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Numbers that have 7 terms in their Zeckendorf representation.
11

%I #15 Feb 13 2021 01:09:26

%S 609,842,931,965,978,983,985,986,1219,1308,1342,1355,1360,1362,1363,

%T 1452,1486,1499,1504,1506,1507,1541,1554,1559,1561,1562,1575,1580,

%U 1582,1583,1588,1590,1591,1593,1594,1595,1829,1918,1952,1965,1970,1972,1973

%N Numbers that have 7 terms in their Zeckendorf representation.

%C A007895(a(n)) = 7. - _Reinhard Zumkeller_, Mar 10 2013

%H Reinhard Zumkeller, <a href="/A179247/b179247.txt">Table of n, a(n) for n = 1..10000</a>

%e 609 = 377+144+55+21+8+3+1;

%e 842 = 610+144+55+21+8+3+1;

%e 931 = 610+233+55+21+8+3+1;

%e 965 = 610+233+89+21+8+3+1;

%e 978 = 610+233+89+34+8+3+1.

%p 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(15)-1 to 2100 do if B(i) = 7 then Q := `union`(Q, {i}) else end if end do: Q;

%t zeck = DigitCount[Select[Range[4*10^4], BitAnd[#, 2*#] == 0 &], 2, 1];

%t Position[zeck, 7] // Flatten (* _Jean-François Alcover_, Jan 30 2018 *)

%o (Haskell)

%o a179247 n = a179247_list !! (n-1)

%o a179247_list = filter ((== 7) . a007895) [1..]

%o -- _Reinhard Zumkeller_, Mar 10 2013

%Y Cf. A035517, A007895, A179242, A179243, A179244, A179245, A179246, A179248, A179249, A179250, A179251, A179252, A179253.

%K nonn

%O 1,1

%A _Emeric Deutsch_, Jul 05 2010