|
%I #26 Jan 13 2023 10:52:02
%S 3,5,6,9,10,15,17,25,28,41,46,67,75,109,122,177,198,287,321,465,520,
%T 753,842,1219,1363,1973,2206,3193,3570,5167,5777,8361,9348,13529,
%U 15126,21891,24475,35421,39602,57313,64078,92735,103681,150049,167760,242785
%N Numbers that are the sum of exactly two sets of Fibonacci numbers.
%H Vincenzo Librandi, <a href="/A122194/b122194.txt">Table of n, a(n) for n = 1..1000</a>
%H J. Berstel, <a href="http://www-igm.univ-mlv.fr/~berstel/Articles/2001ExerciceAldo.pdf">An Exercise on Fibonacci Representations</a>, 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.
%H M. Bicknell-Johnson & D. C. Fielder, <a href="http://www.fq.math.ca/Scanned/37-1/bicknell.pdf">The number of Representations of N Using Distinct Fibonacci Numbers, Counted by Recursive Formulas</a>, Fibonacci Quart. 37.1 (1999) pp. 47 ff.
%H Ron Knott <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibrep.html#sumoffib">Sumthing about Fibonacci Numbers</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1,1,-1).
%F a(2n-1) = A000032(n+2) - 1,
%F a(2n) = 2*A000045(n+3) - 1.
%F a(2n-1) = A001610(n+2), a(2n) = A001595(n+2).
%F a(1)=3, a(2)=5, a(3)=6, a(4)=9, a(n) = a(n-2) + a(n-4) + 1, n > 4.
%F G.f.: (3 + 2*x - 2*x^2 + x^3 - 3*x^4)/(1-x-x^2+x^3-x^4+x^5).
%F a(n) = A272632(n)-1. - _R. J. Mathar_, Jan 13 2023
%e a(1)=3 as 3 is the sum of just 2 Fibonacci sets {3=Fibonacci(4)} and {1=Fibonacci(2), 2=Fibonacci(3)};
%e a(2)=5 as 5 is sum of Fibonacci sets {5} and {2,3} only.
%p fib:= combinat[fibonacci]:
%p lucas:=n->fib(n-1)+fib(n+1):
%p a:=n -> if n mod 2 = 0 then 2 *fib(n/2+3) -1 else lucas((n+1)/2+2)-1 fi:
%p seq(a(n), n=1..50);
%t LinearRecurrence[{1, 1, -1, 1, -1}, {3, 5, 6, 9, 10, 15}, 40] (* _Vincenzo Librandi_, Jul 25 2017 *)
%t 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 *)
%o (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
%o (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
%o (Sage)
%o def a(n):
%o if (mod(n,2)==0): return 2*fibonacci((n+6)/2) - 1
%o else: return lucas_number2((n+5)/2, 1,-1) -1
%o [a(n) for n in (1..50)] # _G. C. Greubel_, Jul 13 2019
%o (GAP)
%o a:= function(n)
%o if n mod 2=0 then return 2*Fibonacci(Int((n+6)/2)) -1;
%o else return Lucas(1,-1, Int((n+5)/2))[2] -1;
%o fi;
%o end;
%o List([1..50], n-> a(n) ); # _G. C. Greubel_, Jul 13 2019
%Y Cf. A000032, A000045, A000071, A000119, A013583, A122195.
%K nonn,easy
%O 1,1
%A _Ron Knott_, Aug 25 2006
|
