

A078642


Numbers with two representations as the sum of two Fibonacci numbers.


5



4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194, 5168, 8362, 13530, 21892, 35422, 57314, 92736, 150050, 242786, 392836, 635622, 1028458, 1664080, 2692538, 4356618, 7049156, 11405774, 18454930, 29860704, 48315634, 78176338, 126491972
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OFFSET

1,1


COMMENTS

A positive integer n has exactly two representations as the sum of two Fibonacci numbers if and only if n is twice a Fibonacci number and n >= 4. Conjectured by John W. Layman, Dec 20 2002. Proved by Max Alekseyev, Mar 02 2007.
From Max Alekseyev, Mar 02 2007: (Start)
Suppose that the number m has exactly two representations as the sum of two Fibonacci numbers. There are three types of representations possible:
(I) the sum of two equal Fibonacci numbers
(II) the sum of two consecutive Fibonacci numbers
(III) the sum of two distinct nonconsecutive Fibonacci numbers
Lemma. The two representations of m > 2 must be of different types.
Proof. Two representations of m > 2 both of type (I) are not possible as 2*F(n) is a strictly increasing function for n >= 2. Similarly, two representations of m both of type (II) are not possible as F(n) + F(n+1) is a strictly increasing function for n >= 0. Finally, two representations of m both of type (III) are not possible as that would violate the property of the Fibonacci numeral system (the uniqueness of representation of all nonnegative integers).
Consider all possible pairs of representation types:
(I) and (II) are possible only for m = 2: 2 = 2*F(1) = 2*F(2) = F(1) + F(2) but m = 2 has more than two different representations.
(II) and (III) are not possible together as that would again violate the property of the Fibonacci numeral system.
Finally, (I) and (III) gives rise to the sequence of a(n) = 2 * F(n) = F(n+1) + F(n1). QED (End)


LINKS

Colin Barker, Table of n, a(n) for n = 1..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1)


FORMULA

a(n) = 2F(n + 2), where F(n) is the nth Fibonacci number.
a(n) = a(n  1) + a(n  2), n > 2 ; a(1) = 4, a(2) = 6 . G.f.: 2x*(2+x)/(1xx^2).  Philippe Deléham, Nov 19 2008
a(n) = 2F(n + 2) = F(n) + F(n + 3), where F(1) = F(2) = 1.  Alonso del Arte, Jul 07 2013
a(n) = (2^(n)*((1r)^n*(3+r) + (1+r)^n*(3+r))) / r where r=sqrt(5).  Colin Barker, Jan 29 2017


EXAMPLE

16 has exactly two representations as the sum of Fibonacci numbers: 16 = 3 + 13 and 16 = 8 + 8. Hence 16 belongs to the sequence.


MATHEMATICA

t = Split@ Sort@ Flatten@ Table[Fibonacci[i] + Fibonacci[j], {i, 2, 39}, {j, 2, i}]; Take[ t[[ # ]][[1]] & /@ Select[ Range@Length@t, Length[ t[[ # ]]] > 1 &], 36] (* Robert G. Wilson v *)


PROG

(PARI) a(n)=([0, 1; 1, 1]^(n1)*[4; 6])[1, 1] \\ Charles R Greathouse IV, Oct 07 2015
(PARI) Vec(2*x*(2 + x) / (1  x  x^2) + O(x^60)) \\ Colin Barker, Jan 29 2017


CROSSREFS

Essentially the same as A006355 = number of binary vectors of length n containing no singletons; and as A055389: a(0)=1, then twice the Fibonacci sequence.
Sequence in context: A027689 A023501 A050887 * A032416 A283496 A117149
Adjacent sequences: A078639 A078640 A078641 * A078643 A078644 A078645


KEYWORD

nonn,easy


AUTHOR

Joseph L. Pe, Dec 12 2002


STATUS

approved



