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A179243 Numbers that have three terms in their Zeckendorf representation.

%I A179243 #22 Apr 09 2025 14:12:12
%S A179243 12,17,19,20,25,27,28,30,31,32,38,40,41,43,44,45,48,49,50,52,59,61,62,
%T A179243 64,65,66,69,70,71,73,77,78,79,81,84,93,95,96,98,99,100,103,104,105,
%U A179243 107,111,112,113,115,118,124,125,126,128,131,136,148,150,151,153,154,155
%N A179243 Numbers that have three terms in their Zeckendorf representation.
%H A179243 Reinhard Zumkeller, <a href="/A179243/b179243.txt">Table of n, a(n) for n = 1..10000</a>
%F A179243 A007895(a(n)) = 3. - _Reinhard Zumkeller_, Mar 10 2013
%e A179243 12 = 1+3+8;
%e A179243 17 = 1+3+13;
%e A179243 19 = 1+5+13;
%e A179243 20 = 2+5+13;
%e A179243 25 = 21+3+1;
%p A179243 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
%t A179243 zeck = DigitCount[Select[Range[2000], BitAnd[#, 2*#] == 0 &], 2, 1];
%t A179243 Position[zeck, 3] // Flatten (* _Jean-François Alcover_, Jan 30 2018 *)
%o A179243 (Haskell)
%o A179243 a179243 n = a179243_list !! (n-1)
%o A179243 a179243_list = filter ((== 3) . a007895) [1..]
%o A179243 -- _Reinhard Zumkeller_, Mar 10 2013
%o A179243 (Python)
%o A179243 from math import comb, isqrt
%o A179243 from sympy import fibonacci, integer_nthroot
%o A179243 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
%Y A179243 Cf. A035517, A007895, A179242, A179244, A179245, A179246, A179247, A179248, A179249, A179250, A179251, A179252, A179253.
%K A179243 nonn,look
%O A179243 1,1
%A A179243 _Emeric Deutsch_, Jul 05 2010