A179245 Numbers that have 5 terms in their Zeckendorf representation.

88, 122, 135, 140, 142, 143, 177, 190, 195, 197, 198, 211, 216, 218, 219, 224, 226, 227, 229, 230, 231, 266, 279, 284, 286, 287, 300, 305, 307, 308, 313, 315, 316, 318, 319, 320, 334, 339, 341, 342, 347, 349, 350, 352, 353, 354, 360, 362, 363, 365, 366, 367

Offset: 1

Author

Emeric Deutsch, Jul 05 2010

Keywords

• nonn
• easy

Comments

A007895(a(n)) = 5. - Reinhard Zumkeller, Mar 10 2013

Numbers that are the sum of five non-consecutive Fibonacci numbers. Their Zeckendorf representation thus consists of five 1's with at least one 0 between each pair of 1's; for example, 122 is represented as 1001010101. - Alonso del Arte, Nov 17 2013

Examples

			88  = 55 + 21 + 8 + 3 + 1.
122 = 89 + 21 + 8 + 3 + 1.
135 = 89 + 34 + 8 + 3 + 1.
140 = 89 + 34 + 13 + 3 + 1.
142 = 89 + 34 + 13 + 5 + 1.
81 is not in the sequence because, although it is the sum of five Fibonacci numbers (81 = 5 + 8 + 13 + 21 + 34), its Zeckendorf representation only has three terms: 81 = 55 + 21 + 5.
		

Links

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

Crossrefs

Cf. A035517, A007895. Numbers that have m terms in their Zeckendorf representations: A179242 (m = 2), A179243 (m = 3), A179244 (m = 4), A179246 (m = 6), A179247 (m = 7), A179248 (m = 8), A179249 (m = 9), A179250 (m = 10), A179251 (m = 11), A179252 (m = 12), A179253 (m = 13).

Programs

Haskell

a179245 n = a179245_list !! (n-1)
a179245_list = filter ((== 5) . 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(11)-1 to 400 do if B(i) = 5 then Q := `union`(Q, {i}) else end if end do: Q;

Mathematica

zeck = DigitCount[Select[Range[3000], BitAnd[#, 2*#] == 0 &], 2, 1];
Position[zeck, 5] // Flatten (* Jean-François Alcover, Jan 30 2018 *)

PARI

A048680(n)=my(k=1,s);while(n,if(n%2,s+=fibonacci(k++)); k++; n>>=1); s
[A048680(n)|n<-[1..100],hammingweight(n)==5] \\ Charles R Greathouse IV, Nov 17 2013

Formula

a(n) = A048680(A014313(n)). - Charles R Greathouse IV, Nov 17 2013