A091484 Recamán's Fibonacci variation : a(1)=a(2)=1 then a(n) = a(n-1)+a(n-2)-F(n) if that number is >0 and not already in the sequence; a(n) = a(n-1)+a(n-2)+F(n) otherwise where F(n) denotes the n-th Fibonacci number.

1, 1, 4, 2, 11, 5, 3, 29, 66, 40, 17, 201, 451, 275, 116, 1378, 3091, 1885, 795, 9445, 21186, 12920, 5449, 64737, 145211, 88555, 37348, 443714, 995291, 606965, 255987, 3041261, 6821826, 4160200, 1754561, 20845113, 46757491, 28514435, 12025940

Offset: 1

Author

Benoit Cloitre, Mar 03 2004

Keywords

• nonn
• easy

Comments

Unlike Recamán's sequence, this one has "simple" behavior since the sequence of signs of a(n+1)-a(n) becomes 4-periodic sequence -1,-1,1,1,-1,-1,1,1,-1,-1,...

Examples

			a(6)+a(5)=5+11=16 and F(7)=13. Since 16-13=3 is not already in the sequence, a(7)= a(6)+a(5)-F(7)=3.
		

Crossrefs

Cf. A005132, A079053 (also Fibonacci variation but starting with 1, 2).

Programs

PARI

m=200; a=vector(m); a[1]=1; a[2]=1; for(n=3, m, a[n]=if(n<0, 0, if(abs(sign(a[n-1]+a[n-2]-fibonacci(n))-1)+setsearch(Set(vector(n-1, i, a[i])), a[n-1]+a[n-2]-fibonacci(n)), a[n-1]+a[n-2]+fibonacci(n), a[n-1]+a[n-2]-fibonacci(n)))); a

Formula

With phi=(1+sqrt(5))/2 for n>1 : a(4n)=floor(r*phi^(4n)) where r = (-675+327*sqrt(5))/90; a(4n+1)=floor(s*phi^(4n+1)) where s=(-132+66*sqrt(5))/18; a(4n+2)=floor(s*phi^(4n+2)) where t= (-15+7*sqrt(5))/2; a(4n+3)=floor(u*phi^(4n+3)) where u=(-115+52*sqrt(5))/15.

For n>=0, a(4n+5) = 11/3*Luc(4n+2), a(4n+6) = 5*Fib(4n+2).

For n>=3, the sequence satisfies the order 8 linear recurrence: a(n+8)-7*a(n+4)+a(n)=0. - Benoit Cloitre, Apr 30 2006

Empirical g.f.: x*(6*x^7-4*x^6+17*x^5-12*x^4+3*x^3-5*x^2-1) / ((x^2+x-1)*(x^4+3*x^2+1)). - Colin Barker, Jun 26 2013

Extensions

PARI code corrected by Colin Barker, Jun 26 2013