A122194 Numbers that are the sum of exactly two sets of Fibonacci numbers.

3, 5, 6, 9, 10, 15, 17, 25, 28, 41, 46, 67, 75, 109, 122, 177, 198, 287, 321, 465, 520, 753, 842, 1219, 1363, 1973, 2206, 3193, 3570, 5167, 5777, 8361, 9348, 13529, 15126, 21891, 24475, 35421, 39602, 57313, 64078, 92735, 103681, 150049, 167760, 242785

Offset

1,1

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

J. Berstel, An Exercise on Fibonacci Representations, RAIRO/Informatique Theorique, Vol. 35, No 6, 2001, pp. 491-498, in the issue dedicated to Aldo De Luca on the occasion of his 60th anniversary.

M. Bicknell-Johnson & D. C. Fielder, The number of Representations of N Using Distinct Fibonacci Numbers, Counted by Recursive Formulas, Fibonacci Quart. 37.1 (1999) pp. 47 ff.

Ron Knott Sumthing about Fibonacci Numbers

Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,-1).

Formula

a(2n-1) = A000032(n+2) - 1,

a(2n) = 2*A000045(n+3) - 1.

a(2n-1) = A001610(n+2), a(2n) = A001595(n+2).

a(1)=3, a(2)=5, a(3)=6, a(4)=9, a(n) = a(n-2) + a(n-4) + 1, n > 4.

G.f.: (3 + 2*x - 2*x^2 + x^3 - 3*x^4)/(1-x-x^2+x^3-x^4+x^5).

a(n) = A272632(n)-1. - R. J. Mathar, Jan 13 2023

Example

a(1)=3 as 3 is the sum of just 2 Fibonacci sets {3=Fibonacci(4)} and {1=Fibonacci(2), 2=Fibonacci(3)};
a(2)=5 as 5 is sum of Fibonacci sets {5} and {2,3} only.

Maple

fib:= combinat[fibonacci]:
lucas:=n->fib(n-1)+fib(n+1):
a:=n -> if n mod 2 = 0 then 2 *fib(n/2+3) -1 else lucas((n+1)/2+2)-1 fi:
seq(a(n), n=1..50);

Mathematica

LinearRecurrence[{1, 1, -1, 1, -1}, {3, 5, 6, 9, 10, 15}, 40] (* Vincenzo Librandi, Jul 25 2017 *)
Table[If[Mod[n, 2]==0, 2*Fibonacci[(n+6)/2]-1, LucasL[(n+5)/2]-1], {n, 50}] (* G. C. Greubel, Jul 13 2019 *)

Prog

(PARI) vector(50, n, f=fibonacci; if(n%2==0, 2*f((n+6)/2)-1, f((n+7)/2) + f((n+3)/2)-1)) \\ G. C. Greubel, Jul 13 2019
(Magma) f:=Floor; [(n mod 2) eq 0 select 2*Fibonacci(f((n+6)/2))-1 else Lucas(f((n+5)/2))-1: n in [1..50]]; // G. C. Greubel, Jul 13 2019
(Sage)
def a(n):
    if (mod(n, 2)==0): return 2*fibonacci((n+6)/2) - 1
    else: return lucas_number2((n+5)/2, 1, -1) -1
[a(n) for n in (1..50)] # G. C. Greubel, Jul 13 2019
(GAP)
a:= function(n)
    if n mod 2=0 then return 2*Fibonacci(Int((n+6)/2)) -1;
    else return Lucas(1, -1, Int((n+5)/2))[2] -1;
    fi;
  end;
List([1..50], n-> a(n) ); # G. C. Greubel, Jul 13 2019

Crossrefs

Cf. A000032, A000045, A000071, A000119, A013583, A122195.

Sequence in context: A367334 A329003 A263654 * A225005 A053091 A367499

Adjacent sequences: A122191 A122192 A122193 * A122195 A122196 A122197

Keyword

nonn,easy

Author

Ron Knott, Aug 25 2006

Status

approved