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 *)

Prog

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

Crossrefs

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

Sequence in context: A087657 A248478 A059390 * A350686 A064825 A162918

Adjacent sequences: A179240 A179241 A179242 * A179244 A179245 A179246

Keyword

nonn,look

Author

Emeric Deutsch, Jul 05 2010

Status

approved