A255978 a(n) = a(n-1) + a(n-2) + (1 + (-1)^(a(n-1) + a(n-2))) with a(0)=0, a(1)=1.

0, 1, 1, 4, 5, 9, 16, 25, 41, 68, 109, 177, 288, 465, 753, 1220, 1973, 3193, 5168, 8361, 13529, 21892, 35421, 57313, 92736, 150049, 242785, 392836, 635621, 1028457, 1664080, 2692537, 4356617, 7049156, 11405773, 18454929, 29860704, 48315633, 78176337, 126491972, 204668309, 331160281, 535828592

Offset: 0

Author

Waldemar Puszkarz, Mar 12 2015

Keywords

• nonn
• easy

Comments

This is a minimally modified Fibonacci sequence (A000045) in that it preserves characteristic properties of the original sequence: a(n) is a function of the sum of the preceding two terms, the ratio of two consecutive terms tends to the Golden Mean, and the initial two terms are the same as in the Fibonacci sequence. See A253197 and A253198 for other members of this family.

Examples

			For n = 2, a(2) = 0 + 1 + (1 + (-1)^(0 + 1)) = 1.
For n = 3, a(3) = 1 + 1 + (1 + (-1)^(1 + 1)) = 4.
For n = 4, a(4) = 1 + 4 + (1 + (-1)^(1 + 4)) = 5.
For n = 5, a(5) = 4 + 5 + (1 + (-1)^(4 + 5)) = 9.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• W. Puszkarz, A Note on Minimal Extensions of the Fibonacci Sequence, viXra:1503.0113, 2015.
• Index entries for linear recurrences with constant coefficients, signature (1,1,1,-1,-1).

Crossrefs

Cf. A000045, A253197, A253198.

Programs

Magma

[n le 2 select (n-1) else Self(n-1)+Self(n-2)+(1+(-1)^(Self(n-1)+Self(n-2))): n in [1..45] ]; // Vincenzo Librandi, Mar 24 2015

Mathematica

RecurrenceTable[{a[n]==a[n-1]+a[n-2] +(1+(-1)^(a[n-1]+a[n-2])), a[0]==0, a[1]==1}, a, {n, 0, 50}]
CoefficientList[Series[x (1 + 2 x^2 - x^3) / ((1 - x) (1 + x + x^2) (1 - x - x^2)), {x, 0, 70}], x] (* Vincenzo Librandi, Mar 24 2015 *)
LinearRecurrence[{1,1,1,-1,-1},{0,1,1,4,5},50] (* Harvey P. Dale, Mar 26 2019 *)

PARI

concat(0, Vec(x*(1+2*x^2-x^3)/((1-x)*(1+x+x^2)*(1-x-x^2)) + O(x^30))) \\ Michel Marcus, Mar 23 2015

Formula

a(n) = a(n-1) + a(n-2) + (1 + (-1)^(a(n-1) + a(n-2))), a(0)=0, a(1)=1.

G.f.: x*(1+2*x^2-x^3)/((1-x)*(1+x+x^2)*(1-x-x^2)). - Joerg Arndt, Mar 16 2015

a(n) = a(n-1) + a(n-2) + a(n-3) - a(n-4) - a(n-5) for n>4. - Colin Barker, Mar 28 2015

a(n) = 2*Fibonacci(n) - (1 if n != 0 (mod 3)). - Nicolas Bělohoubek, Sep 29 2021