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

1,3

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,...

Links

Table of n, a(n) for n=1..39.

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

Example

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.

Prog

(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

Crossrefs

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

Sequence in context: A215500 A188128 A336914 * A163544 A191728 A191434

Adjacent sequences: A091481 A091482 A091483 * A091485 A091486 A091487

Keyword

nonn,easy

Author

Benoit Cloitre, Mar 03 2004

Extensions

PARI code corrected by Colin Barker, Jun 26 2013

Status

approved