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
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
Programs
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Haskell
a179245 n = a179245_list !! (n-1) a179245_list = filter ((== 5) . a007895) [1..] -- Reinhard Zumkeller, Mar 10 2013
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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;
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Mathematica
zeck = DigitCount[Select[Range[3000], BitAnd[#, 2*#] == 0 &], 2, 1]; Position[zeck, 5] // Flatten (* Jean-François Alcover, Jan 30 2018 *)
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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
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